Exam (elaborations) TEST BANK FOR Problem Solution Manual For Fundamentals of Nuclear Science and Engineering 3rd Edition By J. Kenneth Shultis and Richard E. Faw (Solution Manual)

1. Both the hertz and the curie have dimensions of s−1. Explain the difference between these two units. Solution: The hertz is used for periodic phenomena and equals the number of “cycles per second.” The curie is used for the random or stochastic rate at which a radioactive source decays, specifically, 1 Ci = 3.7 × 1010 decays/second. 2. Advantages of SI units are apparent when one is presented with units of barrels, ounces, tons, and many others. (a) Compare the British and U.S. units for the gallon and barrel (liquid and dry measure) in SI units of liters (L). (b) Compare the long ton, short ton, and metric ton in SI units of kg. Solution: Unit conversions are taken from the handbook Conversion Factors and Tables, 3d ed., by O.T. Zimmerman and I. Lavine, published by Industrial Research Service, Inc., 1961. (a) In both British and U.S. units, the gallon is equivalent to 4 quarts, eight pints, etc. However, the quart and pint units differ in the two systems. The U.S. gallon measures 3.7853 L, while the British measures 4.546 L. Note that the gallon is sometimes used for dry measure, 4.405 L U.S. measure. The barrel in British units is the same for liquid and dry measure, namely, 163.65 L. The U.S. barrel (dry) is exactly 7056 in3, 115.62 L. The U.S. barrel (liq) is 42 gallons (158.98 L) for petroleum measure, but otherwise (usually) is 31.5 gallons (119.24 L). (b) The common U.S. unit is the short ton of 2000 lb, 907.185 kg, 20 short hundredweight (cwt). The metric ton is exactly 1000 kg, and the long ton is 20 long cwt, 22.4 short cwt, 2240 lb, or 1016 kg. 1-1 1-2 Fundamental Concepts Chap. 1 3. Compare the U.S. and British units of ounce (fluid), (apoth), (troy), and (avdp). Solution: The U.S. and British fluid ounces are, respectively, 1/32 U.S. quarts (0.02957 L) and 1/40 British quarts (0.02841 L). The oz (avdp.) is exactly 1/16 lb (avdp), i.e., 0.02834 kg. Avdp., abbreviation for avoirdupois refers to a system of weights with 16 oz to the pound. The apoth. apothecary or troy ounce is exactly 480 grains, 0.03110 kg. 4. Explain the SI errors (if any) in and give the correct equivalent units for the following units: (a) mgrams/kiloL, (b) megaohms/nm, (c) N·m/s/s, (d) gram cm/(s−1/mL), and (e) Bq/milli-Curie. Solution: (a) Don’t mix unit abbreviations and names; SI prefixes only in numerator: correct form is μg/L. (b) Don’t mix names and abbreviations and don’t use SI prefixes in denominator: correct form nohm/m. (c) Don’t use hyphen and don’t use multiple solidi: correct form Nms−2. (d) Don’t mix names and abbreviations, don’t use multiple solidi, and don’t use parentheses: correct form g cmsmL or better 10 μgms L. (e) Don’t mix names with abbreviations, and SI prefix should be in numerator: correct form kBq/Ci. 5. Consider H2, D2, and H2O, treated as ideal gases at pressures of 1 atm and temperatures of 293.2K . What are the molecular and mass densities of each. Solution: According to the ideal gas law, molar densities are identical for ideal gases under the same conditions, i.e., m = p/RT. From Table 1.5, R = 8. Pa m3/K. For p = 0. MPa= 1 atm., and T = 293.2K , m = 41.56 mol/m3. Multiplication by molecular weights yield, respectively, 83.78 , 167.4, and 749.0 g/m3 for the three gases. 6. In vacuum, how far does light move in 1 ns? Solution:
x = c
t = (3 × 108 m/s) × (10−9 s) = 3 × 10−4 m = 30 cm. Fundamental Concepts Chap. 1 1-3 7. In a medical test for a certain molecule, the concentration in the blood is reported as 57 mcg/dL. What is the concentration in proper SI notation? Solution: 123 mcg/dL = 10−310−2 g/10−1 L = 1.23 × 10−4 g/L = 57 μg/L. 8. How many neutrons and protons are there in each of the following nuclides: (a) 11B, (b) 24Na, (c) 60Co, (d) 207Pb, and (e) 238U? Solution: Nuclide neutrons protons 11B 6 5 24Na 13 11 60Co 33 27 207Pb 125 82 238U 146 92 9. Consider the nuclide 71Ge. Use the Chart of the Nuclides to find a nuclide (a) that is in the same isobar, (b) that is in the same isotone, and (c) that is an isomer. Solution: (a) 71As, (b) 59Ga, and (c) 71mGe 10. Examine the Chart of the Nuclides to find any elements, with Z less that that of lead (Z = 82), that have no stable nuclides. Such an element can have no standard relative atomic mass. Solution: Promethium (Z = 61) and Technetium (Z = 43) 11. What are the molecular weights of (a) H2 gas, (b) H2O, and (c) HDO? Solution: From Table A.3, A(O) = 15.9994 g/mol; from Table B.1 A(H) = 1. g/mol and A(D) = 2. g/mol. (a) A(H2) = 2 A(H) = 2 × 1. = 2.01565 g/mol (b) A(H2O) = 2 A(H) + A(O) = 2 × 1.+ 15.9994 = 18.0151 g/mol (c) A(HDO) = A(H) + A(D) + A(O) = 1.+ 2.+ 15.9994 = 19.0213 g/mol 1-4 Fundamental Concepts Chap. 1 12. What is the mass in kg of a molecule of uranyl sulfate UO2SO4? Solution: From Table A.3, A(U) = 238.0289 g/mol, A(O) = 15.9994 g/mol, and A(S) = 32.066 g/mol. The molecular weight of UO2SO4 is thus A(UO2SO4) = A(U) + 6A(O) + A(S) = 238.0289+ 6(15.994)+ 32.066 = 366.091 g/mol = 0. kg/mol. Since one mol contains Na = 6.022× 1023 molecules, the mass of one molecule of UO2SO4 = A(UO2SO4)/Na = 0./6.002 × 1023 = 6.079 × 10−25 kg/molecule. 13. Show by argument that the reciprocal of Avogadro’s constant is the gram equivalent of 1 atomic mass unit. Solution: By definition one gram atomic weight of 12C is 12 g/mol. Thus the mass of one atom of 12C is M(12 6C) = 12 g/mol Na atoms/mol = 12 Na g/atom. But by definition, one atom of 12C has a mass of 12 u. Therefore, 1 u = 1 u 12 u/(12C atom)  12 Na g/(12C atom)  = 1 Na g. 14. Prior to 1961 the physical standard for atomic masses was 1/16 the mass of the 16 8O atom. The new standard is 1/12 the mass of the 12 6C atom. The change led to advantages in mass spectrometry. Determine the conversion factor needed to convert from old to new atomic mass units. How did this change affect the value of the Avogadro constant? Solution From Table B.1, the 16 8O atom has a mass of 15. amu. Thus, the pre- 1961 atomic mass unit was 15./16 post-1961 units, and the conversion factor is thus 1 amu (16O) = 0. amu (12C). The Avogadro constant is defined as the number of atoms in 12 g of unbound carbon-12 in its rest-energy electronic state, i.e., the number of atomic mass units per gram. Using data from Table 1.5, one finds that Na is given by the reciprocal of the atomicmass unit, namely, [1.×10−24]−1 = 6.× 1023 mol−1. Pre-1961, the Avogadro constant was more loosely defined as the number of atoms per mol of any element, and had the best value 6.02486×1023. Fundamental