Solution Manual For Introduction to Nuclear Engineering A Study Guide 1st Edition By Supathorn Phongikaroon

Solution Manual For
Introduction to Nuclear Engineering A Study Guide 1st Edition By Supathorn
Phongikaroon
Chapters 1-11


Chapter 1 – BASIC UNITS and THE ATOM


A. True or False
1. False. No electron
2. True.
3. True.
4. True
5. False. Same mass number.
6. True.
7. False. It should be 1.00 x 103 m
8. True
9. False. It should be 1.8 x 1014. How?
m m (12 amu)(1.66 10 27 kg/amu)
  4 3 15
 1.8 1017 kg / m 3
V 3 r 4
3 (3  10 m ) 3


 1.8 1017
  1.8 1014
 water 1000
10. False. It is number of atoms in 12 grams of C-12.

B. Problems


1. Explain the difference and similarity between these two units—the hertz and the curie.

SOLUTION:

Both have units of 1/sec. But Hertz describes frequency and curie decribes the activity of

radiation source.




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,2. What is the number of neutrons in 240Pu? 208Pb?

SOLUTION:

You can use the chart of nuclides or periodic table.

Pu has 94 protons. So, # n = 240 – 94 = 146

Pb has 82 protons. So, # n = 208 – 82 = 126


3. What is the mass in kg of a molecule of uranium oxide U3O8?

SOLUTION:

We can approximate this MW of U3O8 is 3(238) + 8(16) = 842 g/mol or 0.842 kg/mol

So, mass of one molecule = 0.842 (1 mol/6.022E23 molecules) = 1.398x10-24

kg/molecule




4. How many atoms of 235U are there in 1 g of natural uranium?

SOLUTION:

1 kg U x 1000 g/1 kg x 1 mol/238 g x 6.022E23 atoms/1mole = 2.53 x 1024 atoms U

Isotopic abundance U-235 = 0.72%. Thus, 2.53 x 1024 x 0.72/100 = 1.8216 x 1022 atom

U-235.


5. The density of natural uranium oxide (UO2) is about 10.50 g/cm3. Natural uranium is

composed of three isotope 234U, 235U, and 238U.

a. Compute the atomic weight of natural uranium (g/mol).

b. What is the molecular weight of UO2 fuel (g/mol)?

c. What is the atomic density of 234U in the UO2 fuel (atoms/cm3)?

SOLUTION:



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, Part a.

Isotope Isotopic Atomic
Abundance Weight
(atom %) (g/mol)
234
U 0.0057 234.0409
235
U 0.72 235.0439
238
U 99.27 238.0508


MW of uranium =

(0.0057/100)(234.0409)+(0.72/100)(235.0439)+(99.27/100)(238.0508) = 238.0186

g/mol.

Part b:

MW of UO2 is 238.0186 + 2(16) = 270.0186 g UO2/mol.

Part c:

Atomic density of UO2 is (10.5 g/cc)/(270.0186 g/mol)*6.022E23 molecules/mol =

2.341E22 molecules UO2/cm3

So, for U-234 = (0.0057 atom U-234/100 atom U) * 2.341E22

= 1.334 x 1018 atoms U-234/cm3.


6. How many atoms of deuterium are there in 2 kg of water assuming the natural

abundance of deuterium in atom percent is 0.0115%?

SOLUTION:
Molecular weight of H2O (assuming all the hydrogen is 1H1, that is, ignoring the
deuterium contribution) = 2AH + AO = 2(1 g/mol) + 16 g/mol = 18 g/mol
Natoms/m = NA/A
Therefore,
Nwater = mwaterNA/Awater = (2000 g)( 0.60221024 molecules/mol)/(18 g/mol)
Nwater = 6.691025 molecules of water
NH = 2Nwater = (2 atoms/molecule)(6.691025 molecules) = 1.341026 H atoms


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, ND = D NH = (0.000115 deuterium atom/H atom)(1.341026 H atoms) = 1.541022 atoms




7. Mars has a radius of about 3396 km and a mass of 6.417×1023 kg. Determine the

radius of Mars (m) when it has the same mass density as matter in a nucleus.

SOLUTION:

From the definition of mass density of a sphere,
 = m/(4r3/3)
Solving for the radius r cubed gives
r3 = 3m/(4)
Substituting for the mass of the earth and the mass density of a nucleus
r3 = (3)(6.4171023 kg)(1000 g/kg)/[4(2.41014 g/cm3)]
r3 = 6.383 1011 cm3
Taking the cube root gives the equivalent radius of the earth
r = 8610.15 cm or 86.1 m


8. A uranium sample is enriched to 3.2 atom-percent in 235U. Determine the enrichment
of 235U in weight-percent. Consider 234U to be very small that can be ignored for the
calculation. [ Ans. 3.16 w/o ]
SOLUTION:
w235 = m235/mU
where mi = NiAi/NA
N 235 A253
w235 = where Avogadro‘s number, NA, has cancelled.
N 235 A235  N 238 A238
Dividing the numerator and denominator by N235,
A235
w235 = Eq. 1
N
A235  ( 238 ) A238
N 235
Now N235/(N235 + N238) = 0.032 (235U atom fraction)
Therefore
0.032N238 = (1 – 0.032)N235
N238/N235 = 0.968/0.032


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