DOE-HDBK-1122-2009 Chg Notice 2, Radiological Control Technician Training (Part 4 of 9)

DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Part 4 of 9 Radiological Control Technician Training Fundamental Academic Training Study Guide Phase I Coordinated and Conducted for the Office of Health, Safety and Security U.S. Department of Energy DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide This page intentionally left blank. 1.01-ii DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Table of Contents Page Module 1.01 Basic Mathematics and Algebra........................................................................1.01-1 Module 1.02 Unit Analysis and Conversion Module 1.06 Radioactivity and Radioactive Decay .................................................................... 1.06-1 ...........................................................................1.02-1 Module 1.03 Physical Sciences...............................................................................................1.03-1 Module 1.04 Nuclear Physics .................................................................................................1.04-1 Module 1.05 Sources of Radiation..........................................................................................1.05-1 Module 1.07 Interaction of Radiation with Matter .................................................................1.07-1 Module 1.08 Biological Effects of Radiation .........................................................................1.08-1 Module 1.09 Radiological Protection Standards ....................................................................1.09-1 Module 1.10 ALARA .............................................................................................................1.10-1 Module 1.11 External Exposure Control ................................................................................1.11-1 Module 1.12 Internal Exposure Control .................................................................................1.12-1 Module 1.13 Radiation Detector Theory ................................................................................1.13-1 1.01-iii DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide This page intentionally left blank. 1.01-iv DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Course Title: Radiological Control Technician Module Title: Basic Mathematics and Algebra Module Number: 1.01 Objectives: 1.01.01 Add, subtract, multiply, and divide fractions. 1.01.02 Add, subtract, multiply, and divide decimals. 1.01.03 Convert fractions to decimals and decimals to fractions. 1.01.04 Convert percent to decimal and decimal to percent. 1.01.05 Add, subtract, multiply, and divide signed numbers. 1.01.06 Add, subtract, multiply, and divide numbers with exponents. 1.01.07 Find the square roots of numbers. 1.01.08 Convert between numbers expressed in standard form and in scientific notation. 1.01.09 Add, subtract, multiply, and divide numbers expressed in scientific notation. 1.01.10 Solve equations using the "Order of Mathematical Operations." 1.01.11 Perform algebraic functions. 1.01.12 Solve equations using common and/or natural logarithms. Introduction Radiological control operations frequently require the RCT to use arithmetic and algebra to perform various calculations. These include scientific notation, unit analysis and conversion, radioactive decay calculations, dose rate/distance calculations, shielding calculations, stay- time calculations. A good foundation in mathematics and algebra is important to ensure that the data obtained from calculations is accurate. Accurate data is crucial to the assignment of proper radiological controls.

References: 1. DOE-HDBK-1014/1-92 (June 1992) "Mathematics: Volume 1 of 2"; DOE Fundamentals Handbook Series. 1.01-1 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide SYMBOLS FOR BASIC OPERATIONS The four basic mathematical operations are addition, subtraction, multiplication, and division. Furthermore, it is often necessary to group numbers or operations using parentheses or brackets. In writing problems in this course the following notation is used to denote the operation to be performed on the numbers. If a and b represent numbers or variables, the operations will be denoted as follows: Table 1. Symbols for Basic Mathematical Operations Operation Notation Addition: Subtraction Multiplication Division: a + b a - b a × b a ÷ b a · b a/b a(b) b a ab ab Grouping: Equality: Inequality: Less than: Greater than: ( ) [ ] = ≠ < Less than or equal to: > Greater than or equal to: 1.01.01 Add, subtract, multiply, and divide fractions. FRACTIONS Whole numbers consist of the normal counting numbers and zero, e.g., {0, 1, 2, 3, 4...} A fraction is part of a whole number. It is simply an expression of a division of two whole numbers. A fraction is written in the format: a or a/b b 1.01-2 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide The number above the bar a is called the numerator and the number below the bar b is called the denominator. A proper fraction is a fraction in which the number in the numerator is less than the number in the denominator. If the numerator is greater than the denominator then it is an 7 25 15 61improper fraction. For example, ½ and ¼ are proper fractions, while , , , or or are 5 5 7 27 improper fractions. Any whole number can be written as a fraction by letting the whole number be the numerator and 1 be the denominator. For example: 5 2 05 = 2 = 0 = 1 1 1 Five can be written as 10/2, 15/3, 20/4, etc. Similarly, the fraction ¼ can be written as 2/8, 3/12, 4/16, etc. These are called equivalent fractions. An equivalent fraction is built up, per se, by multiplying the numerator and the denominator by the same non-zero number. For example: 3 3 ⋅ 2 6 3 3 ⋅ 5 15 = = = = 4 4 ⋅ 2 8 4 4 ⋅ 5 20 A fraction is reduced by dividing the numerator and the denominator by the same nonzero number. For example: 12 12 ÷ 2 6 6 6 ÷ 3 2 = = = = 18 18 ÷ 2 9 9 9 ÷ 3 3 A fraction is reduced to lowest terms when 1 is the only number that divides both numerator and denominator evenly. This is done by finding the greatest common multiple between the numerator and denominator. In the previous example, two successive reductions were performed. For the fraction 12/18, the greatest common multiple would be 6, or (2 × 3), which results in a reduction down to a denominator of 3. A whole number written with a fraction is called a mixed number. Examples of mixed numbers would be 1½, 3¼, 5¾, etc. A mixed number can be simplified to a single improper fraction using the following steps: 1. Multiply the whole number by the denominator of the fraction. 2. Add the numerator of the fraction to the product in step 1. 3. Place the sum in step 2 as the numerator over the denominator. 1.01-3 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide For example: 3 (5 ⋅ 4) + 3 235 = = 4 4 4 Adding and Subtracting Fractions Fractions With the Same Denominator To add two fractions which have the same denominator:

1. Add the numerators. 2. Place the sum of step 1 over the common denominator. 3. Reduce fraction in step 2 to lowest terms (if necessary). For example: 1 3 1 + 3 4 + = = 5 5 5 5 Subtraction of two fractions with the same denominator is accomplished in the same manner as addition. For example, 5 3 5 − 3 2 2 ÷ 2 1 − = = = = 8 8 8 8 8 ÷ 2 4 Fractions With Different Denominators To add two fractions with different denominators requires that the fractions be built up so that they have the same denominator. This is done by finding the lowest common denominator. Once a common denominator is obtained, the rules given above for the same denominator apply. For example, 1/3 + 2/5. The fraction 1/3 could be built up to 2/6, 3/9, 4/12, 5/15, 6/18, 7/21, etc. The fraction 2/5 could be built up to 4/10, 6/15, 8/20, 10/25, etc. The lowest common denominator for the two fractions would be 15. The problem would be solved as follows: 1 2 1 ⋅ 5 2 ⋅ 3 5 6 5 + 6 11 + = + = + = = 3 5 3 ⋅ 5 5 ⋅ 3 15 15 15 15 1.01-4 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Subtraction of fractions with different denominators is accomplished using the same steps as for addition. For example: 3 2 3 ⋅ 3 2 ⋅ 4 9 8 9 − 8 1 − = − = − = = 4 3 4 ⋅ 3 3 ⋅ 4 12 12 12 12 Multiplying and Dividing Fractions Multiplication of fractions is much easier than addition and subtraction, especially if the numbers in the numerators and denominators are small. Fractions with larger numerators and/or denominators may require additional steps. In either case, the product of the multiplication will most likely need to be reduced in order to arrive at the final answer. To multiply fractions: 1. Multiply the numerators. 2. Multiply the denominators. 3. Place product in step 1 over product in step 2. 4. Reduce fraction to lowest terms. For example: 5 3 5 ⋅ 3 15 15 ÷ 3 5 ⋅ = = = = 6 4 6 ⋅ 4 24 24 ÷ 3 8 A variation on the order of the steps to multiply fractions is to factor the numerators and denominators first, reduce and cancel, and then multiply. For example: 3 20 3 2 ⋅ 2 ⋅ 5 3/ ⋅ 2/ ⋅ 2/ ⋅ 5 5 5 ⋅ = ⋅ = = = 8 9 2 ⋅ 2 ⋅ 2 3 ⋅ 3 2 ⋅ 2/ ⋅ 2/ ⋅ 3/ ⋅ 3 2 ⋅ 3 6 Reciprocals Two numbers whose product is 1 are called reciprocals, or multiplicative inverses. For example: a. 5 and 1 are reciprocals because 5 · 1 = 1. 5 5 4 5b. 4 and 5 are reciprocals because ⋅ = 1 5 4 5 4 c. 1 is its own reciprocal because 1 · 1 = 1. d. 0 has no reciprocal because 0 times any number is 0 not 1. 1.01-5 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide The symbol for the reciprocal, or multiplicative inverse, of a non-zero real number a is 1 . a Every real number except 0 has a reciprocal. Therefore, for every non-zero real number a, there is a unique real number 1 such that: a 1 a ⋅ = 1 a Now, look at the following product: ⎛ 1 1 ⎞ ⎛ 1 ⎞⎛ 1 ⎞⎛ 1 ⎞( ) ⋅ = ⎜a ⋅ ⎜b b = 1 ⋅1 = 1ab ⎜ ⎟ ⎟ ⋅ ⎟⎜ ⋅ ⎟ ⎝ a b ⎠ ⎝ a ⎠⎝ a ⎠⎝ b ⎠ Relationship of multiplication to division The operation of division is really just inverted multiplication (reciprocals). Notice from the above examples that the reciprocal of a fraction is merely "switching" the numerator 1and denominator. The number 5 is really 5 , and the reciprocal of 5 is . 1 5 3Likewise, the reciprocal of 2 is . 3 2 Fractions are a division by definition. Division of fractions is accomplished in two steps: 1. Invert the second fraction, i.e., change it to its reciprocal, and change the division to multiplication.

2. Multiply the two fractions using the steps stated above. For example: 4 2 4 3 12 12 ÷ 2 6 ÷ = ⋅ = = = 7 3 7 2 14 14 ÷ 2 7 1 2 2 3 = 1 ÷ = 1 ⋅ = 2 3 3 23 1.01-6 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Practice Problems Solve the following problems involving fractions. Answers should be reduced to lowest terms. 1. 1/3 + 2/3 2. 5/7 - 3/7 3. 5/9 + 2/3 4. 6/7 - 1/2 5. 2 - 1/3 6. 3/8 + 15/16 7. 25/32 - 3/4 8. 15/21 - 4/7 9. 13/20 - 2/5 10. 7/18 + 5/9 11. 2/3 × 1/5 12. 4/7 × 3/4 13. 1/2 × 2 14. 3/5 × 4 15. 4/9 ÷ 2/3 16. 8/13 × 2/3 17. 12/15 × 3/5 18. 20/25 ÷ 4/5 19. 7/8 × 2/5 20. 14/21 ÷ 2/7 1.01.02 Add, subtract, multiply and divide decimals. DECIMALS A decimal is another way of expressing a fraction or mixed number. It is simply the numerical result of divison (and fractions are division). Recall that our number system is based on 10 ("deci" in "decimal" means ten) and is a place-value system; that is, each digit {i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9} in a numeral has a particular value determined by its location or place in the number. For a number in decimal notation, the numerals to the left of the decimal point comprise the whole number, and the numerals to the right are the decimal fraction, (with the denominator being a power of ten). Figure 1. Decimal Place Value System 1.01-7 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide For example, the numeral 125.378 (decimal notation) represents the expanded numeral 3 7 8100 + 20 + 5 + + + 10 100 1000 If this numeral were written as a mixed number, we would write it as: 378125 1000 Addition and Subtraction of Decimals In order to add or subtract decimals use the following steps: 1. Arrange the numbers in a column so that the decimal points are aligned. 2. Add or subtract in columns from right to left. Additional zeros may need to be added to the right of the subtrahend (a number that is to be subtracted from a minuend). 3. Place the decimal point in the answer in line with the other decimal points. For example: 21.3 654.200 + 4.2 - 26.888 25.5 627.312 Multiplying Decimals To multiply decimal numbers, do the following: 1. Multiply the numbers as if there were no decimal points. 2. Count the number of decimal places in each number and add them together. 3. Place the decimal point in the product so that it has the same number of decimal places as the sum in step 2. For example: 5.28 0.04 ¯ 3.7 ¯ 0.957 3696 028 1584 020 19.536 036 000 0.03828 1.01-8 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Division of Decimals The steps for division of decimals are as follows: 1. Move the decimal point of the divisor (the number by which a dividend is divided) to the right until it becomes a whole number. 2. Move the decimal point of the dividend to the right the same number of places moved in step 1. 3. Divide the numbers as if they were whole numbers. For example: 7 ÷ 0.25 28 0.25 7.00 50 200 200 0 Decimal Forms As we have learned, decimals are the result of division (or a fraction). When the remainder of the division is zero, the resulting decimal is called a terminating or finite decimal. For example, fractions like 2/5, 3/8, and 5/6 will all result in finite decimals. These fractions and the resulting decimals are known as rational numbers.

On the other hand, fractions like 1/3, 2/7, 5/11, and 7/13 result in a non-terminating or infinite decimal. For example, 2/7 results in the decimal 0.285714286 . . . , the dots meaning that the decimal continues without end. These numbers are known as irrational numbers. Note that even though irrational numbers are non-terminating, (e.g., 1/3 and 5/11) are repeating or periodic decimals because the same digit or block of digits repeats unendingly. For example: 1 5 = 0.3333... = 0.454545... 3 11 A bar is often used to indicate the block of digits that repeat, as shown below: 51 = 0.45= 0.3 3 11 1.01-9 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Practice Problems Solve the following problems involving decimals. 1. 0.23 + 3.4 6. 2.3 × 3.2 2. 5.75 - 2.05 7. 0.007 × 2.18 3. 6.1 - 1.6 8. 5.2 ÷ 1.4 4. 0.018 + 0.045 9. 12.26 ÷ 0.04 5. 468.75 - 192.5 10. 4.0 × 0.25 1.01.03 Convert fractions to decimals and decimals to fractions. FRACTION TO DECIMAL CONVERSION To convert a fraction to a decimal we simply perform the operation of division that the fraction represents. For example, the fraction 3/4 represents "3 divided by 4," and would be converted as follows: 0 . 75 00.34 2 . 8 20 20 0 Practice Problems Convert the following fractions to decimals. 1. 1/2 4. 12/25 2. 2/5 5. 13/39 3. 5/8 6. 7/16 Convert the following decimals to fractions. Reduce answers to lowest terms. 7. 0.125 9. 4.25 8. 0.6666 10. 0.2 1.01-10 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide 1.01.04 Convert percent to decimal and decimal to percent. PERCENT Percentage is a familiar and widely used concept for expressing common and decimal fractions. Most people know the meaning of terms such as 100 percent and 50 percent. The word percent actually means "out of a hundred." (Consider that there are 100 "cents" in a dollar, and that a "century" is 100 years.) A percent is simply a fraction whose denominator is 100. Thus, 50 percent means 50/100, or 0.50, while 25 percent means 25/100, or 0.25. Percent is abbreviated by the symbol %. So, 75 percent is written 75%. Converting Decimal to Percent A decimal fraction is changed to a percent by moving the decimal point two places to the right and adding a percent sign. For example, 1/8 equals 0.125. Therefore: 1 = 0.125 = 12.5% 8 rr A percent is changed to a common fraction by omitting the percent sign, placing the number over 100, and reducing the resulting fraction if possible. For example, 32% equals 32/100 which reduces to 8/25. When the percent consists of a mixed decimal number with a percent sign, the resulting fraction will contain a mixed decimal numerator. This can be changed to a whole number by multiplying the numerator and the denominator by 10, 100, 1,000, etc. For example: 40.25 40.25×100 4025 80540.25% = = = = 100 100×100 10,000 2000 Percentage is most frequently used to indicate a fractional part. Thus 20% of the total power output for 75% of the employees refer to fractional parts of some total number. To perform arithmetic operations with a percent, it is normally changed to a common or decimal fraction. Usually, a decimal fraction is more convenient. 1.01-11 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Converting Percent to Decimal A percent is changed to a decimal fraction by omitting the percent sign and moving the decimal point two places to the left. For example: 48% = 0.48

Thus, 92% equals 0.92, 8% equals 0.08, and so on. Practice Problems Convert the following decimals to percent. 1. 0.5 4. 0.06 2. 0.782 5. 0.049 3. 1.1 6. 0.0055 Convert the following percentages to decimals. 7. 65% 9. 300% 8. 0.25% 10. 0.09% 1.01.05 Add, subtract, multiply and divide signed numbers. SIGNED NUMBERS The numbers that are used to quantify the number of objects in a group, the "counting numbers", are always positive numbers; that is, they are always greater than zero. However, there are many occasions when negative numbers (numbers less than zero) must be used. These numbers arise when we try to describe measurement in a direction opposite to the positive numbers. For example, if we assign a value of +3 to a point which is 3 feet above the ground, what number should be assigned to a point which is 3 feet below the ground? Perhaps the most familiar example of the use of negative numbers is the measurement of temperature, where temperatures below an arbitrary reference level are assigned negative values. Every number has a sign associated with it. The plus (+) sign indicates a positive number, whereas the minus ( - ) sign indicates a negative number. When no sign is given, a plus (+) sign is implied. The fact that the plus and minus signs are also used for the arithmetic operations of addition and subtraction should not be a cause for confusion, for we shall see that they have equivalent meanings. 1.01-12 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Every number has an absolute value, regardless of its sign. The absolute value indicates the distance from zero, without regard to direction. The number 5 is 5 units from zero, in the positive direction. The number -5 is also 5 units from zero, but in the negative direction. The absolute value of each of these numbers is 5. The absolute value of a number is indicated by a pair of vertical lines enclosing the number: |5|. Operations with Signed Numbers The arithmetic operations of addition, subtraction, multiplication, and division of signed numbers can be more easily visualized if the numbers are placed on a number line (see Figure 2). The positive numbers are greater than zero, and they lie to the right of zero on the number line. The negative numbers are less than zero, and lie to the left of zero on the number line. FIGURE 2. Number Line The number line extends an infinite distance in each direction and therefore includes all numbers. The process of addition can be considered as counting in one direction or the other from a starting point on the number line. For example, let us add 1 + 2. We locate +1 on the number line and then count 2 units to the right, since we are adding +2. The result will be +3. To illustrate further, let us add +2 and -4. We first locate +2 on the number line and then count 4 units to the left. We end up at -2. The number line is useful for illustrating the principles of addition, but it clearly would be inconvenient to use in the case of large numbers. Consequently, the following rules were developed to govern the addition process: Adding and Subtracting Signed Numbers. To add two numbers with the same signs, add their absolute values and attach the common sign. For example: (-3) + (-2) = -5 To add two numbers with opposite signs, find the difference of their absolute values, then attach the sign of the original number which had the greater absolute value. For example: (-2) + 3 = 1 1.01-13 DOE-HDBK-1122-2009

Module 1.01 Basic Mathematics and Algebra Study Guide Notice that -3 and +3 are the same distance but in opposite directions from 0 on the number line. What happens when you add two numbers like 3 and -3? 3 + (-3) = 0 -7 + 7 = 0 If the sum of two signed numbers is 0, the numbers are called additive inverses or opposites. For example: 7 - 3 = 4 is the same as: 7 + (-3) = 4 8 - 2 = 6 is the same as: 8 + (-2) = 6 It can be seen that subtracting a number is equivalent to adding its additive inverse or opposite. To subtract a signed number, add its opposite or additive inverse. In other words, change the subtraction symbol to addition and change the sign of the second signed number. For example: 5 - (-8) = 5 + (+8) ⇐ add +8 (Answer = 13) 6 - 11 = 6 + ( -11) ⇐ add -11 (Answer = -5) -4 - ( -7) = -4 + (+7) ⇐ add +7 (Answer = 3) Multiplying and Dividing Signed Numbers. The product of two numbers with like signs is a positive number. The product of two numbers with unlike signs is a negative number. In symbols: (+) × (+) = (+) (+) × ( - ) = ( - ) ( - ) × ( - ) = (+) ( - ) × (+) = ( - ) For example: (-4) × (-3) = (+12) (-4) × (+3) = (-12) The division of numbers with like signs gives a positive quotient. The division of numbers with unlike signs gives a negative quotient. In symbols: (+)/(+) = (+) (+)/( - ) = ( - ) ( - )/( - ) = (+) ( - )/(+) = ( - ) 1.01-14 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide For example: (-24)/(-6) = (+4) (-24)/(+6) = (-4) Remember that multiplication is really a short form of addition. When we say +4 × ( - 3), we are adding the number - 3 four times, that is, ( - 3) + ( - 3) + ( - 3) + ( - 3) = - 12. Also, since division is a short form of subtraction, when we say - 24 ÷ ( - 6), we subtract the number - 6 from the number - 24 four times in order to reach 0, i.e., - 24 - ( - 6) - ( - 6) - ( - 6) - ( - 6) = 0. Although we could repeat the process for the multiplication and division of any signed numbers, usage of the two rules will produce equivalent results. Practice Problems Solve the following problems involving signed numbers. 1. (- 28) + ( - 51) 9. (- 4)(5) 2. (- 2) + ( - 5) 10. 6/(- 3) 3. 40 + ( - 21) 11. 4 ( - 5) 4. (- 87) + 50 12. (- 6)/(- 3) 5. 48 + ( - 27) 13. (- 6)/3 6. 56 - ( - 5) 14. (- 8)(- 5) 7. 81 - 4 15. (- 7)(6) 8. - 48 - ( - 2) 1.01.06 Add, subtract, multiply, and divide numbers with exponents. EXPONENTS An exponent is a small number placed to the right and a little above another number, called the base; to show how many times the base is to be multiplied by itself. Thus, 34 (read "three to the fourth power") means 3 used as a factor four times or 3 × 3 × 3 × 3. In this case, 4 is the exponent, and 3 is the base. In general, if b is any real number and n is any positive integer, the nth power of b is written bn (where b is the base and n is the exponent) is read as "b to the nth power." This tells you that b is used as a factor n times. Thus, 52 is called "5 raised to the second power" (or 5 "squared"), and 23 is called "2 raised to the third power" (or 2 "cubed"). When no exponent is shown for a number or no power is indicated, the exponent or power is understood to be 1. Thus, 7 is the same as 71. Any number raised to the power of zero equals one; e.g., 70 = 1. Normally, exponents of zero and one are not left as the final value, but are changed to the simpler form of the base. 1.01-15 DOE-HDBK-1122-2009

Module 1.01 Basic Mathematics and Algebra Study Guide Exponents can be expressed as integers, as in the examples above, or as fractions or decimals such as 91/2 or 103.2 . They may also be positive or negative. Exponents and powers are particularly useful in mathematics not only because they shorten the writing of mathematical expressions, but also because they simplify many mathematical operations. However, there are several special rules which govern mathematical operations involving numbers with exponents. Addition and Subtraction The addition or subtraction of numbers with exponents can be performed only if both the bases and the exponents of the numbers are the same. When the bases of the exponents are different, the multiplication indicated by the exponent must be performed and the numbers then added or subtracted. Thus, 25 and 24 cannot be added directly because their exponents are different. They can be added only by carrying out the indicated multiplication first. Thus, 25 equals 2 × 2 × 2 × 2 × 2 which equals 32, and 24 equals 2 × 2 × 2 × 2 which equals 16. Therefore, 25 + 24 equals 32 + 16, which equals 48. When the bases and the exponents are the same, the numbers can be added or subtracted directly. For example: 35 + 35 = 2(35 ) = 2(243) = 486 Multiplication The multiplication of numbers with exponents of the same base is performed by adding the exponents. The general form is as follows: (am )(an ) = a(m+n) It is important to remember that the bases of the numbers must be the same before they can be multiplied by adding their exponents. The base of the product is the same as the base of the two factors. Thus, 32 × 33 = 35 = 243 Division The division of numbers with exponents of the same base is performed by subtracting the exponent of the divisor (denominator) from the exponent of the dividend (numerator). The general form is: ma (m−n)= a na 1.01-16 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Again, it is important to remember that the bases of the numbers must be the same before they can be divided by subtracting their exponents. The base of the quotient is the same as the base of the number divided. Thus, 52 (5−2) 3= 2 = 2 = 8 22 Division of numbers with exponents can be used to show why any number raised to the power of zero equals one. We know that any fraction in which the numerator equals the denominator can be reduced to 1; e.g., 2/2 = 1. Similarly: 32 (3−3) 0= 2 = 2 = 1 32 Exponent Raised to a Power Raising a number with an exponent to a power is performed by multiplying the exponent by the power. The general form is: (am)n = amn The base of the answer is the same as the base of the number raised to the power. Thus: (52)3 = 5(2¯3) = 56 = 15,625 Product Raised to a Power Raising a product of several numbers to a power is performed by raising each number to the power. The general form is as follows: n bn (ab)n = a For example: [(2)(3)(4)]2 = (22)(32)(42) = (4)(9)(16) = 576 This same result can also be obtained like this: [(2)(3)(4)]2 = 242 = 576 1.01-17 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Mixed Product and Exponents Raised to a Power The same rule can be used to raise a product of several numbers with exponents to a power. The general form looks like this: (Aa Bb Cc )n = A(a¯n) B(b¯n) C(c¯n) For example: [(2)4 (3)3 (4)2 ] 2 = (24¯2) (33¯2) (42¯2) = (28) (36) (44) = (256) (729) (256) Fraction Raised to a Power

Raising a fraction to a power is performed by raising both numerator and denominator to the power. It should be remembered that with a proper fraction (i.e., numerator is less than the denominator) the resulting number must be less than one. Also, the resulting number will be less than the value of the original fraction. Thus, (2/3)3 equals 23/33, which equals 8/27, which is less than one and less than the original fraction, 2/3. Negative Exponents and Powers A negative exponent or power has a special meaning. Any number, except 0, with a negative exponent equals the reciprocal of the same number with the same positive exponent. For example: The same rules for addition, subtraction, multiplication, division, and raising to a power apply to negative exponents that apply to positive exponents. However, in adding, subtracting, or multiplying the exponents, the rules for signed numbers must also be observed. Fractional Exponents Fractional exponents are used to represent roots (see next section). The general form of a fractional exponent is a m n , which reads "the nth root of am ." For example, a 1 2 , means a the 2square root of a1 , or a. In other words, a 1 = a 1.01-18 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Calculator Method To raise a number to a power using a scientific calculator, use the following steps: 1) Enter the number. (Fractions must first be converted to decimal form.) 2) Press the key. yx 3) Enter the power. (Fractions must first be converted to decimal form.) 4) Press the = key. The number displayed will be the number entered in step 1 raised to the power entered in step 3. Practice Problems 1. (32)(33) 4. (106)(10-4) 2. 75/73 5. 6-4/6-3 3. (105)(106) 6. 6-8/63 1.01.07 Find the square roots of numbers. SQUARE ROOTS To square a number means to multiply the number by itself, i.e., raise it to the second power. (Consider that a square is two-dimensional.) For example, 2 squared is 4, since 2 × 2 = 4. The square of 3 is 9, 4 squared is 16, and so on. Just as subtraction "undoes" addition and division "undoes" multiplication, squaring a number can be "undone" by finding the square root. The general definition is as follows: If a2 = b, then a is a square root of b. Be careful not to confuse the terms square and square root. For example, if 52 = 25, this indicates that 25 is the square of 5, and 5 is the square root of 25. To be explicit, we say that it is a perfect square because 5 times itself is 25. All perfect squares other than 0 have two square roots, one positive and one negative. For example, because 72 = 49 and (-7)2 = 49, both 7 and -7 are square roots of 49. The symbol , referred to as the radical, is used to write the principal, or positive, square root of a positive number. 49 = 7 is read "The positive square root of 49 equals 7." A negative square root is designated by the symbol − . 1.01-19 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide ­ 49 = -7 is read "The negative square root of 49 equals -7." It is often convenient to use plus-or-minus notation: ± 49 means the positive or negative square root of 49. Therefore, the rule is that every positive real number a has two square roots: a and - a . 2It follows from the definition of square root that ( a ) = a, and that a 2 = a. Because the square of every real number is either positive or zero, negative numbers do not have square roots in the set of real numbers.

Notice that 4 ⋅ 25 = 100 = 10 , and 4 ⋅ 25 = 2 ⋅5 = 10 . Therefore, 4 ⋅ 25 = 4 ⋅ 25 Therefore, in general, we can say: For any nonnegative real numbers a and b: a ⋅b = a ⋅ b a aIt also follows that = b b Calculator Method To calculate the square root of any number using a scientific calculator, follow these steps: 1) Enter the number. (Fractions must first be converted to decimal form.) 2) Press the x key. The number displayed will be the square root of the number entered in step 1. An alternate method is to press the yx key and then type 0.5. This raises the number in step 1 to the power of 0.5, or ½. Other roots For informational purposes only, we mention the fact that other roots may be found for a number. One of these is the cube root. To cube a number means to multiply the number by itself three times, i.e., raise it to the third power. (Consider that a cube is three-dimensional.) For example, 2 cubed is 8, since 2 × 2 × 2 = 8. The cube root (or third root) of a number, then, is the number that, when raised to the third power (cubed), equals the first number. The notation for a cube root is 3 a . Note that any root may be taken from a number to "undo" an exponent, such as the fourth or fifth root. The general definition for a root is: If an = b, then a is the nth root of b. 1.01-20 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide The notation for the nth root is n a . These roots follow the same general rules as the square root. Practice Exercises Find the indicated square roots. 1. 16 2. 144 3. 2)6( 4. 64 1 5. 36 81 6. 81− 7. 215 8. 400− 9. 499 ⋅ 10. 625− 1.01.08 Convert between numbers expressed in standard form and in scientific notation. SCIENTIFIC NOTATION The difficulty in writing very large or very small numbers in the usual manner is that a large number of zeros are required to write these numbers. This difficulty is overcome by using scientific notation, in which integral powers of ten are used instead of a large number of zeros to indicate the position of the decimal point. In addition to simplifying the writing of very large or very small numbers, scientific notation clearly identifies the number of significant digits in a number and simplifies arithmetic calculations involving multiplication, division, or raising to a power. For these reasons, it is good practice to write numbers in scientific notation when these operations are involved. The following demonstrates how the number 1 million can be represented by various factors of 10: 1.01-21 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Converting From Standard Form To Scientific Notation There are two steps involved in writing a number in scientific notation. 1) Move the decimal point just to the right of the first significant digit. The first significant digit is the first non-zero digit counting from the left. 2) Indicate multiplication of the resulting number by a power of ten that makes its value equal to the original value. The power of ten is found by counting the number of places the decimal point was moved from its original position. If counted to the left, the power is positive; if counted to the right, it is negative. For example: Suppose you want to express a number such as 700 in scientific notation. Suppose you want to express 0.0014 in scientific notation. Converting from Scientific Notation to Standard Form To transform from scientific notation to standard form, follow the opposite procedure.

1.96 × 105 = 196000 2.27 × 10-2 = 0.0227 There are two parts of a number written in scientific notation, the significant digits and the power of ten. Thus, in the number 3.21 × 106, 3, 2, and 1 are the significant digits and106 is the power of ten. The ability to clearly see the number of significant digits can be helpful in performing arithmetic calculations. For example, the number of significant digits which should be reported in the product of two numbers can be readily determined if the two numbers are first written in scientific notation. When numbers are expressed in scientific notation, calculations can be more easily visualized. This is because they involve only numbers between 1 and 10 and positive and negative integral powers of ten which can be treated separately in the calculations using the rules for numbers with exponents. 1.01-22 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide 1.01.09 Add, subtract, multiply, and divide numbers expressed in scientific notation. Addition and Subtraction Using Scientific Notation Addition and subtraction cannot normally be performed directly using scientific notation because they require adding or subtracting digits of equal place value. Thus, when numbers expressed in scientific notation are to be added or subtracted, they must first be converted to forms having equal place value. This is commonly done by expressing them as numbers which are multiplied by the same integral power of ten. The sum or difference of these significant digits, multiplied by their common power of ten, is the sum or difference of the original numbers. For example: (3.54 × 105) + (2.51 × 104) 3.54 × 105 is first changed to 35.4 × 104 35.4 ¯ 104 + 2.41 ¯ 104 37.91¯ 104 = 3.79 ¯ 105 Multiplication and Division Using Scientific Notation Multiplication or division of numbers using scientific notation is performed by multiplying or dividing the significant digits and the powers of ten separately. The significant digits are multiplied or divided in the same manner as other mixed decimals. The powers of ten are multiplied or divided by adding or subtracting their exponents using the rules for multiplication and division of numbers with exponents. For example: (2.7 × 102)(3.1 × 10-3) = (2.7)(3.1) × (102)(10-3) = 8.37 × 10-1 which should be rounded off to 8.4 × 10-1 One of the most useful applications of scientific notation is in arithmetic calculations which involve a series of multiplications and divisions. The use of scientific notation permits accurate location of the decimal point in the final answer. For example: Perform the following calculation using scientific notation: (219)(0.00204) (21.2)(0.0312) 1.01-23 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide 1) Write each term in scientific notation: (2.19 ×102)(2.04 ×10-3) (2.12 ×101)(3.12 ×10-2) 2) Multiply and divide the significant digits: (2.19)(2.04) −4.46 = = 0.675 (2.12)(3.12) 6.61 3) Multiply and divide the powers of ten by adding and subtracting exponents: 2 −3 −1(10 )(10 ) 10 −1+1 0= = 10 = 10 1 −2 −1(10 )(10 ) 10 4) Combine the results: 0.675 × 100 = 0.675 = 6.75 × 10-1 "E" Notation An alternate method for annotating scientific notation is often used by pocket calculators, computers, and some references. The method uses an E in place of the "× 10," and the number written after the E is the exponent of 10. The standard and alternate methods for scientific notation are equivalent and can be converted from one form to another without a change in value. The examples below use both methods in equivalent expressions:

3.79 ×105 = 3.79E5 4.02 ×10-6 = 4.02E-6 5.89 ×100 = 5.89E0 Using "E" Notation with a Calculator Numbers in scientific notation are entered into a scientific calculator as follows: 1) Enter the significant digits. 2) Press the E or EXP key. (Actual key label may vary.) 3) Enter the power of 10. If the power is negative press the +/- key in conjunction with entering the power. 1.01-24 http:2.12)(3.12 http:2.19)(2.04 http:101)(3.12 http:102)(2.04 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Practice Problems 1. (2 × 10-2 )(3 × 102) 2. (6 × 10-8)/(3 × 103) 3. (9 × 104)(- 1 × 10-2) 4. (3 × 10-7)(9 × 102) 5. (7E2)(6E4) 6. (5E - 3)/(5E - 2) 1.01.10 Solve equations using the "Order of Mathematical Operations. ORDER OF MATHEMATICAL OPERATIONS In solving any equation it is necessary to perform the operations in the equation in accordance with a certain hierarchy or order of operations. Equations are solved by simplifying operations of higher order first, according to group, left to right. The order for solving equations is as follows: 1) Simplify expressions within grouping symbols, beginning with the innermost set if more than one set is used. 2) Simplify all powers. 3) Perform all multiplications and divisions in order from left to right. 4) Perform all additions and subtractions in order from left to right. For example: (3 + 1)2 × 3 - 14 ÷ 2 1) Simplify parentheses: (3 +1)2 ×3 −14 ÷ 2 123 2) Simplify powers: (4)2 ×3 −14 ÷ 2 4 2 41 4 4 3 3) Perform multiplication and 16×3 −14 ÷ 2 division left to right:123 123 4) Perform subtraction: 48 - 7 = 41 (final answer) 1.01-25 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Practice Problems 1. 5 + ( - 3) - ( - 2) - 6 6. 33 + 10 ÷ 5 2. 18 - [52 ÷ (7 + 6)] 7. (8 + 4 – 3)2 3. 19 - 7 + 12 - 2 ÷ 8 8. 7(42 - 10) ÷ (12 - ¾) 4. 23 - 20 ÷ 4 + 4 – 3 9. 7(6 – 22 ) 5. 10 + 6(0.5)3 10. (57 – 25 )1/2 1.01.11 Perform algebraic functions. ALGEBRA Algebra is the branch of mathematics which deals with the manipulation of words and letters, generically called symbols, which represent numbers. Two factors contribute to the widespread use of algebra in scientific calculations. First, by using words and letters to represent the values of physical quantities, physical relationships can be expressed in clear, concise, and completely generalized form. Second, by using words and letters in place of numbers, combinations of physical relationships may be simplified to yield results applicable to any set of numbers. For example, the area of a rectangle equals the product of the length of the rectangle multiplied by its width. In generalized terms, this statement can be written as: Area = Length × Width. This expression is a simple rule which tells the relationship between the area and the length and width of a rectangle. It does not mean that words are multiplied together but rather that numbers are inserted for the length and the width to obtain the area. For example, if the length is 4 feet and the width is 2 feet, the area is 2 feet × 4 feet or 8 square feet. This expression can be further simplified by using symbols or letters instead of words. For example, if area is designated by the letter A, length designated by the letter l, and width designated by the letter w, the following expression results: A = l × w or A = lw In algebraic expressions, when two or more letters representing numbers are written next to each other without a symbol between them, multiplication is indicated.

Variables vs. Numbers When words or letters are used to represent numbers, they are called variables. Thus, when letters like x, y, z, f, or k are used to represent the values of physical quantities, they are called variables because their value varies with the actual numbers they may be chosen to represent. In the area calculation above, A, l, and w are variables used to represent the numerical values of area, length and width, respectively. 1.01-26 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Properties of Variables Recall that every number has a sign and an exponent associated with it. Recall also that any number can be written as a fraction by putting that number as the numerator and 1 as the denominator, e.g., 5 = 5/1. These properties also apply to any symbols that we might use to represent numbers. Additionally, a symbol by itself stands for one of whatever the variable represents. That is to say, the symbol a by itself means "one of the variable represented by the letter a", or 1a. Combining this with the other "invisible" properties mentioned, the symbol a is understood to represent "positive one of the variable represented by the letter a to the power of one, over 1," which would be expressed as: +1a1 = a 1 An expression that is either a numeral, a variable or the product of a numeral and one or more variables is called a monomial. A combination or sum of monomials is called a polynomial. Examples of each are: Monomials: 12 z br -4x3 Polynomials: 3x + 9 6a2 – 15 Equations An equation is a statement that indicates how two quantities or expressions are equal. The two quantities are written with an equal sign (=) between them. For example, 1 + 1 = 2 10 = 6 - ( - 4) 5 × 3 = 15 18 ÷ 2 = 9 are all equations because in each case the quantity on the left side is equal to the quantity on the right side. In algebra we use variables to represent numbers in equations. In this lesson we will manipulate and solve equations involving more than one variable, but we will find the solution, i.e., the final answer, to equations having only one variable. Algebraic Manipulation The basic principle, or axiom, used in solving any equation is: whatever operation is performed on one side of an equation - be it addition, subtraction, multiplication, division, raising to an exponent, taking a root - must also be performed on the other side if the equation is to remain true. This principle MUST be adhered to in solving all types of equations. 1.01-27 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide This axiom can be thought of by visualizing the balancing of a scale. If the scale is initially balanced, it will remain balanced if the same weight is added to both sides, if the same weight is removed from both sides, if the weights on both sides are increased by the same factor, or if the weights on both sides are decreased by the same factor. Here are the general forms for algebraic manipulation of equations. For the real numbers a, b, c and n: Manipulating and Solving Linear Equations The addition or subtraction of the same quantity from both sides of an equation may be accomplished by transposing a quantity from one side of the equation to the other. Transposing is a shortened way of applying the addition or subtraction axioms. Any term may be transposed or transferred from one side of an equation to the other if its sign is changed. Thus, in the equation below the +4 can be transposed to the other side of the equation by changing its sign:

5x + 4 = 14 (5x + 4) - 4 = (14) - 4 5x = 14 - 4 5x = 10 Transposing also works with multiplication and division. Remembering that any number can be expressed as a fraction we can rewrite the last line of the equation above. We can then move the 5 in the numerator of the left side to the denominator of the right side: 1.01-28 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Some linear equations may contain multiple terms (monomials) involving the same variable. In order to simplify the equation like terms must be combined. Don't forget those "invisible properties of variables." Here's an example: Quadratic Equations In manipulating an equation involving multiple variables, the "variable of interest" (or the variable to be solved for) must be moved to one side of the equal sign and all other variables must be moved to the other side. In order to accomplish this, operations must be performed on both sides of the equation that will result in a variable or group of variables, to be canceled out from one side. This cancellation can only occur if the "opposite function" is performed on a function that already exists on that side of the equation. This means that a variable that is being multiplied can be canceled by dividing by the same variable. Addition can be canceled with subtraction, multiplication with division, etc. For example: Another example: 1.01-29 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Do not forget that the order of operations must be observed when manipulating equations. Otherwise a completely different solution may result. The key is to do the opposite function in reverse order. Here is an example which shows how this is done. Once the order of the arithmetic functions has been established, manipulation of the formula can begin. In the example above, if the values for a, b, and c were known, the first step would be to add b to a. The second step would be to divide by c. Therefore, in order to solve it, we do the opposite functions in reverse order. So, we first multiply by c. Then, we would subtract b. It is a good idea to rewrite the equation each time so that the operations can be reevaluated before the next step. Substitution Linear equations are solved by combining like terms and reducing to find the solution. Quadratic equations are solved by substituting given values into the equation for all but one of the variables, thus making it a linear equation. The best approach is to first solve for the variable of interest by algebraic manipulation. Then find the solution by substituting the given values into the equation for the respective variables. The single, unknown variable, or the variable of interest, will be left on one side, being set equal to the solution. For example: Given the equation: 2x – y2 = 3a - b; where x = 5, y = (-4) and a = 3; solve for b: 1.01-30 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Practice Problems Solve for the unknown variable: Substitute the values given and simplify the expression. 1.01.12 Solve equations using common and/or natural logarithms. LOGARITHMS

In many cases, arithmetic operations can be performed much more quickly if the numbers involved are numbers with exponents to the same base. For example, the multiplication or division of numbers with exponents to the same base can be performed by merely adding or subtracting the exponents. Raising to a power or taking a root can be performed by merely multiplying or dividing the exponents by the power or root. It is this feature of numbers with exponents which led to the development of logarithms. If all numbers could be readily written as numbers with exponents to the same base, multiplication, division, raising to powers and taking roots could be performed much more quickly. 1.01-31 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Any number can be expressed as a power of any other number. For example, 64 equals 26, 43, or 82. 64 also equals 72.137 or 101.806 . The use of logarithms involves expressing numbers as powers of a common number, such as 10, so that arithmetic operations with these numbers can be performed more quickly. Simply put, a logarithm is an exponent. More explicitly, the logarithm of a number n is the exponent x of a given base b that is required to produce that number. The symbol log is used to denote taking a logarithm. The base is usually indicated by a small number written to the right and slightly below the symbol log (e.g., logb). The general relationship and form are as follows: If n = bx; where b > 0 and b ≠ 1, then: logb n = x. For example: 1000 = 103 – log10 1000 = 3 This says that the base ten logarithm of 1000 is 3, which means that the base number, 10, must be raised to the power of 3 to equal 1000. Here are some additional examples: Before the development of the scientific calculator, the use of logarithms saved considerable computation time. For example, the evaluation of the following expression by hand would take a very long time. However, using logarithms the above expression could be evaluated in a matter of minutes. Thus, logarithms, or logs, became one of the most useful tools in mathematics. In addition to simplifying arithmetic calculations and shortening computation time, logs are also important in engineering applications. The relationship between a number and its logarithm is used frequently to assist in measuring physical quantities when they vary over a wide range. For example, logarithmic scales are used to measure the neutron flux in nuclear reactors. Logarithms are also used for scales on charts and meters. 1.01-32 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Properties of Logarithms Since logarithms are exponents, the basic rules of exponents can be used to develop several useful properties of logarithms. Suppose that a, x, and y are numbers, and a is a suitable base for a logarithm (a > 0, a ≠ 1). The product rule for exponents says: x y (x+ y)a ⋅ a = a Let us say that: u = ax and v = ay If we write each of these in logarithmic form we would have: x = loga u and y = loga v x y (x+ y)Then: u ⋅ v = a ⋅ a = a If we write this in logarithmic form it would be: log (u ⋅ v) = x + y a If we substitute the values for x and y from above we have: log a (u ⋅v) = log u + log v a a This results in one of the rules for logarithms, the product rule. Using similar methods, we could also prove the other two rules that have been developed for logarithms. 1.01-33 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide

Base Ten Logarithms Logs with the base of 10 are the most commonly used logarithms because of their relationship to the place values in the decimal system. Because of their wide use, base ten logarithms are often referred to as common logarithms. Observe the patterns in the number line and table below. Notice the relationship between the power of ten and the logarithm. Any number can be expressed as a power of ten. Thus, 10 equals 101, 1,000 equals 103, 64 equals 101.806 and 527.3 equals 102.7221. Once a number has been expressed as a power of ten, the base ten logarithm of the number is known-it is the exponent of 10. Thus log10 10 equals 1, log10 1000 equals 3, log10 64 equals 1.806 and log10 527.3 equals 2.722. Since base ten logarithms are so commonly used, the subscript 10 is often omitted after the symbol log. Thus, log 27.3 means the logarithm of 27.3 to the base 10. A common logarithm is most often a mixed number consisting of a whole number part and a decimal fraction part. The whole number part is called the characteristic of the logarithm. The decimal fraction part is called the mantissa. For example, in the logarithm of 527.3, which equals 2.7221, the characteristic is 2 and the mantissa is 0.7221. The mantissas of most logarithms are rounded off to a specified number of significant digits, typically four. 1.01-34 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide 1.01-35 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide 1.01-36 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Log-Table Method To find the common logarithm of a number using four-place log tables, use the following steps: 1) Write the number in scientific notation with up to four significant digits. 2) Using the product rule for logarithms, write the log of the product as the sum of the logs of the factors. 3) Determine the mantissa as follows: a) Find the row in the table corresponding to the first two digits of the number, then move over to the column corresponding to the third digit of the number. b) Find the number under the proportional parts section corresponding to the fourth digit of the number, and add it to the last digit of the decimal obtained in step 3.a. 4) Determine the characteristic by using the power of 10 written in step 1 and write the logs in the form of a sum. NOTE: If the power of 10 is negative, the log may be left in this form. 5) Using the product rule, write the sum of the logs as the log of a product. NOTE: If the power of 10 is negative, the mantissa will be changed because of the subtraction of a whole number. Example: log 45,830 1) Write the number in scientific notation. log (4.583 × 104 ) 2) Use the product rule to write the product as the sum of the logs. log 4.583 + log 104 3) a) Find the 4.5 row of the log table. Move over to the 8 column to find the mantissa. 0.6609 log 4.583 + 4 b) Find the 3 column under the proportional parts section of the log table. Add this to the last digit of the mantissa in step 3.a. + 0.0003 0.6612 4) Write the mantissa and characteristic as a sum. 0.6612 + 4 5) Add the characteristic to the mantissa. 4.6612 1.01-37 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Example: log 0.004583 1) First, we write the number in scientific notation. log (4.583 × 10-3) 2) Next, we use the product rule to write the product as log 4.583 + log 10-3

the sum of the logs. log 4.583 + ( - 3) 3) The mantissa will be the same as in the previous example because the significant digits are the same. 0.6612 4) Now, we write the mantissa and characteristic as a 0.6612 + ( - 3) sum. Since the characteristic is negative we could leave the logarithm in this form. 0.6612 - 3 5) Add the characteristic to the mantissa. Note that the - 2.3388 mantissa has changed. Also note that this is not the same as - 3.6612. As you may have observed, the mantissa of the base ten logarithm of a number depends only on the succession of significant digits in the number. The position of the decimal point in the number does not affect the mantissa. Of course, the characteristics are different for each of these numbers. Note that any time the logarithm of a number is rounded off it would be considered an approximate answer since each digit is necessary to exactly duplicate the number when the base is raised to that exponent. Since each significant digit of the logarithm affects the actual value of the number, a standard of four significant digits should be maintained to ensure appropriate accuracy in the answers. 1.01-38 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Calculator Method Since hand-held scientific calculators are readily available today, it is impractical to use log tables. To find the logarithm of a number with a calculator: 1. Enter the number. 2. Press the log key. The number displayed is the logarithm of the number entered in step 1. Natural Logarithms A logarithm can be written to any base. For most practical computations, base ten logarithms are used because of their relationship to the place values in the decimal system. However, in many scientific and engineering problems, it is convenient to use another base, symbolized by the letter e. e is an irrational number whose value is 2.71828 . . . The actual value of e is the limiting value of (1 + 1/n)n as n gets larger and larger. Although it is an irrational number, it can still be used as the base for logarithms in the same way as 10 is used for base ten logarithms. e is the basis for many laws of nature, such as the laws of growth and decay of physical quantities, including the decay of radioactive substances and the growth and decay of neutron population in a nuclear reactor. Because of the relationship of e to natural phenomena, logarithms to the base e are called natural logarithms. The natural logarithm of a number is the exponent to which e must be raised in order to get that number. The symbol ln is used to denote a natural logarithm which is the same as saying loge. The relationship is expressed as follows: If ex = n then ln n = x For example: 0.693147... = 2. ln 2 = 0.693147 . . . which means that e 2.302585... = 10ln 10 = 2.302585 . . . which means that e ln e = 1 which means that e1 = e Natural logarithms are not often used for computations. However, they appear frequently in decay and shielding calculations problems because of the relationship of e to natural phenomena. As a result, it is important to know how to determine the natural logarithms of numbers. Tables of natural logarithms are available in several standard handbooks. However, there are several important differences between natural logarithms and base ten logarithms which must be understood to use natural logarithms. A natural logarithm is not separated into a characteristic and a mantissa. This is because the whole number part of a natural logarithm

1.01-39 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide ln does not relate to the position of the decimal point. Therefore, tables of natural logarithms give the entire logarithm, not just the decimal fraction part. Moreover, if a natural logarithm is negative, the entire logarithm is negative and is shown as such in a table of natural logarithms. Further, there is no part of the natural logarithm of a number which is not affected by the position of the decimal point. For all these reasons, tables of natural logarithms cannot be made concise. To find the natural log of a number using a hand-held calculator: 1. Enter the number. 2. Press the key. The number displayed is the natural logarithm of the number entered in step 1. Antilogarithms An antilogarithm, usually shortened to "antilog," is the opposite of a logarithm and is much easier to do. The antilog of a given number is the value obtained by raising the base to that number. Finding antilogs is an important part in the overall use of logarithms in computations. If numbers are converted to logarithms to perform calculations, the answer must be converted back from logarithms once the calculations have been performed. The symbol log -1 is used in calculations to indicate the antilog is going to be taken. The base of 10 is assumed unless otherwise noted. The general form is: log -1 x = n which means 10 x = n For example: log -1 3 which means 103 = 1000 To find the antilog of a number using log tables: log -1 2.7832 1) Write the number in log long form. log -1 (0.7832 + 2) 2) Find the mantissa in the table. Take the 6.0 value at the head of the row and attach to it 7 the value at the head of the column. 6.07 3) Write this value in scientific notation, 6.07 × 10 2 putting the characteristic of the original number as the power of ten. 4) Write this number in standard form. 607 1.01-40 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide On a scientific calculator the antilog of a number is obtained by raising the base (10) to that number. 1) Enter the number. 2) Press the log-1 or 10 x key. The number displayed is the antilog of the number log of entered in step 1. In other words, 10 raised to that power. The symbol ln-1 is used to denote the inverse natural log, i.e. the antilog of base e. xln-1 x = n which means e = n For example: 0.693... = 2ln-1 0.693 . . . which means e On a scientific calculator the inverse natural log of a number is obtained as follows: 1) Enter the number. ln-1 x2) Press the or e key. The number displayed is the inverse natural log of the number of entered in step 1. In other words, e raised to that power. Solving for Variables as Exponents One of the useful applications for logarithms is to solve algebraic equations with unknown exponents. In the following example, for instance, if the exponent is not known, it would be difficult to determine the correct value of x in order to make the statement (or equation) true. 2356 = 3x With the use of logarithms, however, this type of problem can be easily solved. The steps for solving an equation of this type are: 1.01-41 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide 1) Make sure the base raised to the unknown exponent is isolated on one side of the equation (this may involve some manipulation of the formula in more complicated equations). 2) Take the log of both sides of the equation:

log 2356 = log 3x 3) The right side of the equation can be rewritten using the power rule: log 2356 = x (log 3) 4) Divide both sides by log 3 which moves it to the right side of the equation: log 2356 x(log 3) = log 3 log 3 5) Cancel terms and rewrite the equation: log 2356 = x log 3 6) Perform the operations and solve: 7) This answer can now be checked by substituting it back into the original equation to see if it makes the statement true: 2356 = 37.068 Some problems may involve the base of the natural logarithm e raised to an unknown power. This exponent can be determined by isolating e on one side of the equation and then taking the natural log of both sides. This is done because the natural log of e is 1. For example: n125 = 1000e 125 n= e 1000 125 nln = ln e 1000 ln 0.125 = (ln e)(n) ln 0.125 = (1)(n) -2.0794 = n 1.01-42 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Practice Problems Perform the following operations: 1. log 6.40 5. ln 86 2. log 0.5 6. ln 0.5 3. log-1 16 7. ln-1 0.695 4. log-1 3.7846 8. ln-1 1 Use logarithms to solve for the unknown variable: 9. 5 = 4x 11. 9 = 22(3)y 10. 23 = 6t 12. 50 = 2000(½)n SUMMARY A good foundation in mathematics is essential for the RCT. Calculations of various types are performed routinely in radiological control operations. The skills learned in this lesson will be applied in many of the lessons that follow as well as in the workplace. 1.01-43 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide ANSWERS TO PRACTICE PROBLEMS: Fractions 1. 1/3 + 2/3 = 3/3 = 1 2. 5/7 _ 3/7 = 2/7 3. 5/9 + 2/3 = 5/9 + 6/9 = 11/9 = 1 2/9 4. 6/7 _ 1/2 = 12/14 _ 7/14 = 5/14 5. 2 _ 1/3 = 6/3 _ 1/3 = 5/3 = 1 2/3 6. 3/8 + 15/16 = 6/16 + 15/16 = 21/16 = 1 5/16 7. 25/32 _ 3/4 = 25/32 _ 24/32 = 1/32 8. 15/21 _ 4/7 = 5/7 _ 4/7 = 1/7 9. 13/20 _ 2/5 = 13/20 _ 8/20 = 5/20 = 1/4 10. 7/18 + 5/9 = 7/18 + 10/18 = 17/18 11. 2/3 × 1/5 = 2/15 12. 4/7 × 3/4 = 3/7 13. 1/2 × 2 = 1 14. 3/5 × 4 = 12/5 = 2 2/5 15. 4/9 ÷ 2/3 = 4/9 × 3/2 = 2/3 16. 8/13 × 2/3 = 16/39 17. 12/15 × 3/5 = 36/75 18. 20/25 ÷ 4/5 = 20/25 × 5/4 = 4/5 × 5/4 = 1 19. 7/8 × 2/5 = 7/4 × 1/5 = 7/20 20. 14/21 ÷ 2/7 = 14/21 × 7/2 = 2/3 × 7/2 = 7/3 = 2 1/3 Decimals 1. 0.23 + 3.4 = 3.63 2. 5.75 _ 2.05 = 3.7 3. 6.1 _ 1.6 = 4.5 4. 0.018 + 0.045 = 0.063 5. 468.75 _ 192.5 = 276.25 6. 2.3 × 3.2 = 7.36 7. 0.007 × 2.18 = 0.01526 8. 5.2 ÷ 1.4 = 3.7143 9. 12.26 ÷ 0.04 = 306.5 10. 4.0 × 0.25 = 1 1.01-44 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Fraction/Decimal Conversions 1. 1/2 = 0.5 2. 2/5 = 0.4 3. 5/8 = 0.625 4. 12/25 = 0.48 5. 13/39 = 0.¯3 6. 7/16 = 0.4375 7. 0.125 = 125/1000 = 1/8 8. 0.6666 = 2/3 9. 4.25 = 4 25/100 = 4¼ 10. 0.2 = 2/10 =1/5 Percent 1. 0.5 = 50% 2. 0.782 = 78.2% 3. 1.1 = 110% 4. 0.06 = 6% 5. 0.049 = 4.9% 6. 0.0055 = 0.55% 7. 65% = 0.65 8. 0.25% = 0.0025 9. 300% = 3 10. 0.09% = 0.0009 Signed Numbers 1. (-28) + (-51) = -79 2. (-2) + (-5) = -7 3. 40 + (-21) = 19 4. -87 + 50 = -37 5. 48 + (-27) = 21 6. 56 - (-5) = 61 7. 81 - 4 = 77 8. -48 - (-2) = -46 9. -4(5) = -20 10. 6/(-3) = -2 11. 4 (-5) = -20 12. (-6)/(-3) = 2 13. (-6)/3 = -2 14. (-8)(-5) = 40 15. (-7)(6) = -42 1.01-45 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Exponents 1. (32)(33) = 35 2. 75/73 = 72 3. (105)(106) = 1011 4. (106)(10-4) = 102 5. 6-4/6-3 = 6-1 6. 6-8/63 = 6-11 Square Roots 1. 2. 3. 4. 5. 6. 7. 8. 416 = 981 = −−

12144 = 15152 = 66)( 2 = 20400 = −− 8 1 64 1 = 739 49 22 ⋅=⋅ = 32 ⋅ 7 2 213 7 =⋅= 281 81 9 9 39. = = = = 2 6 236 36 6 210. − 625 = − 25 = −25 Scientific Notation 1. (2 × 10-2)(3 × 10-2) = 6 × 10-4 2. (6 × 10-8)/(3 × 103) = 2 × 10-11 3. (9 × 104)(-1 × 10-2) = -9 × 102 4. (3 × 10-7)(9 × 102) = 2.7 × 10-4 5. (7E2)(6E4) = 4.2E7 6. (5E-3)/(5E-2) = 1E-1 Order of Operations 1. 5 + (-3) - (-2) - 6 = 5 + 6 - 6 = 5 2. 18 - [52 ÷ (7 + 6)] = 18 - (52 ÷ 13) = 18 - 4 = 14 3. 19 - 7 + 12 - 2 ÷ 8 = 12 + 24 ÷ 8 = 12 + 3 + 15 4. 23 - 20 ÷ 4 + 4 - 3 = 8 - 5 + 12 = 3 + 12 = 15 5. 10 + 6(0.5)3 = 10 + 6(0.125) = 10 + 0.75 = 10.75 6. 33 + 10 ÷ 5 = 27 + 2 = 29 7. (8 + 4 - 3)2 = (8 + 12)2 = 202 = 400 1.01-46 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide 8. 7(42 - 10) ÷ (12 - ¾) = 7(16 - 10) ÷ 9 = 7(6) ÷ 9 = 42 ÷ 9 = 4.667 9. 7(6 - 22) = 7(6 - 4) = 7(2) = 14 10. (57 - 25)½ = (57 - 32)½ = 25½ = 5 Algebra 1. x + 3 = 10 x + 3 – 3 = 10 – 3 x = 7 2. 5 + z = 8 5 + z – 5 = 8 – 5 z = 3 y3. = 2 4 y4 ⋅ = 2 ⋅ 4 4 y = 8 4. r = 6 2 ⋅ r2 = 6 ⋅ 2 2 r = 12 5. 5(j + 5) = 45 5(j + 5) = 45 5 5 j + 5 = 9 j + 5 – 5 = 9 – 5 j = 4 6. 5d = 25 5d 25 = 5 5 d = 5 7. 9n 2 = 81 29n 81 = 9 9 2n = 9 2n = 9 n = 3 1.01-47 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide 8. a + 6 − 6a = −14 6 − 5a = −14 6 − 5a − 6 = −14 −5a = −20 −5a −20 = − 5 − 5 a = 4 9. a 2b = c 2 d 2 2a b c d = b b 2 c 2 d a = b 2c d2a = b 2c d a = b b c10. = a d b c a ⋅ = ⋅ a a d cab = d cad ⋅b = ⋅ d d db = ca db ca = c c db = a c 11. 3x +14 y − xy where x = 5, y = 2 3(5) + 14(2) – 5(2) 15 + 28 – 10 43-10 33 1.01-48 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide 4s +112. where s = -3, t = 10 6t + s 4(−3) +1 6(10) + (−3) −12 +1 60 + (−3) −11 57 16a + 5b13. where a = -1, b = 5 1− 4a 16(−1) + 5(5) 1− 4(−1) −16 + 25 1+ 4 9 5 41 5 j14. 3i + = 0 where k = 2, j = -12, solve for i 1+ k −123i + = 0 1+ 2 3i − 4 = 0 3i = 4 4i = 3 1 i = 1 3 15. 7(2m - ¾n) - l - 10 where m - ½ n - 8, solve for l 7[2(½) - ¾(8)] - l - 10 7(1 - 6) - l - 10 7(-5) - l - 10 -35 - l - 10 l - 45 1.01-49 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Logarithms 1. log 6.40 = 0.8062 2. log 0.5 = _0.301 3. log-1 16 = 1 × 1016 4. log-1 3.7846 = 6089.76 5. ln 86 = 4.45 6. ln 0.5 = _0.693 7. ln-1 0.695 = 2.004 8. ln-1 1 = e = 2.7182 . . . 9. 10. 11. 1.01-50 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide 12. 1.01-51 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide This page intentionally left blank. 1.01-52 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Course Title: Radiological Control Technician Module Title: Unit Analysis & Conversion Module Number: 1.02 Objectives : 1.02.01 Identify the commonly used unit systems of measurement and the base units for mass, length, and time in each system. 1.02.02 Identify the values and abbreviations for SI prefixes. 1.02.03 Given a measurement and the appropriate conversion factor(s) or conversion factor table, convert the measurement to the specified units. 1.02.04 Using the formula provided, convert a given temperature measurement to specified units. INTRODUCTION

A working knowledge of the unit analysis and conversion process is necessary for the Radiological Control Technician. It is useful for air and water sample activity calculations, contamination calculations, and many other applications. This lesson will introduce the International System of Units (SI), the prefixes used with SI units, and the unit analysis and conversion process. Many calculations accomplished in radiological control are actually unit conversions, not complex calculations involving formulas that must be memorized. REFERENCES: 1. "Health Physics and Radiological Health Handbook"; Shleien; 1992. 2. DOE-HDBK-1010-92 (June 1992) "Classical Physics" DOE Fundamental Handbook; US Department of Energy. 3. "Chart of the Nuclides"; Sixteenth Edition, Knolls Atomic; 2003. UNITS AND MEASUREMENTS Units are used in expressing physical quantities or measurements, i.e., length, mass, etc. All measurements are actually relative in the sense that they are comparisons with some standard unit of measurement. Two items are necessary to express these physical quantities: a number which expresses the magnitude and a unit which expresses the dimension. A number and a unit must both be present to define a measurement. Measurements are algebraic quantities and as such may be mathematically manipulated subject to algebraic rules. 1.02-1 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Fundamental Quantities All measurements or physical quantities can be expressed in terms of three fundamental quantities. They are called fundamental quantities because they are dimensionally independent. They are: � Length (L) � Mass (M) (not the same as weight) � Time (T) Figure 1. Fundamental Units Derived Quantities Other quantities are derived from the fundamental quantities. These derived quantities are formed by multiplication and/or division of fundamental quantities. For example: � Area is the product of length times length (width), which is L × L, or L2. � Volume is area times length, which is length times length times length, or L3. � Velocity is expressed in length per unit time, or L/T. � Density is expressed in mass per unit volume, or M/L3. 1.02-2 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide 1.02.01 Identify the commonly used unit systems of measurements and the base units for mass, length, and time in each system. SYSTEMS OF UNITS The units by which physical quantities are measured are established in accordance with an agreed standard. Measurements made are thereby based on the original standard which the unit represents. The various units that are established, then, form a system by which all measurements can be made. English System The system that has historically been used in the United States is the English System, sometimes called the English Engineering System (EES). Though no longer used in England, many of the units in this system have been used for centuries and were originally based on common objects or human body parts, such as the foot or yard. Though practical then, the standards for these units were variable as the standard varied from object to object, or from person to person. The base units for length, mass, and time in the English system are the foot, pound, and second, respectively.

Even though fixed standards have since been established for these antiquated units, no uniform correlation exists between units established for the same quantity. For example, in measuring relatively small lengths there are inches, feet, and yards. There are twelve inches in a foot, and yet there are only three feet in a yard. This lack of uniformity makes conversion from one unit to another confusing as well as cumbersome. However, in the U.S., this system is still the primary system used in business and commerce. Table 1. English System Base Units Physical Quantity Unit Abbr. Length: foot ft. Mass: pound lb. Time: second sec. 1.02-3 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide International System of Units (SI) Since the exchange of scientific information is world-wide today, international committees have been set up to standardize the names and symbols for physical quantities. In 1960, the International System of Units (abbreviated SI from the French name Le Système Internationale d'Unites) was adopted by the 11th General Conference of Weights and Measures (CGPM). The SI, or modernized metric system, is based on the decimal (base 10) numbering system. First devised in France around the time of the French Revolution, the metric system has since been refined and expanded so as to establish a practical system of units of measurement suitable for adoption by all countries. The SI system consists of a set of specifically defined units and prefixes that serve as an internationally accepted system of measurement. Nearly all countries in the world use metric or SI units for business and commerce as well as for scientific applications. 1.02.02 Identify the values and abbreviations for SI prefixes. SI Prefixes The SI system is completely decimalized and uses prefixes for the base units of meter (m) and gram (g), as well as for derived units, such as the liter (l) which equals 1000 cm3. SI prefixes are used with units for various magnitudes associated with the measurement being made. Units with a prefix whose value is a positive power of ten are called multiples. Units with a prefix whose value is a negative power of ten are called submultiples. For example, try using a yard stick to measure the size of a frame on film for a camera. Instead you would use inches, because it is a more suitable unit. With the metric system, in order to measure tiny lengths, such as film size, the prefix milli- can be attached to the meter unit to make a millimeter, or 1/1000 of a meter. A millimeter is much smaller and is ideal in this situation. On the other hand, we would use a prefix like kilo- for measuring distances traveled in a car. A kilometer would be more suited for these large distances than the meter. 1.02-4 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Table 2. SI Prefixes PREFIX FACTOR SYMBOL PREFIX FACTOR SYMBOL yotta 1024 Y deci 10-1 d zetta 1021 Z centi 10-2 c exa 1018 E milli 10-3 m peta 1015 P micro 10-6 µ tera 1012 T nano 10-9 n giga 109 G pico 10-12 p mega 106 M femto 10-15 f kilo 103 k atto 10-18 a hecto 102 h zepto 10-21 z deka 101 da yocto 10-24 y Prior to the adoption of the SI system, two groups of units were commonly used for the quantities length, mass, and time: MKS (for meter-kilogram-second) and CGS (for centimeter-gram-second). Table 3. Metric Subsystems Physical Quantity CGS MKS Length: centimeter meter Mass: gram kilogram Time: second second

SI Units There are seven fundamental physical quantities in the SI system . These are length, mass, time, temperature, electric charge, luminous intensity, and molecular quantity (or amount of substance). In the SI system there is one SI unit for each physical quantity. The SI system base units are those in the metric MKS system. Table 4 lists the seven fundamental quantities and their associated SI unit. The units for these seven fundamental quantities provide the base from which the units for other physical quantities are derived. For most applications the RCT will only be concerned with the first four quantities as well as the quantities derived from them. 1.02-5 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Radiological Units In the SI system, there are derived units for quantities used for radiological control. These are the becquerel, the gray, and the sievert. The SI unit of activity is the becquerel, which is the activity of a radionuclide decaying at the rate of one spontaneous nuclear transition per second. The gray is the unit of absorbed dose, which is the energy per unit mass imparted to matter by ionizing radiation, with the units of one joule per kilogram. The unit for dose equivalence is the sievert, which has the units of joule per kilogram. These quantities and their applications will be discussed in detail in Lesson 1.06. Other units There are several other SI derived units that are not listed in Table 4. It should be noted that the SI system is evolving and that there will be changes from time to time. The standards for some fundamental units have changed in recent years and may change again as technology improves our ability to measure even more accurately. 1.02-6 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Table 4. International System (SI) Units Physical Quantity Unit Symbol Dimensions Base units Length: meter m m Mass: kilogram kg kg Time: second s or sec. s Temperature: kelvin K K or °K Electric current: ampere A or amp A or (C/s) Luminous intensity: candela cd cd Molecular quantity: mole mol mol Selected derived units Volume: cubic meter m3 m3 Force: newton N kg·m/s2 Work/Energy: joule J N·m Power: watt W J/s Pressure: pascal Pa N/m2 Electric charge: coulomb C A·s Electric potential: volt V J/C Electric resistance: ohm Σ V/A Frequency: hertz Hz s-1 Activity: becquerel Bq disintegration/s Absorbed dose: gray Gy J/kg Equivalent dose: sievert Sv Gy·WR 1.02-7 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide 1.02.03 Given a measurement and the appropriate conversion factor(s) or conversion factor table, convert the measurement to the specified units. UNIT ANALYSIS AND CONVERSION PROCESS Units and the Rules of Algebra Remember that a measurement consists of a number and a unit. When working problems with measurements, it should be noted that the measurement units are subject to the same algebraic rules as the values. Some examples are provided below. ( ) ( )× cm = cm2cm ft3 2= ft ft 1 = yr−1 yr As a result, measurements can be multiplied, divided, etc., in order to convert to a different system of units. Obviously, in order to do this, the units must be the same. For example, a square measures one foot in length and 18 inches in width. To find the area of the square in square inches we must multiply the length by the width. However, when the measurements are in different units, and cannot be multiplied directly.

We can convert feet to inches. We know that there are 12 inches in one foot. We can use this ratio to convert 1 foot to 12 inches. Then we can then calculate the area as 12 inches × 18 inches, which equals 216 in2, which is a valid measurement. Steps for Unit Analysis and Conversion 1) Determine given unit(s) and desired unit(s). 2) Build (or obtain) conversion factor(s) -- see Conversion Tables at end of lesson A conversion factor is a ratio of two equivalent physical quantities expressed in different units. When expressed as a fraction, the value of all conversion factors is 1. Because a conversion factor equals 1, it does not matter which value is placed in the numerator or denominator of the fraction. Examples of conversion factors are: 1.02-8 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide 365days 12inches 1foot 3 1year 1 foot 2.832E4cm 3 Building conversion factors involving the metric prefixes for the same unit can be tricky. This involves the conversion of a base unit to, or from, a subunit or superunit. To do this, use the following steps: Example: 1 gram to milligrams a) Place the base unit in the numerator g and the subunit/ superunit in the mg denominator (or vice versa): b) Place a 1 in front of the g subunit/superunit: 1mg c) Place the value of the prefix on the m (milli-) = 10-3 or 1E subunit/ superunit in front of the 1E − 3g base unit: 1mg Also remember that algebraic manipulation can be used when working with metric prefixes and bases. For example, 1 centimeter = 10-2 meters. This means that 1 meter = 1/10-2 centimeters, or 100 cm. Therefore, the two conversion factors below are equal: 1E − 2m 1m = 1cm 100 cm 3) Set up an equation by multiplying the given units by the conversion factor(s) to obtain desired unit(s). When a measurement is multiplied by a conversion factor, the unit(s) (and probably the magnitude) will change; however, the actual measurement itself does not change. For example, 1 ft and 12 inches are still the same length; only different units are used to express the measurement. 1.02-9 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide By using a "ladder" or "train tracks," a series of conversions can be accomplished in order to get to the desired unit(s). By properly arranging the numerator and denominator of the conversion factor(s), given and intermediate units will cancel out by multiplication or division, leaving the desired units. Some examples of the unit analysis and conversion process follow: EXAMPLE 1. Convert 3 years to seconds. Step 1 - Determine given and desired unit(s): Given units: years Desired units: seconds. Step 2 - Build/obtain conversion factor(s): We can use multiple conversion factors to accomplish this problem: 1 year = 365.25 days 1 day = 24 hours 1 hour = 60 minutes 1 minute = 60 seconds Step 3 - Analyze and cancel given and intermediate units. Perform multiplication and division of numbers: ⎛ 3years ⎞⎛ 365.25days ⎞⎛ 24hours ⎞⎛ 60minutes ⎞⎛ 60seconds ⎞ = 94,672,800sec. ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ 1year ⎠⎝ 1day ⎠⎝ 1hour ⎠⎝ 1minute ⎠ 1.02-10 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Unit analysis and conversion process follow: EXAMPLE 2. μCi dpmWhat is the activity of a solution in if it has 2000 ? ml gallon Step 1 - Determine given and desired unit(s): dpmGiven units: gallon μCi Desired units: ml Step 2 - Build conversion factor(s):

1 liter = 0.26418 gallons 1 dpm = 4.5 E-07 µCi 1 liter = 1000 ml Step 3 - Analyze and cancel given and intermediate units. Perform multiplication and division of numbers. 2000dpm ⎞ 4.5E − μCi ⎞⎛ 0.26418 ⎛ l⎛ ⎛ 7 gal ⎞ 1 ⎞ μCi = 2.38E − 7⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ gal ⎠⎝ 1dpm ⎠⎝ 1l ⎠⎝1,000 ml ⎠ ml Practical exercises and their solutions are provided at the end of this lesson. 1.02-11 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide 1.02.04 Using the formula provided, convert a given temperature measurement to specified units. TEMPERATURE MEASUREMENTS AND CONVERSIONS Temperature measurements are made to determine the amount of heat flow in an environment. To measure temperature it is necessary to establish relative scales of comparison. Three temperature scales are in common use today. The general temperature measurements we use on a day-to-day basis in the United States are based on the Fahrenheit scale. In science, the Celsius scale and the Kelvin scale are used. Figure 2 shows a comparison of the three scales. The Fahrenheit scale, named for its developer, was devised in the early 1700's. This scale was originally based on the temperatures of human blood and salt-water, and later on the freezing and boiling points of water. Today, the Fahrenheit scale is a secondary scale defined with reference to the other two scientific scales. The symbol °F is used to represent a degree on the Fahrenheit scale. About thirty years after the Fahrenheit scale was adopted, Anders Celsius, a Swedish astronomer, suggested that it would be simpler to use a temperature scale divided into one hundred degrees between the freezing and boiling points of water. For many years his scale was called the centigrade scale. In 1948 an international conference of scientists re-named it the Celsius scale in honor of its inventor. The Celsius degree, °C, was defined as 1/100 of the temperature difference between the freezing point and boiling point of water. In the 19th century, an English scientist, Lord Kelvin, established a more fundamental temperature scale that used the lowest possible temperature as a reference point for the beginning of the scale. The lowest possible temperature, sometimes called absolute zero, was established as 0 K (zero Kelvin). This temperature is 273.15°C below zero, or -273.15°C. Accordingly, the Kelvin degree, K, was chosen to be the same as a Celsius degree so that there would be a simple relationship between the two scales. Note that the degree sign (°) is not used when stating a temperature on the Kelvin scale. Temperature is stated simply as Kelvin (K). The Kelvin was adopted by the 10th Conference of Weights and Measures in 1954, and is the SI unit of thermodynamic temperature. Note that the degree Celsius (°C) is the SI unit for expressing Celsius temperature and temperature intervals. The temperature interval one degree Celsius equals one kelvin exactly. Thus, 0°C = 273.15 K by definition. 1.02-12 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide To convert from one unit system to another, the following formulas are used: Table 5. Equations for Temperature Conversions ° − C ( F 32) or ° =C (°F )⎛ ⎞5 ° = − 32 ⎜ ⎟°F to °C 1.8 9⎝ ⎠ ⎛ ⎞F 1.8 ° +C F ⎜ ⎟ ° + 32 ° = 32 or ° = C( ) 9 ( )°C to °F 5⎝ ⎠ 1.02-13 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide K = C° + 273.15°C to K EXAMPLE 3. Convert 65° Fahrenheit to Celsius. (65° − 32)F C° = 1.8

33 C° = 1.8 ° =C 18.3°C 1.02-14 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide PRACTICAL EXERCISES: Convert the following measurements: 1. 67 mm = __________ feet. 2. 1843 ounces = __________ kg. 3. 3500 microsieverts (µSv) = __________ millirem (mrem). 4. 0.007 years = __________ minutes. 5. 5000 disintegrations per minute (dpm) = __________ millicuries (mCi). 1.02-15 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide 6. 2350 micrometer (µm) = __________ inches. 7. 2.5E-4 ergs = __________ keV. 8. 205 °F = __________ K. 9. 2E-3 rad = __________ milligray (mGy). 10. −25 °C = __________ °F. 1.02-16 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Use unit analysis and conversion to solve the following problems. 11. Light travels at 186,000 miles per second. How many feet will light travel in one minute? 12. A worker earns a monthly salary of $2500. If the worker gets paid every two weeks and works no overtime, what will be the gross amount for a given pay period? 13. An air sampler has run for 18 hours, 15 minutes at 60 liters per minute. When collected and analyzed the sample reads 7685 disintegrations per minute (dpm). What is the concentration of the sample in microcuries/cm3? 1.02-17 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide PRACTICAL EXERCISE SOLUTIONS: 1. 67 mm = __________ feet. ⎛ 67mm ⎞⎛ 1m ⎞⎛ 3.2808 ft ⎞ = 0.22 ft⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 1 ⎠⎝ 1 3 E mm ⎠⎝ 1m ⎠ 2. 1843 ounces = __________ kg. ⎛1843oz ⎞⎛ 28.35g ⎞⎛ 1kg ⎞ oz = 52.25kg ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 1 ⎠⎝ 1 ⎠⎝ E g ⎠1 3 3. 3500 microsieverts (µSv) = __________ millirem (mrem). 3.5 3μ ⎞ 1Sv ⎞ 1 2rem ⎞⎛ 1 3 mrem ⎞⎛ E Sv ⎛ ⎛ E E = 3.5E2mrem = 350mrem ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎠⎝ E Sv 1⎝ 1 1 6μ ⎠⎝ Sv ⎠⎝ 1rem ⎠ 4. 0.007 years = __________ minutes. ⎛ 0.007 year ⎞⎛ 365.25 ⎞⎛ 24hours ⎞⎛ 60minutes ⎞ = 3681.72minutes ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 1 ⎠⎝ 1year ⎠⎝ 1day ⎠⎝ 1hour ⎠ 5. 5000 dis./min. = __________ millicuries (mCi). ⎛ ⎞ ⎛ 5 3E dis ⎞⎜ 1Ci ⎟⎛ 1 3 E mCi ⎞ = 2.25E − 6mCi ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ min ⎠ dis ⎝ 1ci ⎠⎜ 2.22 12 ⎟E ⎝ min ⎠ 1.02-18 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide 6. 2350 micrometer (µm) = __________ inch. ⎛ 2350μ ⎞⎛ 3.937E − 5inches ⎞ = 0.0925inches = 9.25E − 2inch ⎜ ⎟⎜ ⎟1 1μ⎝ ⎠⎝ ⎠ 7. 2.5E-4 ergs = __________ keV. ⎛ 2.5E − 4ergs ⎞⎛ E ⎛ 1 ⎞6.2148 11 ⎞ keV =1.55 5E keV ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 1 ⎠⎝ 1erg ⎠⎝1 3 ⎠E eV 8. 205 ΕF = __________ K. (205° − 32)F ° =C = 96.1°C 1.8 K = 96.1 C 273.16 = 369.27K ° + 9. 2E-3 rad = __________ milligray (mGy). ⎛ 2E − 3rad ⎞⎛ 0.01Gy ⎞⎛1 3 ⎞E mGys ⎜ ⎟⎜ ⎟⎜ ⎟ = 2E − 2mGy = 0.02mGy ⎝ 1 ⎠⎝ 1rad ⎠⎝ 1Gy ⎠ 10. −25 °C = __________ °F. ° = ( 25 C )1.8 + 32 = 13 F − ° − °F 1.02-19 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Use unit analysis and conversion to solve the following: 11. Light travels at 186,000 miles per second. How many feet will light travel in one minute? ⎛186,000 miles ⎞⎛ 5280 ft ⎞⎛ 60sec ⎞ ft = 5.89 10 E⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ sec ⎠⎝ 1mile ⎠⎝ 1minute ⎠ min 12. A worker earns a monthly salary of $2500. If the worker gets paid every two weeks and works no overtime, what will be the gross amount for a given pay period? ⎛ 2500dollars ⎞⎛12months ⎞⎛ 1year ⎞⎛ 2weeks ⎞ dollars = 1153.85⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ months ⎠⎝ 1year ⎠⎝ 52 weeks ⎠⎝ payperiod ⎠ payperiod

13. An air sampler has run for 18 hours, 15 minutes at 60 liters per minute. When collected and analyzed the sample reads 7685 dis./min. What is the concentration of the sample in microcuries/cm3? ⎛15minutes ⎞⎛ 1hour ⎞ = 0.25hour⎜ ⎟⎜ ⎟ ⎝ 1 ⎠⎝ 60 minutes ⎠ 18 hours + 0.25 hours = 18.25 hours ⎛18.25hours ⎞⎛ 60minutes ⎞⎛ 60l ⎞ = 65,700 l⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 1 ⎠⎝ 1hour ⎠⎝ 1minute ⎠ ⎛ 7685dpm ⎞⎛ 1Ci ⎞⎛1 6E μCi ⎞⎛ 1l ⎞⎛ 0.99997 ml ⎞ μCi = 5.26E −11⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟E ⎠⎝ 1 E ml cc ⎝ 6.57E4l ⎠⎝ 2.22 12 dpm Ci ⎠⎝ 1 3 ⎠⎝ 1 ⎠ cc 1.02-20 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide INSTRUCTIONS FOR USING CONVERSION FACTOR TABLES The tables that follow include conversion factors that are useful to the RCT. They are useful in making a single conversion from one unit to another by using the guide arrows at the top of the page in accordance with the direction of the conversion. However, when using the tables to develop equivalent fractions for use in unit analysis equations, a better understanding of how to read the conversion factors given in the table is required. The conversions in the table have been arranged by section in the order of fundamental units, followed by derived units: Length Mass Time Area Volume Density Radiological Energy Fission Miscellaneous (Temperature, etc.) The easiest way to read a conversion from the table is done as follows. Reading left to right, "one (1) of the units in the left column is equal to the number in the center column of the unit in the right column." For example, look at the first conversion listed under Length. This conversion would be read from left to right as "1 angstrom is equal to E-8 centimeters," or Another conversion would be read from left to right as "1 millimeter (mm) is equal to 1E-1 centimeters," or 1 mm = 0.1 cm. This method can be applied to any of the conversions listed in these tables when reading left to right. If reading right to left, the conversion should be read as "one (1) of the unit in the right column is equal to the inverse of (1 over) the number in the center column of the unit in the left column." For example, using the conversion shown previously, the conversion reading right to left would be "1 inch is equal to the inverse of 3.937E-5 (1/3.937E-5) micrometers," or 11inch = = 2.54 E4 μm 3.937E − 5μm 1.02-21 DOE-HDBK-1122-2009 Module 1.01 Basic Mathematics and Algebra Study Guide Multiply # of by to obtain # of to obtain # of by Divide # of Length angstroms (Å) 10-8 Cm 10-10Å M micrometer (µm) 10-3 Mm µm 10-4 Cm µm 10-6 M µm 3.937 × 10-5 in. mm 10-1 Cm cm 0.3937 in. cm 3.2808 × 10-2 Ft cm 10-2 M m 39.370 in. m 3.2808 Ft m 1.0936 Yd m 10-3 Km m 6.2137 × 10-4 Miles km 0.62137 Miles mils 10-3 in. mils 2.540 × 10-3 Cm in. 103 Mils in. 2.5400 Cm ft 30.480 Cm rods 5.500 Yd miles 5280 Ft miles 1760 Yd miles 1.6094 Km 1.02-23 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Multiply # of by to obtain # of to obtain # of by Divide # of Mass mg 10-3 G mg 3.527 × 10-5 oz avdp mg 1.543 × 10-2 Grains g 3.527 × 10-2 oz avdp g 10-3 Kg g 980.7 Dynes g 2.205 × 10-3 Lb kg 2.205 Lb kg 0.0685 Slugs kg 9.807 × 105 Dynes lb 4.448 × 105 Dynes lb 453.592 G lb 0.4536 Kg lb 16 oz avdp lb 0.0311 Slugs dynes 1.020 × 10-3 G dynes 2.248 × 10-6 Lb u (unified--12C scale) 1.66043 × 10-27 Kg amu (physical--160 scale) 1.65980 × 10-27 Kg oz 28.35 G oz 6.25 × 10-2 Lb Note: Mass to energy conversions under miscellaneous

1.02-24 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Multiply # of by to obtain # of to obtain # of by Divide # of Time days 86,400 Sec days 1440 Min days 24 Hours years 3.15576 × 107 Sec years 525,960 Min years 8766 Hr years 365.25 Days Area 10-24 2barns cm circular mils 7.854 × 10-7 in.2 cm2 1024 Barns cm2 0.1550 in.2 cm2 1.076 × 10-3 ft2 cm2 10-4 m2 ft2 929.0 cm2 ft2 144 in2 ft2 9.290 × 10-2 m2 in.2 6.452 cm2 in.2 6.944 × 10-3 ft2 in.2 6.452 × 10-4 m2 m2 1550 in.2 m2 10.76 ft2 m2 1.196 yd2 m2 3.861 × 10-7 sq mi 1.02-25 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Multiply # of by to obtain # of to obtain # of by Divide # of Volume cm3 (cc) 0.99997 Ml cm3 6.1023 × 10-2 in.3 cm3 10-6 m3 cm3 9.9997 × 10-4 Liters cm3 3.5314 × 10-5 ft3 m3 35.314 ft3 m3 2.642 × 102 Gal m3 9.9997 × 102 Liters in.3 16.387 cm3 in.3 5.787 × 10-4 ft3 in.3 1.639 × 10-2 Liters in.3 4.329 × 10-3 Gal ft3 2.832 × 10-2 m3 ft3 7.481 Gal ft3 28.32 Liters ft3 1728 in.3 gal (U.S.) 231.0 in.3 gal 0.13368 ft3 liters 33.8147 fluid oz liters 1.05671 Quarts liters 0.26418 Gal gm moles (gas) 22.4 liters (s.t.p.) 1.02-26 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Multiply # of to obtain # of cm3/g ft3/lb g/cm3 lb/ft3 lb/in.3 lb/gal becquerel curies curies curies curies curies curies curies dis/min dis/min dis/sec dis/sec kilocuries microcuries microcuries millicuries millicuries R R R by by Density 1.602 × 10-2 62.43 62.43 1.602 × 10-2 27.68 0.1198 Radiological Units 2.703 × 10-11 3.700 × 1010 2.220 × 1012 103 106 1012 10-3 3.700 × 1010 4.505 × 10-10 4.505 × 10-7 2.703 × 10-8 2.703 × 10-5 103 3.700 × 104 2.220 × 106 3.700 × 107 2.220 × 109 2.58 × 10-4 1 2.082 × 109 to obtain # of Divide # of ft3/lb cm3/g lb/ft3 g/cm3 g/cm3 g/cm3 Curies dis/sec dis/min Millicuries Microcuries Picocuries Kilocuries Becquerel Millicuries Microcuries Millicuries Microcuries Curies dis/sec dis/min dis/sec dis/min C/kg of air esu/cm3 of air (s.t.p.) ion prs/cm3 of air (s.t.p.) 1.02-27 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Multiply # of to obtain # of R R (33.7 eV/ion pr.) R (33.7 eV/ion pr.) R (33.7 eV/ion pr.) R (33.7 eV/ion pr.) R (33.7 eV/ion pr.) rads rads rads rads rads rads rads (33.7 eV/ion pr.) gray rem sievert µCi/3 (µCi/ml) µCi/cm3 dpm/m3 Btu Btu Btu Btu Btu/lb eV by by Radiological Units (continued) 1.610 × 1012 7.02 × 104 5.43 × 107 86.9 2.08 × 10-6 .98 0.01 0.01 100 8.071 × 104 6.242 × 107 10-5 2.39 × 109 100 0.01 100 2.22 × 1012 2.22 × 109 0.4505 Energy 1.0548 × 103 0.25198 1.0548 × 1010 2.930 × 10-4 0.556 1.6021 × 10-12 to obtain # of Divide # of ion prs/g of air MeV/cm3 of air (s.t.p.) MeV/g of air ergs/g of air g-cal/g of air ergs/g of soft tissue Gray J/kg ergs/g MeV/cm3 or air (s.t.p.) MeV/g watt-sec/g ion prs/cm3 of air (s.t.p.) Rad Sievert Rem dpm/m3 dpm/liter pCi/m3 joules (absolute) kg-cal Ergs kW-hr g-cal/g Ergs 1.02-28 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Multiply # of by to obtain # of to obtain # of by Divide # of Energy (continued) eV 1.6021 × 10-19 joules (abs) eV 10-3 keV eV 10-6 MeV ergs 10-7 joules (abs) ergs 6.2418 × 105 MeV ergs 6.2418 × 1011 eV ergs 1.0 dyne-cm ergs 9.480 × 10-11 Btu ergs 7.375 × 10-8 ft-lb ergs 2.390 × 10-8 g-cal ergs 1.020 × 10-3 g-cm gm-calories 3.968 × 10-3 Btu gm-calories 4.186 × 107 Ergs joules (abs) 107 Ergs joules (abs) 0.7376 ft-lb joules (abs) 9.480 × 10-4 Btu g-cal/g 1.8 Btu/lb kg-cal 3.968 Btu kg-cal 3.087 × 103 ft-lb ft-lb 1.356 joules (abs) ft-lb 3.239 × 10-4 kg-cal kW-hr 2.247 × 1019 MeV kW-hr 3.60 × 1013 Ergs MeV 1.6021 × 10-6 Ergs

Note: Mass to energy conversions under miscellaneous 1.02-29 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Multiply # of by to obtain # of to obtain # of by Divide # of Fission Btu 1.28 × 10-8 grams 235U fissionedb Btu 1.53 × 10-8 grams 235U destroyedb,c Btu 3.29 × 1013 Fissions fission of 1 g 235U 1 megawatt-days fissions 8.9058 × 10-18 kilowatt-hours fissionsb 3.204 × 10-4 Ergs kilowatt-hours 2.7865 × 1017 235U fission neutrons average thermal kilowatts per kilogram 235U 2.43 × 1010 neutron flu× in fuelb,d megawatt-days per ton U 1.174 × 10-4 % U atoms fissionede average thermal megawatts per ton U 2.68 × 1010/Ef neutron flu× in fuelb neutrons per kilobarn 1 × 1021 neutrons/cm2 watts 3.121 × 1010 fissions/sec 1.02-30 DOE-HDBK-1122-2009 Module 1.02 Unit Analysis & Conversion Study Guide Multiply # of by to obtain # of to obtain # of by Divide # of Miscellaneous radians 57.296 Degrees eV 1.78258 × 10-33 Grams eV 1.07356 × 10-9 U erg 1.11265 × 10-21 Grams proton masses 938.256 MeV neutron masses 939.550 MeV electron masses 511.006 keV u (amu on 12C scale) 931.478 MeV Temperature F 32) 5(° − ⎛ ⎞C C ( F 32 ⎜ ⎟° = ° = ° − ) 1.8 9⎝ ⎠ F ° ⎛ ⎞ C ° =1.8( )C + 32 9F ⎜ ⎟ ° + 32° = ( ) 5⎝ ⎠ K C 273.16° = ° + Wavelength to Energy Conversion 1.02-31 DOE-HDBK-1122-2008 Module 1.03Physical Sciences Course Title: Radiological Control Technician Module Title: Physical Sciences Module Number: 1.03 Objectives : 1.03.01 Define the following terms as they relate to physics: a. Work b. Force c. Energy 1.03.02 Identify and describe four forms of energy. 1.03.03 State the Law of Conservation of Energy. 1.03.04 Distinguish between a solid, a liquid, and a gas in terms of shape and volume. 1.03.05 Identify the basic structure of the atom, including the characteristics of subatomic particles. 1.03.06 Define the following terms: a. Atomic number b. Mass number c. Atomic mass d. Atomic weight 1.03.07 Identify what each symbol represents in the X notation. 1.03.08 State the mode of arrangement of the elements in the Periodic Table. 1.03.09 Identify periods and groups in the Periodic Table in terms of their layout. 1.03.10 Define the terms as they relate to atomic structure: a. Valence shell b. Valence electron INTRODUCTION This lesson introduces the RCT to the concepts of energy, work, and the physical states of matter. Knowledge of these topics is important to the RCT as he or she works in environments where materials can undergo changes in state, resulting in changes in the work environment. 1.03-1 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide References: 1. "Chart of the Nuclides"; Sixteenth Edition, Knolls Atomic; 2003. 2. "Modern Physics"; Holt, Rinehart and Winston, Publishers; 1976. 3. "Chemistry: An Investigative Approach"; Houghton Mifflin Co., Boston; 1976. 4. "Chemical Principles with Qualitative Analysis"; Sixth ed.; Saunders College Pub.; 1986. 5. "Introduction to Chemistry" sixth ed., Dickson, T. R., John Wiley & Sons, Inc.; 1991. 6. "Matter"; Lapp, Ralph E., Life Science Library, Time Life Books; 1965. 7. "Physics"; Giancoli, Douglas C., second ed., Prentice Hall, Inc.; 1985. 8. DOE/HDBK-1015 "Chemistry: Volume 1 of 2"; DOE Fundamentals Handbook Series; January 1993. 1.03-2 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide 1.03.01 Define the following terms as they relate to physics: a. Work b. Force c. Energy

WORK & FORCE Physics is the branch of science that describes the properties, changes, and interactions of energy and matter. This unit will serve as a brief introduction to some of the concepts of physics as they apply to the situations that may be encountered by RCTs. Energy can be understood by relating it to another physical concept - work. The word work has a variety of meanings in everyday language. In physics, however, work is specifically defined as a force acting through a distance. Simply put, a force is a push or a pull. A more technical definition of force is any action on an object that can cause the object to change speed or direction. Units Force is derived as the product of mass and acceleration (see equation below). The SI derived unit of force is the newton (N). It is defined as the force which, when applied to a body having a mass of one kilogram, gives it an acceleration of one meter per second squared; that is: kg × mN = 2s As we said before, work is what is accomplished by the action of a force when it makes an object move through a distance. Mathematically, work is expressed as the product of a displacement and the force in the direction of the displacement; that is: W = Fd where: W = Work F = Force (newtons) d = Distance (meters) 1.03-3 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide For example, a horse works by exerting a physical force (muscle movement) to move a carriage. As the horse pulls, the carriage moves forward in the direction that the horse is pulling. Work is also done by an outside force (energy) to remove an electron from its orbit around the nucleus of an atom. The SI derived unit of work is the joule (J). One joule of work is performed when a force of one newton is exerted through a distance of one meter. Thus: J N m= × By this definition, work can only be performed when the force causes an object to be moved. This means that if the distance is zero then no work has been performed, even though a force has been applied. For example, if you stand at rest holding a bag of groceries in your hands, you do no work on it; your arms may become tired (and indeed energy is being expended by your muscles), but because the bag is not moved through a distance (d = 0), no work is performed (W = 0). ENERGY Energy (E) is defined as the ability to do work. Energy and work are closely related, but they are not the same thing. The relationship is that it takes energy to do work, and work can generate energy. This energy will be found in various forms. 1.03.02 Identify and describe four forms of energy. Kinetic Energy Kinetic energy describes the energy of motion an object possesses. For example, a moving airplane possesses kinetic energy. 1 2EK = mv 2 where: m = mass v = velocity 1.03-4 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide Potential Energy Potential energy (gravitational) indicates how much energy is stored as a result of the position or the configuration of an object. For example, water at the top of a waterfall possesses potential energy. EP = mgh where: m = mass g = free fall acceleration h = vertical distance Thermal Energy Thermal energy, or heat, describes the energy that results from the random motion of molecules. (Molecules are groups of atoms held together by strong forces called chemical bonds.) For example, steam possesses thermal energy. Chemical Energy

Chemical energy describes the energy that is derived from atomic and molecular interactions in which new substances are produced. For example, the substances in a dry cell provide energy when they react. Other Forms of Energy Other forms of energy, such as electrical and nuclear, will be described in later lessons. Energy may also appear as acoustical (sound) or radiant (light) energy. 1.03.03 State the Law of Conservation of Energy. Law of Conservation of Energy The Law of Conservation of Energy states that the total amount of energy in a closed system remains unchanged. Stated in other terms, as long as no energy enters or leaves the system, the amount of energy in the system will always be the same, although it can be converted from one form to another. For example, suppose a boulder lies at the bottom of a hill and bulldozer is used to push it to the top. If the dozer puts a certain continuous force on the boulder to keep it moving up the slope and moves it a distance, work has been done. The dozer is able to do this work because its engine burns gasoline, creates heat. The heat is converted into the kinetic energy of the moving bulldozer and the boulder in front of it. Some of this energy 1.03-5 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide is converted into heat and noise. Some is converted into the potential energy that the dozer and the boulder have gained in going to the top of the hill. If the boulder is allowed to roll back down the hill again, its potential energy will be converted partly into kinetic energy and partly into heat. The heat is produced by friction as the boulder rolls. Eventually the boulder will come to a stop, when all of its kinetic energy has been converted into heat. It leaves a trail of heat that is soaked up in the surroundings. Gasoline contains chemical energy that is released in the form of heat when a chemical reaction (burning) with oxygen occurs. This energy comes from the breaking and making of bonds between atoms. New products, carbon dioxide and water, are formed as the gasoline combines with oxygen. The energy of the burning gasoline produces heat energy which causes the gaseous combustion products to do work on the pistons in the engine. The work results in the bulldozer moving, giving it kinetic energy. Units of Energy Energy is expressed in the same units as work, that is, joules (J). The joule is the SI unit of energy. However, because energy can take on many different forms, it is sometimes measured in other units which can be converted to joules. Some of these units are mentioned below. Thermal Energy Thermal energy is often measured in units of calories (CGS) or British Thermal Units or BTUs (English). • A calorie is the amount of heat needed to raise the temperature of 1 gram of water by 1 °C. One calorie is equal to 4.18605 joules. • A BTU is the amount of heat needed to raise the temperature of 1 pound of water by 1 °F. One BTU is equal to 1.055E3 joules. Electrical Energy Electrical energy is sometimes expressed in units of kilowatt-hours. One kw-hr is equal to 3.6E6 joules A very small unit used to describe the energy of atomic and subatomic size particles is the electron volt (eV). One electron volt is the amount of energy acquired by an electron when it moves through a potential of one volt. For example, it takes about 15.8 eV of energy to remove an electron from an argon atom. Superunits such as kiloelectron volt (keV) and megaelectron volt (MeV) are used to indicate the energies of various ionizing radiations.

1.03-6 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide Work-Energy Relationship When work is done by a system or object, it expends energy. For example, when the gaseous combustion products in an automobile engine push against the pistons, the gas loses energy. The chemical energy stored in the gasoline is used to do work so that the car will move. When work is done on a system or object, it acquires energy. The work done on the car by the combustion of the gasoline causes the car to move, giving it more kinetic energy. When energy is converted to work or changed into another form of energy, the total amount of energy remains constant. Although it may appear that an energy loss has occurred, all of the original energy can be accounted for. Consider again the automobile engine. The energy stored in the gasoline is converted to heat energy, some of which is eventually converted to kinetic energy. The remainder of the heat energy is removed by the engine's cooling system. The motion of the engine parts creates friction, heat energy, which is also removed by the engine's cooling system. As the car travels, it encounters resistance with the air. If no acceleration occurs, the car will slow down as the kinetic energy is converted to friction or heat energy. The contact of the tires on the road converts some of the available kinetic energy to heat energy (friction), slowing down the car. A significant amount of the energy stored in the gasoline is dissipated as wasted heat energy. Energy-mass relationship Energy can also be converted into mass and mass converted into energy. This will be discussed further in section 1.04 "Nuclear Sciences." 1.03-7 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide 1.03.03 Distinguish between a solid, a liquid, and a gas in terms of shape and volume. ENERGY AND CHANGE OF STATE Matter is anything that has mass and takes up space. All matter is made up of atoms and molecules which are the building blocks used to form all kinds of different substances. These atoms and molecules are in constant random motion. Because of this motion they have thermal energy. The amount of energy depends on the temperature and determines the state or phase of the substance. There are three states of matter, solid, liquid and gas. Any substance can exist in any of the three states, but there is generally one state which predominates under normal conditions (temperature and pressure). Take water, for example. At normal temperatures, water is in the liquid state. In the solid state, water is called ice. The gaseous state of water is called steam or water vapor. It's all still water, just in different states. Table 1 provides a summary of these three states in terms of shape and volume. Table 1. States of Matter Compared State Shape Volume Solid Liquid Gas definite indefinite indefinite definite definite indefinite 1.03-8 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide Solid State A solid has definite shape and volume. The solid state differs from the liquid and gaseous states in that: • The molecules or ions of a solid are held in place by strong attractive forces. • The molecules have thermal energy, but the energy is not sufficient to overcome the attractive forces. • The molecules of a solid are arranged in an orderly, fixed pattern. The rigid arrangement of molecules causes the solid to have a definite shape and a definite volume.

Liquid State When heat is added to a substance, the molecules acquire more energy, which causes them to break free of their fixed crystalline arrangement. As a solid is heated, its temperature rises until the change of state from solid to liquid occurs. The volume of a liquid is definite since the molecules are very close to each other, with almost no space in between. Consequently, liquids can undergo a negligible amount of compression. However, the attractive forces between the molecules are not strong enough to hold the liquid in a definite shape. For this reason a liquid takes the shape of its container. High energy molecules near the surface of a liquid can overcome the attractive forces of other molecules. These molecules transfer from the liquid state to the gaseous state. If energy (heat) is removed from the liquid, the kinetic energy of the molecules decreases 1.03-9 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide and the attractive forces can hold the molecules in fixed positions. When compared with the kinetic energy, the attractive forces are not strong enough to hold the molecules in fixed positions, forming a solid. Gaseous State If the temperature of a liquid is increased sufficiently, it boils, that is, molecules change to the gaseous state and escape from the surface. Eventually, all of the liquid will become a gas. A gas has both indefinite shape and indefinite volume. A large space exists between gas molecules because of their high thermal energy. This allows for even more compression of a substance in the gaseous state. 1.03.03 Identify the basic structure of the atom, including the characteristics of subatomic particles. THE ATOM The Bohr Model As stated previously, the fundamental building block of matter is the atom. The basic atomic model, as described by Ernest Rutherford and Niels Bohr in 1911, consists of a positively charged core surrounded by negatively-charged shells. The central core, called the nucleus, contains protons and neutrons. Nuclear forces hold the nucleus together. The shells are formed by electrons which exist in structured orbits around the nucleus. Below is a summary of the three primary subatomic particles which are the constituent parts of the atom. 1.03-10 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide Protons • Positively charged (+1) • Mass: 1.6726E-24 gm or 1.007276470 amu • Each element is determined by the number of protons in its nucleus. All atoms of the same element have the same number of protons. Neutrons • Neutrally charged (0) • Mass: 1.6749E-24 gm or 1.008665012 amu • The number of neutrons determines the isotope of an element. Isotopes are atoms which have the same number of protons (therefore, of the same element) but different number of neutrons. This does not affect the chemical properties of the element. Electrons • Negatively charged (-1) • Small mass: 9.1085E-28 gm or 0.00054858026 amu (1/1840 of a proton) Because the mass of an electron is so small as compared to that of a proton or neutron, virtually the entire mass of an atom is furnished by the nucleus. • The number of electrons is normally equal to the number of protons. Therefore, the atom is electrically neutral. • The number of electrons in the outermost shell determines the chemical behavior or properties of the atom. THE ELEMENTS

Even though all atoms have the same basic structure, not all atoms are the same. There are over a hundred different types of atoms. These different types of atoms are known as elements. The atoms of a given element are alike but have different properties than the atoms of other elements. Elements are the simplest forms of matter. They can exist alone or in various combinations. Different elements can chemically combine to form molecules or molecular compounds. For example, water is a compound, consisting of water molecules. These molecules can be 1.03-11 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide decomposed into the elements hydrogen and oxygen. The elements hydrogen and oxygen are fundamental forms of matter. They cannot be further separated into simpler chemicals. Chemical Names Currently, there are over 110 named elements. Table 2 lists the elements and their symbols. Some have been known for many centuries, while others have only been discovered in the last 15 or 20 years. Each element has a unique name. The names of the elements have a variety of origins. Some elements were named for their color or other physical characteristics. Others were named after persons, places, planets or mythological figures. For example, the name chromium comes from the Greek word chroma, which means "color." Chromium is found naturally in compounds used as pigments. The elements curium, einsteinium, and fermium were named after famous nuclear physicists. Germanium, polonium and americium, were named after countries. Uranium, neptunium and plutonium are named in sequence for the three celestial bodies Uranus, Neptune and Pluto. Chemical Symbols For convenience, elements have a symbol which is used as a shorthand for writing the names of elements. The symbol for an element is either one or two letters taken from the name of the element (see Table 2). Note that some have symbols that are based on the historical name of the element. For example, the symbols for silver and gold are Ag and Au respectively. These come from the old Latin names argentum and aurum. The symbol for mercury, Hg, comes from the Greek hydrargyros which means "liquid silver." 1.03-12 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide Element Symbol Z Element Symbol Z Element Symbol Z Actinium Ac 89 Hafnium Hf 72 Promethium Pm 61 Aluminum Al 13 Hassium Hs 108 Protactinium Pa 91 Americium Am 95 Helium He 2 Radium Ra 88 Antimony Sb 51 Holmium Ho 67 Radon Rn 86 Argon Ar 18 Hydrogen H 1 Rhenium Re 75 Arsenic As 33 Indium In 49 Rhodium Rh 45 Astatine At 85 Iodine I 53 Rubidium Rb 37 Barium Ba 56 Iridium Ir 77 Ruthenium Ru 44 Berkelium Bk 97 Iron Fe 26 Rutherfordium Rf 104 Beryllium Be 4 Krypton Kr 36 Samarium Sm 62 Bismuth Bi 83 Lanthanum La 57 Scandium Sc 21 Bohrium Bh 107 Lawrencium Lw 103 Seaborgium Sg 106 Boron B 5 Lead Pb 82 Selenium Se 34 Bromine Br 35 Lithium Li 3 Silicon Si 14 Cadmium Cd 48 Lutetium Lu 71 Silver Ag 47 Calcium Ca 20 Magnesium Mg 12 Sodium Na 11 Californium Cf 98 Manganese Mn 25 Strontium Sr 38 Carbon C 6 Meitnerium Mt 109 Sulfur S 16 Cerium Ce 58 Mendelevium Md 101 Tantalum Ta 73 Cesium Cs 55 Mercury Hg 80 Technetium Tc 43 Chlorine Cl 17 Molybdenum Mo 42 Tellurium Te 52 Chromium Cr 24 Neodymium Nd 60 Terbium Tb 65 Cobalt Co 27 Neon Ne 10 Thallium Tl 81 Copper Cu 29 Neptunium Np 93 Thorium Th 90 Curium Cm 96 Nickel Ni 28 Thulium Tm 69 Dubnium Db 105 Niobium Nb 41 Tin Sn 50 Dysprosium Dy 66 Nitrogen N 7 Titanium Ti 22 Einsteinium Es 99 Nobelium No 102 Tungsten W 74 Erbium Er 68 Osmium Os 76 Uranium U 92 Europium Eu 63 Oxygen O 8 Vanadium V 23 Fermium Fm 100 Palladium Pd 46 Xenon Xe 54 Fluorine F 9 Phosphorus P 15 Ytterbium Yb 70 Francium Fr 87 Platinum Pt 78 Yttrium Y 39 Gadolinium Gd 64 Plutonium Pu 94 Zinc Zn 30 Gallium Ga 31 Polonium Po 84 Zirconium Zr 40 Germanium Ge 32 Potassium K 19 Gold Au 79 Praseodymium Pr 59

1.03-13 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide 1.03.03 Define the following terms: a) Atomic number b) Mass number c) Atomic mass d) Atomic weight Atomic Number The number of protons in the nucleus of an element is called the atomic number. All atoms of a particular element have the same atomic number. Atomic numbers are integers. For example, a hydrogen atom has one proton in the nucleus. Therefore, the atomic number of hydrogen is 1. A helium atom has two protons in the nucleus, which means that its atomic number is 2. Uranium has 92 protons in the nucleus and, therefore, has an atomic number of 92. Atomic number is often represented by the symbol Z. Mass Number The total number of protons plus neutrons in the nucleus of a particular isotope of an element is called the mass number. It is the integer nearest to the mass of the atom of concern. Since a proton has a mass of 1.0073 amu, we will give a proton a mass number of 1. The mass number of a neutron would also be 1, since its mass is 1.0087 amu. So, by adding the number of protons and the number of neutrons we can determine the mass number of the atom of concern. For example, a normal hydrogen atom has 1 proton, but no neutrons. Therefore, its mass number is 1. A helium atom has 2 protons and 2 neutrons, which means that it has a mass number of 4. If a uranium isotope has 146 neutrons then it has a mass number of 238 (92 + 146), while if it only has 143 neutrons its mass number would be 235. The mass number can be used with the name of the element to identify which isotope of an element we are referring to. If we are referring to the isotope of uranium that has a mass number of 238, we can write it as Uranium-238. If we are referring to the isotope of mass number 235, we write it as Uranium-235. Often, this expression is shortened by using the chemical symbol instead of the full name of the element, as in U-238 or U-235. Atomic Mass The actual mass of an atom of a particular isotope is called its atomic mass. The units are expressed in Atomic Mass Units (AMU). AMUs are based on 1/12 of the mass of a Carbon-12 atom (1.660E-24 gm). In other words, the mass of one C-12 atom is exactly 12 amu. 1.03-14 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide For example, the mass of a hydrogen atom is 1.007825 amu (1 proton + 1 electron = 1.00727647 + 0.00054858026) . The mass of a Uranium-238 atom in amu is 238.0508, while the mass of a U-235 atom is only 235.0439. Notice that atomic masses are very accurate and are written as decimals. Atomic Weight The weighted average of the isotopic masses of an element, based on the percent abundance of its naturally occurring isotopes, is called the atomic weight. The atomic weight is expressed in AMU and is used mainly in calculations of chemical reactions. Since AMUs are based on Carbon-12, one may wonder why the Periodic Table (see Figure 5) shows the atomic weight of Carbon as 12.011, and not exactly 12. The explanation is simple and will help to clarify the difference between the atomic weight of an element and the atomic mass of an isotope of that element. Carbon, as it occurs in nature, is a mixture of two isotopes: about 98.9% of all carbon atoms are C-12, while the abundance of C-13 atoms is 1.1% (a total of 100%). The presence of these heavier Carbon atoms explains why the atomic weight of carbon is slightly more than 12. The atomic weight of an element is a "weighted average" (no pun intended). This average is determined by finding the sum of the mass of each isotope multiplied by its percent abundance. If the atomic mass of C-12 is 12.00, and the atomic mass of C-13 is 13.00, we can determine the atomic weight of carbon:

12.00(0.989) + 13.00(0.011) = 11.868 + 0.143 = 12.011 amu With the understanding of these concepts, we can discuss the Periodic Table of the Elements and the information it provides. 1.03.07 Identify what each symbol represents in the Z AX notation. NUCLIDE NOTATION The format for representing a specific combination of protons and neutrons is to use its nuclear symbol. This is done by using the standard chemical symbol, with the atomic number written as a subscript at the lower left of the symbol, and the mass number written as a superscript at the upper left of the symbol: A Z X 1.03-15 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide where: X = Symbol for element Z = Atomic number: number of protons A = Mass number: number of protons (Z) plus number of neutrons (N); therefore: A = Z + N For example, the notation for Uranium-238 would be 238 92 U . 1.03.08 State the mode of arrangement of the elements in the Periodic Table. 1.03.09 Identify periods and groups in the Periodic Table in terms of their layout. MODERN PERIODIC TABLE The modern Periodic Table (see Figure 5) is an arrangement of the elements in order of increasing atomic number. A comparison of the properties for selected elements will illustrate that there is a predictable, recurring pattern, or periodicity. This observation is summarized in the Periodic Law, which states that the properties of the elements are repetitive or recurring functions of their atomic numbers. Data about each element in the Periodic Table are presented in a column and row format. The rows or horizontal sections in the Periodic Table are called periods. The columns or vertical sections are called groups or families because they "behave" chemically similar; that is they have similar chemical properties. Since the number of electrons is equal to the number of protons, the structure of the Periodic Table directly relates to the number and arrangement of electrons in the atom (see Table 3). Figure 4 below gives a simple illustration of the electron shells described in the Bohr model of the atom. Electrons orbit around the nucleus in structured shells, designated sequentially as 1 through 7 (K through Q) from inside out. Shells represent groups of energy states called orbitals. The higher the energy of the orbital the greater the distance from the nucleus. The lowest energy state is in the innermost shell (K). The number of orbitals in a shell is the square of the shell number (n). The maximum number of electrons which can occupy an orbital is 2. Therefore, each shell can hold a maximum of 2n2 electrons. For example, for the L shell the maximum number of electrons would be 8: 1.03-16 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide L-shell: n = 2 →→→→ 2(22) = 8 1.03-17 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide 1.03-18 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide Z Element K L M N O Z Element K L M N O P Q 1 Hydrogen 1 55 Cesium 2 8 18 18 8 1 2 Helium 2 56 Barium 2 8 18 18 8 2 3 Lithium 2 1 57 Lanthanum 2 8 18 18 9 2 4 Beryllium 2 2 58 Cerium 2 8 18 20 8 2 5 Boron 2 3 59 Praseodymium 2 8 18 21 8 2 6 Carbon 2 4 60 Neodymium 2 8 18 22 8 2 7 Nitrogen 2 5 61 Promethium 2 8 18 23 8 2 8 Oxygen 2 6 62 Samarium 2 8 18 24 8 2 9 Fluorine 2 7 63 Europium 2 8 18 25 8 2

10 Neon 2 8 64 Gadolinium 2 8 18 25 9 2 11 Sodium 2 8 1 65 Terbium 2 8 18 27 8 2 12 Magnesium 2 8 2 66 Dysprosium 2 8 18 28 8 2 13 Aluminum 2 8 3 67 Holmium 2 8 18 29 8 2 14 Silicon 2 8 4 68 Erbium 2 8 18 30 8 2 15 Phosphorus 2 8 5 69 Thulium 2 8 18 31 8 2 16 Sulfur 2 8 6 70 Ytterbium 2 8 18 32 8 2 17 Chlorine 2 8 7 71 Lutetium 2 8 18 32 9 2 18 Argon 2 8 8 72 Hafnium 2 8 18 32 10 2 19 Potassium 2 8 8 1 73 Tantalum 2 8 18 32 11 2 20 Calcium 2 8 8 2 74 Tungsten 2 8 18 32 12 2 21 Scandium 2 8 9 2 75 Rhenium 2 8 18 32 13 2 22 Titanium 2 8 10 2 76 Osmium 2 8 18 32 14 2 23 Vanadium 2 8 11 2 77 Iridium 2 8 18 32 15 2 24 Chromium 2 8 13 1 78 Platinum 2 8 18 32 16 2 25 Manganese 2 8 13 2 79 Gold 2 8 18 32 18 1 26 Iron 2 8 14 2 80 Mercury 2 8 18 32 18 2 27 Cobalt 2 8 15 2 81 Thallium 2 8 18 32 18 3 28 Nickel 2 8 16 2 82 Lead 2 8 18 32 18 4 29 Copper 2 8 18 1 83 Bismuth 2 8 18 32 18 5 30 Zinc 2 8 18 2 84 Polonium 2 8 18 32 18 6 31 Gallium 2 8 18 3 85 Astatine 2 8 18 32 18 7 32 Germanium 2 8 18 4 86 Radon 2 8 18 32 18 8 33 Arsenic 2 8 18 5 87 Francium 2 8 18 32 18 8 1 34 Selenium 2 8 18 6 88 Radium 2 8 18 32 18 8 2 35 Bromine 2 8 18 7 89 Actinium 2 8 18 32 18 9 2 36 Krypton 2 8 18 8 90 Thorium 2 8 18 32 18 10 2 37 Rubidium 2 8 18 8 1 91 Protactinium 2 8 18 32 20 9 2 38 Strontium 2 8 18 8 2 92 Uranium 2 8 18 32 21 9 2 39 Yttrium 2 8 18 9 2 93 Neptunium 2 8 18 32 22 9 2 40 Zirconium 2 8 18 10 2 94 Plutonium 2 8 18 32 24 8 2 41 Niobium 2 8 18 12 1 95 Americium 2 8 18 32 25 8 2 42 Molybdenum 2 8 18 13 1 96 Curium 2 8 18 32 25 9 2 43 Technetium 2 8 18 13 2 97 Berkelium 2 8 18 32 27 8 2 44 Ruthenium 2 8 18 15 1 98 Californium 2 8 18 32 28 8 2 45 Rhodium 2 8 18 16 1 99 Einsteinium 2 8 18 32 29 8 2 46 Palladium 2 8 18 18 0 100 Fermium 2 8 18 32 30 8 2 47 Silver 2 8 18 18 1 101 Mendelevium 2 8 18 32 31 8 2 48 Cadmium 2 8 18 18 2 102 Nobelium 2 8 18 32 32 8 2 49 Indium 2 8 18 18 3 103 Lawrencium 2 8 18 32 32 9 2 50 Tin 2 8 18 18 4 104 Rutherfordium 2 8 18 32 32 10 2 51 Antimony 2 8 18 18 5 105 Dubnium 2 8 18 32 32 11 2 52 Tellurium 2 8 18 18 6 106 Seaborgium 2 8 18 32 32 12 2 53 Iodine 2 8 18 18 7 107 Bohrium 2 8 18 32 32 13 2 54 Xenon 2 8 18 18 8 109 Meitnerium 2 8 18 32 32 15 2 1.03-19 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide 1.03.10 Define the terms as they relate to atomic structure: a) Valence shell b) Valence electron The highest occupied energy level in a ground-state atom is called its valence shell. Therefore, the electrons contained in it are called valence electrons. The rows or periods in the Periodic Table correspond to the electron shells. The elements contained in first period have their valence electrons in the first energy level or K-shell. The elements contained in the second period have their outer or valence shell electrons in the second energy level or L-shell, and so on. The pattern continues down the table. The number of electrons in the valence shell determines the chemical properties or "behavior" of the atom. The valence shell can have a maximum of eight electrons, except for the K-shell which can only have two. Atoms are chemically stable when the valence shell has no vacancies; that is, they "prefer" to have a full valence shell. Atoms of elements toward the right of the Periodic Table seem to lack only one or two electrons. These will "look" for ways to gain electrons in order to fill their valence shell. Atoms of elements on the left side of the table seem to have an excess of one or two electrons. These will tend to find ways to lose these excess electrons so that the full lower shell will be the valence shell.

The outcome is that certain atoms will combine with other atoms in order to fill their valence shells. This combination that occurs is called a chemical bond, and results in the formation of a molecule. The bond is accomplished by "sharing" or "giving up" valence electrons, thus forming a molecule whose chemical properties are different than those of the individual element atoms. A good example is table salt. Salt is a 1:1 combination of sodium and chlorine; that is, a salt molecule is formed when one sodium atom bonds with one chlorine atom. If we look at Table 3, we can see that sodium (Na) has 1 electron in its outermost shell. Chlorine (Cl) needs one electron to complete its valence shell. The sodium atom "gives up" its extra electron to the chlorine atom who then "thinks" that its valence shell is full. Because the sodium atom has one less electron, the atom now has a net positive charge; that is, it has one less electron than it has protons. The chlorine atom now has a net negative charge because it has one more electron than it has protons. The opposite charges of the two ions attract and form an ionic bond. The bond results in a sodium chloride molecule (NaCl). However, this is just one type of chemical bond between atoms. There are several other types of chemical bonds that can occur, but which are beyond the scope of this lesson. Note the rightmost column in the Periodic Table. These elements are known as the noble or inert gases because they all have a full valence shell (see also the underlined elements in Table 3). This means that they "feel" no need to bond with other atoms. Noble gases are thus considered chemically inert and very rarely interact with other elements. 1.03-20 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide The Quantum Mechanical Model Over the years, the Bohr model of the atom was found to be inadequate as the principles of quantum mechanics evolved. A newer model, known as the quantum mechanical model, describes the electrons arranged in energy levels corresponding to the "electron shells" of the Bohr model. In the quantum mechanical model the electron is not viewed as particle in a specific orbit, but rather as an electron cloud in which the negative charge of the electron is spread out within the cloud. These energy levels are referred to as orbitals to emphasize that these are not circular "orbits" like those of the Bohr model but rather electron clouds. An electron cloud is a representation of the volume about the nucleus in which an electron of a specific energy is likely to be found. The quantum mechanical model further states that the energy levels are subdivided into sublevels, referred to by the letters s, p, d, f, etc. An energy level can contain one or more sublevels or orbitals, and a maximum of two electrons can reside in each sublevel. For example, the first energy level contains one s sublevel which can accommodate a maximum of two electrons. 1.03-21 DOE-HDBK-1122-2009 Module 1.03 Physical Sciences Study Guide This page intentionally left blank. 1.03-22 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide Course Title: Radiological Control Technician Module Title: Nuclear Physics Module Number: 1.04 Objectives : 1.04.01 Identify the definitions of the following terms: a. Nucleon b. Nuclide c. Isotope 1.04.02 Identify the basic principles of the mass-energy equivalence concept. 1.04.03 Identify the definitions of the following terms: a. Mass defect b. Binding energy c. Binding energy per nucleon

1.04.04 Identify the definitions of the following terms: a. Fission b. Criticality c. Fusion INTRODUCTION Nuclear power is made possible by the process of nuclear fission. Fission is but one of a large number of nuclear reactions which can take place. Many reactions other than fission are quite important because they affect the way we deal with all aspects of handling and storing nuclear materials. These reactions include radioactive decay, scattering, and radiative capture. This lesson is designed to provide an understanding of the forces present within an atom. 1.04-1 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide References: 1. "Basic Radiation Protection Technology"; Gollnick, Daniel; 5th ed.; Pacific Radiation Corporation; 2008. 2. "Introduction to Health Physics"; Cember, Herman; 4nd ed.; McGraw-Hill Medical; 2008. 3. ANL-88-26 (1988) "Operational Health Physics Training"; Moe, Harold; Argonne National Laboratory, Chicago 1.04-2 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide 1.04.01 Identify the definitions of the following terms: a. Nucleon b. Nuclide c. Isotope NUCLEAR TERMINOLOGY There are several terms used in the field of nuclear physics that an RCT must understand. Nucleon Neutrons and protons are found in the nucleus of an atom, and for this reason are collectively referred to as nucleons. A nucleon is defined as a constituent particle of the atomic nucleus, either a neutron or a proton. Nuclide A species of atom characterized by the constitution of its nucleus, which is specified by its atomic mass and atomic number (Z), or by its number of protons (Z), number of neutrons (N), and energy content. A listing of all nuclides can be found on the "Chart of the Nuclides," which will be introduced in a later lesson. Isotope This term was mentioned in Lesson 1.03 when we discussed the concepts of atomic mass and atomic weight. Isotopes are defined as nuclides which have the same number of protons but different numbers of neutrons. Therefore, any nuclides which have the same atomic number (i.e. the same element) but different atomic mass numbers are isotopes. For example, hydrogen has three isotopes, known as Protium, Deuterium and Tritium. Since hydrogen has one proton, any hydrogen atom will have an atomic number of 1. However, the atomic mass numbers of the three isotopes are different: Protium (1H) has an mass number of 1 (1 proton, no neutrons), deuterium (D or 2H) has a mass number of 2 (1 proton, 1 neutron), and tritium (T or 3H) has a mass number of 3 (1 proton, 2 neutrons). 1.04.02 Identify the basic principles of the mass-energy equivalence concept. 1.04-3 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide MASS-ENERGY EQUIVALENCE Of fundamental concern in nuclear reactions is the question of whether a given reaction is possible and, if so, how much energy is required to initiate the reaction or is released when the reaction occurs. The key to these questions lies in the relationship between the mass and energy of an object. The theory that relates the two was proposed by Albert Einstein in 1905. Einstein's special theory of relativity culminated in the famous equation: E = mc2 where: E = Energy m = mass c = speed of light This equation expresses the equivalence of mass and energy, meaning that mass may be transformed to energy and vice versa. Because of this equivalence the two are often referred to collectively as mass-energy.

The mass-energy equivalence theory implies that mass and energy are interchangeable. The theory further states that the mass of an object depends on its speed. Thus, all matter contains energy by virtue of its mass. It is this energy source that is tapped to obtain nuclear energy. The Law of Conservation of Energy applies to mass as well as energy, since the two are equivalent. Therefore, in any nuclear reaction the total mass-energy is conserved i.e. mass-energy cannot be created or destroyed. This is of importance when it becomes necessary to calculate the energies of the various types of radiation which accompany the radioactive decay of nuclei. Pair Annihilation: An Example of Mass to Energy Conversion An interaction which occurs is pair annihilation, where two particles with mass, specifically a positron and an electron (negatron), collide and are transformed into two rays (photons) of electromagnetic energy. A positron is essentially an anti-electron, having a positive charge. When a positron collides with an electron, both particles are annihilated and their mass is converted completely to electromagnetic energy. (This interaction will be discussed in Lesson 1.07 "Interactions of Radiation with Matter.") If the mass of an electron/positron is 0.00054858026 amu, the resulting annihilation energy (radiation) resulting from the collision would be: 2 0.0005485026 ( amu) 931.478MeV × = 1.022MeV 1 amu 1.04-4 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide 1.04.03 Identify the definitions of the following terms: a. Mass defect b. Binding energy c. Binding energy per nucleon MASS DEFECT AND BINDING ENERGY With an understanding of the equivalence of mass and energy, we can now examine the principles which lie at the foundation of nuclear power. Mass Defect As we said earlier, the mass of an atom comes almost entirely from the nucleus. If a nucleus could be disassembled to its constituent parts, i.e., protons and neutrons, it would be found that the total mass of the atom is less than the sum of the masses of the individual protons and neutrons. This is illustrated in Figure 1 below. Figure 1. Atomic Scale This slight difference in mass is known as the mass defect, ∗ (pronounced "delta"), and can be computed for each nuclide, using the following equation. ∗ = (Z)(Mp) + (Z)(Me) + (A-Z)(Mn) - Ma where: ∗ = mass defect Z = atomic number Mp = mass of a proton (1.00728 amu) Me = mass of a electron (0.000548 amu) A = mass number Mn = mass of a neutron (1.00867 amu) Ma = atomic mass (from Chart of the Nuclides) 1.04-5 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide For example, consider an isotope of Lithium, Li: A = 7 Z = 3 M = 7.01600 amu (per Chart of the Nuclides) Therefore: ∗ = (3)(1.00728) + (3)(0.000548) + (7-3)(1.00867) - (7.01600) ∗ = (3.02184) + (0.001644) + (4.03468) - (7.01600) ∗ = (7.058164) - (7.01600) ∗ = 0.042164 amu Binding Energy The mass defect of 2 3Li is 0.042164 amu. This is the mass that is apparently "missing", but, in fact, has been converted to energy, the energy that binds the lithium atom nucleus together, or its binding energy. Binding energy is the energy equivalent of mass defect. Note from the Lesson 1.02 Conversion Tables that: 1 amu = 931.478 MeV So, if we multiply the mass defect by this number we can calculate the binding energy. ⎛ 0.042164amu ⎞⎛ 931.478MeV ⎞BE = = 39.27MeV ⎜ ⎟⎜ ⎟ ⎝ 1 ⎠⎝ amu ⎠

From this we have determined the energy converted from mass in the formation of the nucleus. We will see that it is also the energy that must be applied to the nucleus in order to break it apart. Another important calculation is that of the binding energy of a neutron. This calculation is of significance when the energetics of the fission process are considered. For example, when 235U absorbs a neutron, the compound nucleus 236U is formed (see NUCLEAR FISSION later in this lesson). The change in mass (Δm) is calculated and then converted to its energy equivalent: Δm = (mn + mU235) - mU236 Δm = (1.00867 + 235.0439) - 236.0456 Δm = 0.0070 amu 1.04-6 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide 0.0070 amu × 931.5 MeV/amu = 6.52 MeV Thus, the nucleus possesses an excitation energy of 6.52 MeV when 235U absorbs a neutron. Binding Energy per Nucleon If the total binding energy of a nucleus is divided by the total number of nucleons in the nucleus, the binding energy per nucleon is obtained. This represents the average energy which must be supplied in order to remove a nucleon from the nucleus. For example, using the Li atom as before, the binding energy per nucleon would be calculated as follows: 39.27MeV = 5.61MeVpernucleon 7nucleons If the binding energy per nucleon is plotted as a function of mass number (total number of nucleons) for each element, a curve is obtained (see Figure 2). The binding energy per nucleon peaks at about 8.5 MeV for mass numbers 40 - 120 and decreases to about 7.6 MeV per nucleon for uranium. The binding energy per nucleon decreases with increasing mass number above mass 56 because as more protons are added, the proton-proton repulsion increases faster than the nuclear attraction. Since the repulsive forces are increasing, less energy must be supplied, on the average, to remove a nucleon. That is why there are no stable nuclides with mass numbers beyond that of 208. 1.04-7 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide NUCLEAR TRANSFORMATION EQUATIONS Using the X format, equations can be written which depict a transformation that has occurred in a nucleus or nuclei. Since it is an equation, both sides must be equal. Therefore, the total mass-energy on the left must be equal to the total mass-energy on the right. Keeping in mind that mass and energy are equivalent, any difference in total mass is accounted for as energy released in the transformation. This energy release is called the Q value. For example, the alpha decay of Radium-226 would be depicted as: 226 222 4Ra ⎯⎯→ Rn + α + Q88 86 2 The energy release, represented by Q in this case, is manifest as the kinetic energy of the high-speed alpha particle, as well as the recoil of the Radon-222 atom. 1.04.04 Identify the definitions of the following terms: a. Fission b. Criticality c. Fusion 1.04-8 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide NUCLEAR FISSION As we have shown, the nucleus of a single atom could be the source of considerable energy. If the nucleus could be split so as to release this energy it could be used to generate power. Therefore, if the energy from millions of atoms were released it would be a great source of power. This thinking is the basis for nuclear power.

When a free neutron strikes a nucleus, one of the processes which may occur is the absorption of the neutron by the nucleus. It has been shown that the absorption of a neutron by a nucleus raises the energy of the system by an amount equal to the binding energy of the neutron. Under some circumstances, this absorption may result in the splitting of the nucleus into at least two smaller nuclei with an accompanying release of energy. This process is called fission. Two or three neutrons are usually released during this type of transformation. In order to account for how this process is possible, use is made of the liquid drop model of the nucleus. This model is based on the observation that the nucleus, in many ways, resembles a drop of liquid. The liquid drop is held together by cohesive forces between molecules, and when the drop is deformed the cohesive forces may be insufficient to restore the drop to its original shape. Splitting may occur, although in this case, there would be no release of energy. Figure 3 illustrates this model. The absorption of a neutron raises the energy of the system by an amount equal to the binding energy of the neutron. This energy input causes deformation of the nucleus, but if it is not of sufficient magnitude, the nucleon-nucleon attractive forces will act to return the nucleus to its original shape. If the energy input is sufficiently large, the nucleus may reach a point of separation, and a fission has occurred. The energy required to drive the nucleus to the point of separation is called the critical energy for fission, Ec. The values of Ec for various nuclei can be calculated, based on a knowledge of the forces which act to hold the nucleus together. 1.04-9 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide An example of a fission is shown below, involving neutron absorption by 235U: 235 1 236 138 95 1U + n ⎯⎯ U* ⎯⎯ ( ) + Q → → Ba + Kr + 3 n92 0 92 56 36 0 On the average, approximately 200 MeV of energy is released per fission. The fission process can be "energetically" explained by comparing the critical energy for fission with the amount of energy input, i.e., the neutron binding energy. For 238U and 232Th, the critical energy for fission is greater than the neutron binding energy. Therefore, an additional amount of energy must be supplied in order for fission to occur in these nuclei. This additional energy is in the form of neutron kinetic energy, and confirms the observation that fission occurs in these fissionable nuclei only when the neutron has approximately 1 MeV of kinetic energy. The situation is quite different for 235U, 233U, and 239Pu. In these cases, the neutron binding energy exceeds the critical energy for fission. Thus, these nuclei may be fissioned by thermal, or very low energy, (0.025 eV) neutrons. As mentioned earlier, the new elements which are formed as a result of the fissioning of an atom are unstable because their N/P ratios are too high. To attain stability, the fission fragments will undergo various transformations depending on the degree of instability. Along with the neutrons immediately released during fission, a highly unstable element may give off several neutrons to try to regain stability. This, of course, makes more neutrons available to cause more fissions and is the basis for the chain reaction used to produce nuclear power. The excited fission product nuclei will also give off other forms of radiation in an attempt to achieve a stable status. These include beta and gamma radiation.

1.04-10 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide Criticality Criticality is the condition in which the neutrons produced by fission are equal to the number of neutrons in the previous generation. This means that the neutrons in one generation go on to produce an equal number of fission events, which events in turn produce neutrons that produce another generation of fissions, and so forth. This continuation results in the self-sustained chain reaction mentioned above. Figure 5 gives a simple illustration of the chain reaction that occurs in a criticality. As one can discern, this reaction requires control. If the population of neutrons remains constant, the chain reaction will be sustained. The system is thus said to be critical. However, if too many neutrons escape from the system or are absorbed but do not produce a fission, then the system is said to be subcritical and the chain reaction will eventually stop. On the other hand, if the two or three neutrons produced in one fission each go on to produce another fission, the number of fissions and the production of neutrons will increase exponentially. In this case the chain reaction is said to be supercritical. 1.04-11 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide In nuclear reactors this concept is expressed as the effective multiplication constant or Keff. Keff is defined as the ratio of the number of neutrons in the reactor in one generation to the number of neutrons in the previous generation. On the average, 2.5 neutrons are emitted per uranium fission. If Keff has a value of greater than 1, the neutron flux is increasing, and conversely, if it has a value of less than 1, the flux is decreasing with time. Table 1 illustrates the reactor condition for various values of the multiplication constant. Table 1 - The Effective Multiplication Constant Keff = Effective Multiplication Constant Keff <1 = Subcritical Condition Keff = 1 = Critical Condition Keff >1 = Supercritical Condition In a subcritical reactor, the neutron flux and power output will die off in time. When critical, the reactor operates at a steady neutron and power output. A reactor must be supercritical to increase the neutron flux and power level. 1.04-12 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide Therefore, the population of neutrons must be controlled in order to control the number of fissions that occur. Otherwise, the results could be devastating. Fusion Another reaction between nuclei which can be the source of power is fusion. Fusion is the act of combining or "fusing" two or more atomic nuclei. Fusion thus builds atoms. The process of fusing nuclei into a larger nucleus with an accompanying release of energy is called fusion. Fusion occurs naturally in the sun and is the source of its energy. The reaction is initiated under the extremely high temperatures and pressure in the sun whereby H interacts with 12 6C . Hydrogen is then converted to helium and energy is liberated in the form of heat. 1 4 2+ +4( H) ⎯⎯ 2 ( ) + 24.7MeV 1 → He + 2 e What occurs in the above equation is the combination of 4 hydrogen atoms, giving a total of 4 protons and 4 electrons. 2 protons combine with 2 electrons to form 2 neutrons, which combined with the remaining 2 protons forms a helium nucleus, leaving 2 electrons and a release of energy. 1.04-13 DOE-HDBK-1122-2009 Module 1.04 Nuclear Physics Study Guide This page intentionally left blank. 1.04-14

DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide Course Title: Radiological Control Technician Module Title: Sources of Radiation Module Number: 1.05 Objectives : 1.05.01 Identify the following four sources of natural background radiation including the origin, radionuclides, variables, and contribution to exposure. a. Terrestrial b. Cosmic c. Internal Emitters d. Radon 1.05.02 Identify the following four sources of artificially produced radiation and the magnitude of dose received from each. a. Nuclear Fallout b. Medical Exposures c. Consumer Products d. Nuclear Facilities INTRODUCTION Apart from the amount of radiation a worker may receive while performing work, they will also be exposed to radiation because of the very nature of our environment. All individuals are subject to some irradiation even though they may not work with radioactive substances. This natural source of exposure is often referred to as background radiation. Studies of the nature and origin of this source of exposure to man have revealed three main components: terrestrial radiation (which includes the radioactivities of the earth's surface, air and water), cosmic radiation, and the naturally occurring radionuclides of the human body. One might add that man-made sources influence the contribution from some of these sources. The amount which each of these factors contributes varies with the locale. The study of these factors throughout the world is of value for a number of reasons. Foremost among these is that the use of such data provides a basis or standard from which allowable exposure limits for radiation workers may be developed. In areas where the levels are much higher because of larger concentrations of natural radioactive materials, knowledge may be gained about human hereditary effects at these increased levels. Such data are also needed in assessing the impact on, or contribution of a nuclear facility to the existing concentrations in a given area. In the design of buildings and/or shielding for low-level work, it is of value to know the radioactive contents of the substances used. Often the levels inside a building are higher than those outside of the building because this factor has been neglected. 1.05-1 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide Because of these needs, much data about background levels in many areas has been acquired. This section is devoted to the discussion of these background factors and the relative contribution of man-made radiation. References: 1. "Basic Radiation Protection Technology"; Gollnick, Daniel; 5th ed.; Pacific Radiation Corporation; 2008. 2. ANL-88-26 (1988) "Operational Health Physics Training"; Moe, Harold; Argonne National Laboratory, Chicago. 3. NCRP Report No. 45 "Natural Background Radiation in the United States". 4. NCRP Report No. 56 "Radiation Exposure from Consumer Product Miscellaneous Sources". 5. NCRP Report No. 93 "Ionizing Radiation Exposure of the Population of the United States". 1.05-2 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide NATURAL BACKGROUND RADIATION SOURCES 1.05.01 Identify the following four sources of natural background radiation including the origin, radionuclides, variables and contribution to exposure. a. Terrestrial Terrestrial Radiation Radioactivity of the Earth

The presence of certain small amounts of radioactivity in the soil adds to the background levels to which man is exposed. The amount of radioactive materials found in soil and rocks varies widely with the locale. The main contribution to the background is the gamma ray dose from radioactive elements chiefly of the uranium and thorium series and lesser amounts from radioactive K-40 and Rb-87. Due to the high concentration of monazite, a thorium mineral, some regions in the world have an extremely high background level. The majority of the population of the Kerala region in India receive an annual dose greater than 500 mrem. A small percentage of the inhabitants receive over 2,000 mrem per year and the highest recorded value has been 5,865 mrem in one year. It is interesting to note that this value is more than what is allowed for a DOE radiation worker. The Minas Garais state in Brazil has an average terrestrial background dose rate of 1,160 mrem per year. Their maximum recorded dose rate has been 12,000 mrem per year. In the United States on the average, a square mile of soil, one foot deep, contains one ton of K-40, three tons of U-238 and six tons of Th-232. The amount of exposure one is subjected to depends upon the concentration in the soil and the type of soil. In the U.S., three broad areas have been found. These are: the coastal region along the Atlantic Ocean and the Gulf of Mexico, the Colorado Plateau region, and the remainder of the country. The yearly whole body dose equivalent rates in these areas range from 15-35 mrem, 75-140 mrem, and 35-75 mrem, respectively. When absorbed dose rate measurements are weighted by population, and averaged over the entire U.S., the yearly average is estimated at 28 mrem (280 µSv) in NCRP Report No. 93. Radioactivity of Water Depending upon the type of water supply one is talking about, a number of products may turn up. For example, sea water contains a large amount of K-40. On the other hand, many natural springs show amounts of uranium, thorium, and radium. Almost all water should be expected to contain certain amounts of radioactivity. Since rain water will pick up radioactive substances from air, and 1.05-3 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide ground water will pick up activity present in rocks or soil, one would expect to find some radioactivity in water throughout the world. The U.S. average of alpha emitters in water is <1 pCi/l. Some regions contain significantly higher levels of naturally occurring alpha emitters: 40-50 pCi/l may be found in Colorado; and 200 pCi/l may be found in bottled water from Brazil. The chief source of dose rate from this background factor occurs as the result of uptake of these waters by ingestion. This leads to an internal exposure. Any estimate of the dose rate from this source is thus included in the estimate of the dose rate from radioactivity in the human body. The transfer of radioactive substances to the body seems to be mainly by food intake except in cases of very high water concentrations. 1.05.01 Identify the following four sources of natural background radiation including the origin, radionuclides, variables and contribution to exposure. b. Cosmic Cosmic Radiation

Much work has been carried out in the study of cosmic radiation. This factor in background levels was discovered during attempts to reduce background. Though detection devices showed a response even in the absence of any known sources, it was assumed this background was due entirely to traces of radioactive substances in the air and ground. Thus, if a detector was elevated to a greater height above the earth's surface, the background should be greatly reduced. The use of balloons carrying ion chambers to great heights yielded data which showed the effect increased, rather than decreased. These and other data showed that radiation was really coming from outer space. The name cosmic rays was given to this high energy. Further study has shown that cosmic radiation consists of two parts: primary and secondary. The primary component may be further divided into galactic, geomagnetically trapped radiation, and solar. Primary The galactic cosmic rays come from outside the solar system and are composed mostly of positively charged particles. Studies have shown that outside the earth's atmosphere, cosmic rays consist of 87% protons, of 11% alpha particles, and about 1% each of other heavier nuclei and electrons at latitudes above 55 degrees. These particles may have energies in the range of about 1 GeV and higher. 1.05-4 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide As a charged particle approaches the earth, it is acted upon by the earth's magnetic field. In order to pass on through to the earth, the particle must have a certain momentum. Otherwise, it may be trapped by the earth's magnetic field. This gives rise to the second type of primary cosmic rays, the geomagnetically trapped radiation. Solar cosmic rays are produced following severe solar flares on the surface of the sun. These rays consist of protons. The events are classed as high energy or low energy. The high energy events can be observed by ground-level neutron devices. The low energy events are more frequent but must be detected at high altitude. Since these events produce radiation throughout the solar system, they are of great importance in shielding design for manned space missions. Secondary Secondary cosmic rays result from interactions which occur when the primary rays reach the earth's atmosphere. When the high energy particles collide with atoms of the atmosphere, many products are emitted: pions, muons, electrons, photons, protons, and neutrons. These, in turn, produce other secondaries as they collide with elements or decay on the way toward the earth's surface. Thus, a multiplication or shower occurs in which as many as 108 secondaries may result from a single primary. Most of the primary rays are absorbed in the upper 1/10 of the atmosphere. At about 20 km and below, cosmic rays are almost wholly secondary in nature. The total intensity of cosmic rays shows an increase from the top of the atmosphere down to a height of 20 km. Although the primary intensity decreases, the total effect increases because of the rapid rise in the number of the secondaries. Below 20 km, the total intensity shows a decrease with height because of attenuation of the secondaries without further increase in their number due to primaries. At less than 6 km of altitude, the highly penetrating muons, and the electrons they produce, are the dominant components.

At the earth's surface, the secondary cosmic rays consist mainly of muons (hard component), electrons and photons (soft components), and neutrons and protons (nucleonic component). At sea level about 3/4 of the cosmic ray intensity is due to the hard component. Because of the earth's magnetic field, cosmic ray intensity also varies with latitude. The energy which is needed for a charged particle to reach the earth's atmosphere at the geomagnetic equator is larger than that needed at other latitudes. The effect is greatest for latitudes between 15 and 50 degrees. Above 50 degrees, the intensity remains almost constant. Thus, the lowest value of the intensity occurs at the geomagnetic equator, and the effect is expressed as the percentage increase at 55 degrees over that at the equator. At sea level, the effect is small for the ionizing component (10%) but is larger for the neutron component. 1.05-5 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide The dose rate produced by this source of background may be divided into two parts. The portion caused by the ionizing component is estimated from ion chamber readings. The portion caused by the neutron component is hard to measure because the dose rate depends so much on the energy spectrum of the neutrons. For the neutron dose estimates one must rely on calculations. At sea level and high latitudes, the ionization rate, is about 2.1 x 106 ion pairs per cubic meter. Using a neutron calculation, the sea level dose would be increased by about 5%. Taking into account the dose variation with altitude, and the population distribution with altitude, the average yearly dose equivalent rate to the U.S. population from cosmic radiation is estimated to be 27 mrem (270 µSv). This dose equivalent rate would be expected to decrease slightly with latitude and increase with altitude. For example at Denver, the yearly dose would be about 50 mrem (500 µSv). 1.05.01 Identify the following four sources of natural background radiation including the origin, radionuclides, variables and contribution to exposure. c. Internal Emitters Internal Emitters (Radioactivity of the Human Body) Since small amounts of radioactive substances are found throughout the world in soil and water, some of this activity is transferred to man by way of the food chain cycle. A number of studies have been made to try to find a correlation between the amounts in soil and that in man. Results have not shown a clear-cut relationship as yet. In the human body, K-40, Rb-87, Ra-226, U-238, Po-210, and C-14 are the main radionuclides of concern. Of these, K-40 is the most abundant substance in man. The amount in food varies greatly, so that intake is quite dependent on diet. However, variations in diet seem to have little effect on the body content. The content of K-40 in body organs of man varies widely. Based on an average content of 0.2% by weight in soft tissue, 0.05% in bone, the yearly dose equivalent rate to the gonads is estimated to be 19 mrem (190 µSv); 15 mrem (150 µSv) to bone surfaces; and 15 mrem (150 µSv) to bone marrow. Rb-87 contributes only a few percent of these values. Most of the Ra-226 which is taken into the body will be found in the skeleton. Much data has been gathered on the concentration in humans, and the present assumed average skeletal concentration is taken as about 0.29 Bq/kg. The skeletal content of Ra-228 is taken as 0.14 Bq/kg. The yearly dose rate produced by these components is estimated to be .5 mrem (5 µSv) to the gonads, 14.6 mrem (146 µSv) to bone surfaces and 2.2 mrem (22 µSv) to bone marrow.

1.05-6 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide Based upon an average concentration of U-238 of 0.26 Bq/kg in bone, the estimated doses in man are 4.8 mrem (48 µSv) to bone surfaces and 0.9 mrem (9 µSv) to the marrow. From the estimated content in the gonads, the annual dose equivalent is estimated to be about 1 mrem (10 µSv). Similarly, the Po-210 decay chain contribution is taken as 2.22 Bq/kg, yielding annual dose equivalents of 24 mrem (240 µSv) to bone surfaces and 4.9 mrem (49 µSv) to bone marrow. The soft tissue concentration is taken as 0.111 Bq/kg, but is about twice that in the gonads. This gives an annual gonad dose equivalent of 6 mrem (60 µSv). The average whole body content of carbon is taken as 23%. However, C-14 is present in normal carbon only to a very small extent (C-14/C-12 ~10-12), so that only a small amount of C-14 is present. The annual average dose equivalent turns out to be about 1 mrem (10 µSv) total body. In soft tissue, the annual dose is 0.7 mrem (7 µSv). The annual dose to the bone surfaces is 0.8 mrem (8 µSv), and to the bone marrow, 0.7 mrem (7 µSv). The U.S. annual average dose equivalent for all internal emitters (food chain) in the body is 39 mrem (390 µSv) as listed by the NCRP Report No. 93. 1.05.01 Identify the following four sources of natural background radiation including the origin, radionuclides, variables and contribution to exposure. d. Radon Inhaled Radionuclides (Radioactivity of the Air) The background which is found in air is due mainly to the presence of radon and thoron gas, formed as daughter products of elements of the uranium and thorium series. The decay of U-238 proceeds to Ra-226. When Ra-226 emits an alpha as it decays, the gas Rn-222 is formed, which is called radon. In the thorium chain, the decay of Ra-224 results in the gaseous product Rn-220, which is called thoron. Since uranium and thorium are present to some extent throughout the crust of the earth, these products are being formed all the time. Since they are gases, they tend to diffuse up through the earth's surface to become airborne. In turn, the decay products of these gases attach themselves to dust in the air. The amount of these gases in the air depends upon the uranium and thorium content of a certain area. In any given area, the weather conditions will greatly affect the concentrations of these gases. It is also common to find that the levels indoors are higher than those outdoors. This is a function of the material of the building and the ventilation rate. In mines and other underground caverns, the concentrations have been found to be quite high. 1.05-7 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide Some homes in Grand Junction and Durango, Colorado, have been found to have high radon levels. This was traced to the use of uranium mill tailings, residues rich in radium, as backfill. This discovery has led to radon measurements in homes in other areas of the country. Some homes in Pennsylvania are situated on land with naturally elevated radium concentrations, giving rise to increased indoor radon levels. Investigations have been made of radon levels in homes in the Chicago area. Their results indicated that 6% of the homes studied had radon concentrations comparable to those found at Grand Junction. Because of the potential population dose from this source, much more work on defining this potential problem is being carried out.

The major source of exposure from radon in air occurs when the daughter products attach themselves to aerosols and are inhaled. This leads to an internal dose to the lungs. As for external exposure, the external gamma dose rate from Rn-222 and Rn-220 is estimated to be less than 5% of the total external terrestrial dose rate. The contribution of inhaled radon gas to the annual average effective dose equivalent is included as an inhaled radionuclide. Among other radioactive products which are found in air in measurable amounts are C-14, H-3, Na-22, and Be-7. These are called cosmogenic radionuclides, since they are produced in the atmosphere by cosmic rays. None of these products add a significant amount to the background dose rate. The U.S. annual average dose equivalent for various inhaled radionuclides (primarily radon) is estimated at 200 mrem (2,000 µSv) by the NCRP Report No. 93. MAN-MADE BACKGROUND RADIATION SOURCES 1.05.02 Identify the following four sources of artificially produced radiation and the magnitude of dose received from each. a. Nuclear Fallout Nuclear Fallout The term fallout has been applied to debris which settles to the earth as the result of a nuclear blast. This debris is radioactive and thus a source of potential radiation exposure to man. Radioactive fallout is not considered naturally occurring but is definitely a contributor to background radiation sources. Because of the intense heat produced in a nuclear explosion during a very short time, matter which is in the vicinity of the bomb is quickly vaporized. This includes fission products formed in the fission process, unused bomb fuel, the bomb casing and parts, and, in short, any and all substances which happen to be around. These are caught in the fireball which expands and rises very quickly. As the fireball cools and condensation occurs, a mushroom-shaped cloud is formed, containing small solid particles of debris as well as small drops of water. The cloud continues to rise to a height which is a function of the bomb yield and the meteorological factors of the area. For yields in the megaton 1.05-8 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide range (1 megaton equals an energy release equivalent to one million tons of TNT), the cloud top may reach a height of 25 miles. The fallout which occurs may be described as local or world-wide. The portion of debris which becomes local fallout varies from none (in the case of a high-altitude air burst) to about half (in the case of a contact surface burst). The height at which the bomb goes off is thus quite important in the case of local fallout. If the fireball touches the surface of the earth, it will carry aloft large amounts of surface matter. Also, because of the vacuum effect created by the rapid rise of the fireball, other matter may be taken up into the rising fireball. This leads to the formation of larger particles in the cloud that tend to settle out quickly. If the width is not too great, the fallout pattern will be roughly a circle around ground zero. Ground zero is the point on the surface directly under, at, or above the burst. Other bits of matter will fall out at various stages. the distance from ground zero at which they strike the surface and the time it takes depend upon the height from which they fall, their size, and the wind patterns at all altitudes. This results in a cigar-shaped pattern downwind of the burst point. Local fallout usually occurs within the first 24 hours after the blast.

If the height of the burst is such that the fireball does not touch the surface, then the debris is carried aloft and dispersed into the atmosphere. This matter then descends to earth at a later time and is called world-wide fallout. The residence time of this debris is a function of the bomb yield. For yields in the kiloton range, the debris is not projected into the stratosphere. It is limited to a region called the troposphere, between about 9,000 and 17,000 meters. In this region, there is quite a bit of turbulence as well as precipitation. The debris is removed rather quickly; from about one day to one month. If the burst is in the megaton range, the debris is carried into the stratosphere. In this region little mixing will occur, and the absence of rain or snow prevents this matter from being washed down. The time that it takes for this debris to return to the troposphere and be washed down varies. It is a function of both the height in the stratosphere to which the debris is lifted and the locale at which the burst occurs. It may take up to 5 years or more for this debris to return to earth. On the other hand, for bursts in the northern hemisphere in which the debris is confined to only the lower part of the stratosphere, the half- residence time is thought to be less than one year. Half-residence time is the time for one- half of the debris to be removed from the stratosphere. In all, there are more than 200 fission products which result from a nuclear blast. The half-life of each of these products covers the range from a fraction of a second to millions of years. Local fallout will contain most of these products. Because of the time delay in the appearance of world-wide fallout, only a few of these products are important from that standpoint. Since local fallout is confined to a relatively small area, its effect on the human population can be negated by proper choice of test sites, weather conditions, and type of burst. The fallout of interest from the standpoint of possible effects on man due to testing is the world-wide fallout. 1.05-9 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide A number of factors must be considered when one attempts to assess the hazard from world-wide fallout. Because of the associated time delay before world-wide fallout shows up, many fission products and activation products decay out in transit. Others, because they are produced in such small amounts, are diluted so that they do not produce much of an effect. Also, once the fallout does arrive, to be of importance internally, there must be a transfer to the body and absorption into the body organs. All these factors combine to limit the number of fission products which may have an effect on man. The main contribution comes from Sr-90, Cs-137, I-131, C-14, H-3 with minor contributions from Kr-85, Fe-55, and Pu-239. Although the U.S. ceased atmospheric testing in 1962, the inventory of fission products from previous bursts has committed man to future doses. NCRP Report No. 93 lists the annual average effective dose equivalent from nuclear fallout exposure at less than 1 mrem (10 µSv). However the total dose commitment, to be delivered over many generations, is 140 mrem (1400 µSv). 1.05.02 Identify the following four sources of artificially produced radiation and the magnitude of dose received from each. b. Medical Exposures Medical Exposures

The exposure to the U.S. population from X-rays used in medical and dental procedures is the largest source of man-made radiation. It is estimated that more than 300,000 X-ray units are in use in the U.S., and that about 2/3 of the U.S. population is exposed. In 1970, the estimated annual average bone marrow dose equivalent from dental and medical X-rays to the U.S. population was about 78 mrem (780 µSv). In addition to the exposure from X-rays, nuclear medicine programs use radiopharmaceuticals for diagnostic purposes. Radiologists also use radionuclides for therapy treatment. It has been estimated that more than 10 million doses are administered each year. NCRP Report No. 160 gives the dose equivalent for medical radionuclides as ~ 300 mrem/yr 1) 2) 3) computed tomography (total average dose ~ 150 mrem/yr) nuclear medicine (total average dose ~ 75 mrem/yr) radiography/fluroscopy (total average dose ~ 75 mrem/yr) Diagnostic X-Rays There are many different types and styles of X-ray machines used in the medical field. An X-ray machine generally consists of the X-ray tube, an electrical source of high voltage, a type of filament, and radiation shielding to collimate the beam to some limited size and shape. A diagnostic X-ray machine is used to obtain an image of some part of the body on some type of storage material. There are three general types of diagnostic X-ray equipment: radiographic, fluoroscopic, and photofluorographic. Radiography involves the use of an X-ray tube and a photographic plate. The patient is placed between the two and an image is produced on the film of the area exposed. A common "chest X-ray" is an example of a radiographic X-ray. 1.05-10 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide In a fluoroscopic X-ray machine the film cassette is substituted with an imaging device (image intensifier). This enables the radiologist to observe the part of the body exposed live on a video monitor. A blocking agent, such as barium, is often swallowed by the patient to allow the medical staff to observe internal processes in action. A fluoroscopic examination can be used to locate ulcers in a "GI series." The photofluorographic process utilizes an X-ray tube, a fluorescent screen and a camera. This practice is similar to radiographic X-rays, with the substitution of a fluorescent screen for the film. The radiologist can take several pictures on one roll of film of the image on the fluorescent screen. Photofluorography is used for screening large numbers of individuals such as in the military or prison. As science gradually became more aware of the potential hazards associated with radiation exposures, the doses received from diagnostic X-rays have been closely examined. Medical diagnostic exposures contribute more than 50% of the dose to the U.S. population from artificial sources. The U.S. Public Health Services has been tasked with tracking medical X-ray procedures at ten year intervals. Surveys were conducted in 1961, 1970, and 1980. Due to budget cut backs, the 1980 survey does not contain as much useful data as the one produced in 1970. The concept of "Genetically Significant Dose" (GSD) is used in most publications covering background radiation. The GSD includes only the fraction of the radiation which actually deposits energy in the gonads (ovaries and testes) of persons of childbearing potential. Dose rates that produce a small exposure over a years time cannot be expected to produce any acute somatic radiation injury. Late effects from this exposure are almost negligible at these low dose rates. Genetic mutations are transmitted on to our offspring who will then be exposed during their lifetimes. The cumulative effect on genetic mutations over several generations might show a very slight increase due to background radiation.

Medical Radionuclides Radionuclides are used in medicine by two general classifications: Nuclear Medicine for diagnostic procedures and Radiation Oncology for radiation therapy. Because this science is utilized by a limited portion of the U.S. population, its contribution to the average U.S. dose is not significant. Radionuclides are used to determine the extent of a medical problem in a patient. The radionuclide is "attached" to a pharmaceutical which is administered to the patient. The drug has the properties to deposit the radioisotope in the organ of concern. Then using external radiation detectors, the medical staff can determine abnormalities in the organ. A thyroid scan and lung function test are examples. Since the radioisotope is internally deposited either by mouth or by injection, it should decay by emitting only photons. Isotopes emitting alpha or beta particles would be locally absorbed in the organ and would not contribute to the information signal. Another consideration in radionuclide selection would be the effective half-life. To maintain organ doses ALARA, isotopes with a few hour 1.05-11 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide half-life are optimum. Technetium-99m and Indium-113m are commonly used radiopharmaceuticals. Radiation Oncology (study/treatment of tumors) uses radionuclides for tumor treatment. In the United States Cobalt-60 is generally used for the high activity sealed source. This consists of a mechanical device which moves the source to an opening in a collimator which projects a beam of photons used for treatment. A typical 6,000 curie Cobalt-60 source delivers about 100 rad/minute to a tumor. 1.05.02 Identify the following four sources of artificially produced radiation and the magnitude of dose received from each. c. Consumer Products Consumer Products In NCRP Report 56, a number of consumer products and miscellaneous sources of radiation exposure to the U.S. population are discussed. In general, two groups of sources have been found: 1. Those in which the dose equivalent is relatively large and many people are exposed. 2. Those in which the dose equivalent is small but many people are exposed or the dose equivalent is large but only a few people are exposed. Such products as television sets, luminous-dial watches, smoke detectors, static eliminators, tobacco products, airport luggage inspection systems, building materials and many other sources have been studied. The estimated annual average whole body dose equivalent to the U.S. population from consumer products is approximately 10 mrem (100 µSv). The major portion of this exposure (approximately 70%) is due to radioactivity in building materials. Television Receivers Television receivers have the potential for three X-rays sources: the picture tube, the shunt regulator and the vacuum tube regulator. In 1960 the ICRP and the NCRP recommended limits be established such that receivers produce less than 0.5 mrem/hr at any access point 5 cm from the surface of the set. In May of 1967 a major manufacturer recalled 149 big screen sets. Of this group, two sets were found to produce exposures in excess of 100 mR/hr. Due to the ever increasing improvements in TV manufacturing, solid state, and the use of "hold down" circuits, the annual exposure is being reduced. X- ray emissions can be kept below 0.1 mR/hr with low voltage within manufacture specifications. Higher emissions can result if the voltage is increased by repairman in order to increase picture quality. It is estimated that the U.S. total average exposure from watching TV is between 0.5 and 1.5 mrem/yr.

1.05-12 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide Shoe-Fitting Fluoroscopes In the 1950's the use of fluoroscopes was wide spread in shoe stores. It was estimated that 10,000 of these were in use in 1953. Exposures to the feet ranged from 7 to 14 Roentgens per 20 second exposure. The concurrent exposure to the pelvis ranged between 30 and 170 mR per exposure. Shoe-fitting fluoroscopes have been banned or restricted in most states. Radioluminous Watches Radium-226 was used widely in the earlier part of this century for it's luminescence in watches, clocks and dials. No radium-226 watches have been sold in the U.S. since 1970. It is estimated that 10 million of these watches are still in use. Individual dose can reach 310 mrem/yr for the wearer of a watch containing 4.5 uCi of radium-226. The average dose to radium watch wearers is approximately 3 mrem/yr. The majority of luminescent watches on the market today contain either tritium (H-3) or promethium-147. It's estimated that the 16 million who wear tritium watches receive 0.6 mrem/yr, and the 2 million who wear promethium watches receive 0.25 mrem/yr. Miscellaneous There are many additional consumer products that may be included as a source of radiation. Polonium and lead isotopes have been found in tobacco products and contribute a dose-equivalent to small areas of the bronchial epithelium of up to 8 rem/yr to smokers. Building materials, smoke detectors, lantern mantles, and ceramic glazes are all known sources. Another source of radiation exposure to the public arises from the wide use of coal. Coal contains C-14, K-40, uranium and thorium and when burned, the resulting flyash released to the atmosphere carries some of this radioactivity with it. This leads to inhalation of airborne flyash producing lung exposure. The dose equivalent rate in the vicinity of one of these plants has been estimated to be in the range of 0.25-4 mrem/y (2.5-40 µSv/y). 1.05.02 Identify the following four sources of artificially produced radiation and the magnitude of dose received from each. d. Nuclear Facilities Nuclear Facilities By 1988, 90 nuclear power plants had been licensed in the U.S. In addition, over 300 other reactors, classed as non-power reactors, are being operated. In order to provide fuel for these reactors, mining and milling of uranium ore is carried out and fuel fabrication plants are operating. There are several hundred mines, 20 uranium mills and 21 fuel fabrication facilities. 1.05-13 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide Sources of radiation from nuclear reactors consist of prompt neutrons, gamma rays and possible exposures from contamination or environmental releases. The NRC has been tasked by the federal government to calculate doses for populations living within 50 miles of a nuclear facility. Three radionuclides released during routine operations, which contribute to the population dose, are H-3, C-14, and Kr-85. Current estimates of the yearly average dose equivalent in the U.S. from environmental releases is <1 mrem (10 µSv). Total Background Radiation The average annual total effective dose to the general population (non-smokers) from naturally occurring and manmade sources is about 620 mrem. 1.05-14 DOE-HDBK-1122-2009 Module 1.05 Sources of Radiation Study Guide This page intentionally left blank. 1.05-15