TEST BANK FOR Problem Solution Manual For Nuclear Science 3rd Ed By J. Kenneth Shultis

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 = 8.314472
Pa m3/K. For p = 0.101325 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 = ct = (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) = 1.007825
g/mol and A(D) = 2.014102 g/mol.
(a) A(H2) = 2 A(H) = 2 × 1.007825 = 2.01565 g/mol
(b) A(H2O) = 2 A(H) + A(O) = 2 × 1.007825+ 15.9994 = 18.0151 g/mol
(c) A(HDO) = A(H) + A(D) + A(O) = 1.007825+ 2.014102+ 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 = 0.336091 kg/mol.
Since one mol contains Na = 6.022× 1023 molecules, the mass of one molecule
of UO2SO4 = A(UO2SO4)/Na = 0.366091/6.002 × 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 15.9949146 amu. Thus, the pre-
1961 atomic mass unit was 15.9949146/16 post-1961 units, and the conversion
factor is thus 1 amu (16O) = 0.99968216 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, [1.6605387×10−24]−1 = 6.0221420× 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