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Nuclear Chemistry
Chapter 11
Problem-Set Solutions
11.1
To identify a nucleus or atom uniquely, both its atomic number and its mass number must be
specified. In one notation, a superscript before the symbol for the element is the mass number,
and a subscript is the atomic number. In the second notation, the mass number is placed
immediately after the name of the element (which indicates the atomic number).
a. 104 Be, Be - 10
38
20 Ca,
b. 25
11Na, Na - 25
Ca-38
b.
80
37 Rb,
Rb-80
c. 96
41Nb, Nb - 96
125
51 Sb,
Sb-125
Es-256
a.
11.3
Use the alternate notation that also indicates the atomic number and the mass number.
b. 197
79 Au
17
8O
b.
212
82 Pb
d.
256
99 Es,
11.2
a. 147 N
c.
d. 257
103 Lr, Lr - 257
c. tin–121
d. boron–10
c. rubidium–92
d. bismuth–201
11.4
a.
11.5
a. 24
An alpha particle consists of two protons and two neutrons. The superscript
indicates a mass of 4 amu (two protons + two neutrons); the subscript indicates
that the charge is a +2 (from two protons).
b. -10 
A beta particle has a mass (very close to zero amu) and charge (–1) identical to
those of an electron.
c. 00
A gamma ray is a form of high-energy radiation without mass or charge.
11.6
a. +2 and 4 amu
b. –1 and 0 amu (0.00055 amu)
11.7
An alpha particle consists of two protons and two neutrons.
11.8
They are identical in charge and mass.
11.9
Alpha particle decay is the emission of an alpha particle from a nucleus. The product
nucleus has a mass number that is four less than the original nucleus, and an atomic number
that is two less than the original nucleus. In writing nuclear equations, make sure that the
mass numbers and the atomic numbers balance on the two sides of the equation.
a.
200
84
Po  24 +
c.
244
96
Cm  24 +
11.10 a.
229
90
c.
152
64
418
196
82
b.
240
96
Cm  24 +
Pu
d.
238
92
U  24 +
Pb
240
94
c. 0 and 0 amu
236
94
Pu
234
90
Th
Th  24 +
225
88
Ra
b.
210
83
Bi  24 +
Gd  24 +
148
62
Sm
d.
243
95
Am  24 +
206
81
Tl
239
93
Np
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419
Problem-Set Solutions Chapter 11
11.11 In beta particle decay, the mass number of the new nuclide is the same as that of the original
nuclide; however, the atomic number has increased by one unit. Balance the superscripts and
subscripts on both sides of the equation.
 + 147 N
a.
10
4
Be 
0
-1
 + 105 B
b.
14
6
C 
c.
21
9
F 
21
 + 10
Ne
d.
25
11
Na 
0
-1
11.12 a.
77
32
Ge 
b.
235
92
U 
0
-1
+
c.
16
7
N 
d.
60
26
Fe 
0
-1
 + 60
27 Co
0
-1
0
-1
0
-1
+
77
33
As
 + 168 O
0
-1
 + 25
12 Mg
235
93
Np
11.13 When an alpha particle decay occurs, the product nucleus has an atomic number that is two
less than that of the original nucleus and a mass number that is four less than that of the
original nucleus:
A  A – 4; Z  Z – 2
11.14 A does not change; Z  Z + 1
11.15 We know that the sum of the superscripts (mass numbers) and the sum of the subscripts
(atomic numbers or particle charges) on the two sides of the equation must be equal.

a.
0
-1
b.
28
12 Mg
11.16 a.
200
80 Hg
234
94
Superscripts: The particle has a zero mass number, so the reactant and product
atoms have the same mass number (28). Subscripts: The sum of the atomic
numbers on the product side is 12, so this is also the atomic number of the
reactant (at. no. 12 = Mg).
Superscripts: The mass number of the product is four less than the mass number
of the reactant; the particle has a mass number of four. Subscripts: The atomic
number of the product is two less than that of the reactant; the particle has an
atomic number of two.
c. 24
d.
Superscripts: Reactant and product atoms have the same mass number, so the
particle has a zero mass number. Subscripts: The product has an atomic number
one greater than that of the reactant, so the particle has a charge of –1.
Pu
Superscripts: The mass number of the product must be four less than the mass
number of the reactant (204 – 4 = 200). Subscripts: The atomic number of the
product must be two less than that of the reactant (82 – 2 = 80).
b.
188
76
Os
c.
84
36
Kr
d.
0
-1

11.17 To write each of these decay equations, put the parent nuclide on the reactant side and the
daughter nuclide on the product side. Then put in the correct particle on the product side to
balance the subscripts and the superscripts.
a. 24
b.
0
-1

Superscripts (mass numbers): 190 (parent) – 186 (daughter) = 4
Subscripts (atomic number or charge): 78 (parent) – 76 (daughter) = 2
Superscripts (mass numbers): 19 (parent) – 19 (daughter) = 0
Subscripts (atomic number or charge): 8 (parent) – 9 (daughter) = –1
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420
Problem-Set Solutions Chapter 11
4
11.18 a. 2 
b.
0
-1

11.19 The half-life (t) is the time required for one-half of a given quantity of a radioactive
substance to undergo decay. To calculate the fraction of nuclide remaining after a given time,
first determine the number of half-lives that have elapsed. Then use the equation below
(n is the number of half-lives):
 Amount of radionuclide
  original amount   1 

=
x 
 undecayed after n half-lives   of radionuclide   2 n 
a.
b.
c.
d.
11.20 a.
1
2
2
1
2
6
1
2
3
1
2
6
1
=
=
=
=
of original is undecayed.
4
1
of original is undecayed.
64
1
of original is undecayed.
8
1
of original is undecayed.
64
1
1
=
4
2
16
b.
1
1
=
6
2
64
c.
1
1
=
3
2
8
d.
1
1
=
8
2
256
11.21 First, determine the number of half-lives that have elapsed (using the equation in
Problem 11.19). Then divide the time that has elapsed by the number of half-lives to find the
length of one half-life.
a.
1
1
= 4;
16
2
c.
1
1
= 8;
256
2
11.22 a.
1
1
= 3;
8
2
b.
1
1
= 7;
128
2
c.
1
1
= 5;
32
2
d.
1
1
= 9;
512
2
5.4 days
= 1.4 days
4
5.4 days
= 0.68 day
8
b.
1
1
= 6;
64
2
d.
1
1
= 10 ;
1024
2
5.4 days
= 0.90 day
6
5.4 days
= 0.54 day
10
3.2 days
= 1.1 days
3
3.2 days
= 0.46 day
7
3.2 days
= 0.64 day
5
3.2 days
= 0.36 day
9
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421
Problem-Set Solutions Chapter 11
11.23 First, determine the number of half-lives by dividing the total time elapsed (60.0 hr) by the
length of one half-life (15.0 hr). Then use the number of half-lives to determine the fraction of
undecayed nuclide remaining (equation in problem 11.19).
11.24
60.0 hr
= 4;
15.0 hr
1
1
;
4 =
2
16
112 yr
= 4.0;
28 yr
1
1
=
;
4
2
16
1
x 4.00 g = 0.250 g
16
1
x 4.00 g = 0.250 g
16
11.25 Over 2000 bombardment-produced radionuclides that do not occur naturally are now known.
11.26 7 times greater
11.27 Uranium, element 92, has the highest atomic number of any naturally occurring element.
11.28 83, bismuth
11.29 To write each of these equations for bombardment reactions, put the parent nuclide and the
bombardment particle on the reactant side and the daughter nuclides and decay particles on the
product side. Balance the superscripts (mass numbers) and the subscripts (atomic numbers or
charges).
a. 24
b.
25
12 Mg
Superscripts: 24 + x (reactant side) = 27 + 1 (product side); x = 4
Subscripts: 12 + y (reactant side) = 14 + 0 (product side); y = 2
Superscripts: 27 + 2 (reactant side) = x + 4 (product side); x = 25
Subscripts: 13 + 1 (reactant side) = y + 2 (product side); y = 12
c. 24
Superscripts: 9 + x (reactant side) = 12 + 1 (product side); x = 4
Subscripts: 4 + y (reactant side) = 6 + 0 (product side); y = 2
d. 11 p
Superscripts: 6 + x (reactant side) = 4 + 3 (product side); x = 1
Subscripts: 3 + y (reactant side) = 2 + 2 (product side); y = 1
11.30 a.
249
98
Cr
b.
17
8
O
c.
1
0
n
d. 24
11.31 We would expect lead-207 to be a stable nuclide, since it terminates the uranium-235 decay
series; termination of a decay series requires a stable nuclide.
11.32 All isotopes of radon are radioactive.
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422
Problem-Set Solutions Chapter 11
11.33 In each equation in this decay series, a parent nuclide produces a daughter nuclide and an
alpha or beta particle. For each equation (after the first), the parent nuclide is the daughter
nuclide from the previous equation. Balance the superscripts and the subscripts.
232
90
228
88
Superscripts: 232 (reactants) = 4 + x (products); x = 228
Subscripts: 90 (reactant) = 2 + y (product); y = 88
Th  24 +
228
88
Ra
+
228
89
Ac Superscripts: 228 (reactants) = 0 + x (products); x = 228
Subscripts: 88 (reactant) = –1 + y (product); y = 89
Ra 
0
-1
Th Superscripts: 228 (reactants) = 0 + x (products); x = 228
Subscripts: 89 (reactant) = –1 + y (product); y = 90
Superscripts: 228 (reactants) = 4 + x (products); x = 224
228
4
224
90Th  2  + 88 Ra
Subscripts: 90 (reactant) = 2 + y (product); y = 88
11.34
+
228
89
Ac 
235
92
U  24 +
231
90
Th 
0
-1
0
-1
231
90
+
231
91
Pa  24 +
227
89
Ac 
0
-1
228
90
+
Th
231
91
227
89
Pa
Ac
227
90
Th
11.35 An ion pair is the electron and the positive ion that are produced during an ionizing interaction
between a molecule (or an atom) and radiation.
11.36 a chemical species that contains an unpaired electron
11.37 Draw the Lewis structure for each of the species; from the Lewis structure, determine whether
there is an unpaired electron (a free radical).
a.
b.
H O
H O H
H
H
Yes, there is an unpaired electron;
H2O+ is a free radical.
c.
H O
Yes, there is an unpaired electron;
OH is a free radical.
11.38 a. H2O + radiation  H2O+ + e–
No, there is no unpaired electron;
H3O+ is not a free radical.
d.
H O
No, there is no unpaired electron;
OH— is not a free radical.
b. H2O+ + H2O  H3O+ + OH
11.39 The fate of a radiation particle involved in an ion pair formation is to continue on, interacting
with other atoms and forming more ion pairs.
11.40 (1) interaction with another free radical to form a non-free radical molecule; (2) interaction
with another molecule to form another free radical
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Problem-Set Solutions Chapter 11
423
11.41 Alpha particles are the most massive and the slowest particles involved in natural radioactive
decay processes; they are, therefore, less penetrating and are stopped by a thick sheet of paper.
Beta particles and gamma rays, having much greater speed than alpha particles, go through a
thick sheet of paper.
11.42 Alpha and beta are stopped; gamma goes through.
11.43 Alpha particle velocities are on the order of 0.1 the speed of light; beta particle velocities are
up to 0.9 the speed of light; gamma rays have a velocity equal to the speed of light.
11.44 An alpha produces 40,000 ion pairs, and a beta produces 2000 ion pairs.
11.45 Table 11.3 gives the effects of short-term whole-body radiation exposure on humans.
a. Whole-body radiation exposure of 10 rems has no detectable effects.
b. Whole-body radiation exposure of 150 rems can cause nausea, fatigue, and lowered blood
cell count.
11.46 a. short-term reduction in blood cell count
b. appetite loss, general malaise, sore throat, pallor, diarrhea
11.47 Figure 11.11 shows the estimated annual radiation exposure of an average American. From
this graph we can see that 19% of the radiation exposure that an average American receives is
human-made, and 81% comes from natural sources.
11.48 radon seepage, cosmic radiation, rocks and soil, minerals in the body, medical x-rays
11.49 Film badges are used to monitor the extent of radiation exposure.
11.50 Radiation ionizes atoms and molecules; this produces a pulse of electricity which is detected.
11.51 Nearly all diagnostic radionuclides are gamma emitters because the penetrating power of
alpha and beta particles is too low; it is necessary to be able to detect the radiation externally
(outside the body).
11.52 to minimize exposure and increase radiation intensity to detectable levels
11.53 Table 11.4 lists selected radionuclides used in diagnostic procedures.
a. Barium-131 is used to detect bone tumors.
b. Sodium-24 is used to detect circulatory problems.
c. Iron-59 is used in the evaluation of iron metabolism in blood.
d. Potassium-42 is used to determine intercellular spaces in fluids.
11.54 a.
b.
c.
d.
fluid buildup in brain, location of cysts and blood clots, thyroid problems
blood studies, breast carcinoma
brain tumors, spleen and thyroid problems, location of blood clots
blood studies, kidney activity
11.55 Radionuclides used for therapeutic purposes are usually  or  emitters instead of  emitters.
Therapeutic radionuclides are used to selectively destroy abnormal (usually cancerous) cells;
an intense dose of radiation in a small localized area is needed.
11.56 cobalt-60, external radiation source; yttrium-90, implantation therapy
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424
Problem-Set Solutions Chapter 11
11.57 In each of these nuclear fission equations, the difference in mass numbers between product
and reactant sides gives the number of neutrons produced.
a. Reactant side (235 + 1 = 236) – product side (135 + 97 = 232) = 4 neutrons
b. Reactant side (235 + 1 = 236) – product side (72 + 160 = 232) = 4 neutrons
c. Reactant side (235 + 1 = 236) – product side (90 + 144 = 234) = 2 neutrons
d. Reactant side (235 + 1 = 236) – product side (142 + 91 = 233) = 3 neutrons
11.58 a. 3
b. 3
c. 2
d. 2
11.59 The mass number of the nuclide that undergoes nuclear fission plus the mass of the initiating
neutron is equal to the sum of the mass numbers of the products:
x + 1 = 143 + 94 + 3(1); x = 239
The atomic number (and thus the atomic identity) of the reacting nuclide is equal to the sum of
the atomic numbers of the fission products:
y = 56 + 37 = 93
Element number 93, with a mass number of 239, is neptunium-239.
11.60
234
92
U
11.61 a. Balance the mass numbers: products [2(3)] = reactants [2(1) + x] = 6; x = 4
Balance the atomic numbers: products (2 + x) = reactants [2(2)] = 4; x = 2
The particle having a mass number of 4 and an atomic number of 2 is:

b. Balance the mass numbers: products [2(4) + 1] = reactants [7 + x] = 9; x = 2
Balance the atomic numbers: products [2(2) + 0] = reactants [3 + x] = 4; x = 1
The particle having a mass number of 2 and an atomic number of 1 is:
4
2
2
1
11.62 a.
2
1
H
H
b.
3
2
He
11.63 a. Nuclear fusion is a nuclear reaction in which two small nuclei are put together to make a
larger one. For fusion to occur, an extremely high temperature is required.
b. The process of nuclear fusion occurs on the sun.
c. During transmutation a nuclide of one element is changed into another nuclide of another
element. Transmutation occurs in both fission and fusion reactions.
d. Bombardment of a nuclide with neutrons may cause it to split into two fragments; this
process is called nuclear fission.
11.64 a. both
b. fission
c. fission
d. fusion
11.65 During a nuclear fission reaction, a large nucleus splits into two medium-sized nuclei. During
a nuclear fusion reaction, two small nuclei are put together to make a larger one.
a. This is a fusion reaction because two small nuclei come together to form a larger one.
b. This is a fission reaction because one large nucleus splits into two medium sized nuclei.
c. This reaction is neither fission nor fusion because the large nucleus does not split into two
medium-sized nuclei.
d. This reaction is neither fission nor fusion because the large nucleus does not split into two
medium-sized nuclei.
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425
Problem-Set Solutions Chapter 11
11.66 a. neither
b. neither
c. fusion
d. fission
11.67 Write these products and/or reactants of some radioactive decay processes in the form of
equations. Balance the mass numbers and the atomic numbers to identify the missing parts of
the equations.
11.68
Tl
Superscripts: x (reactant side) = 0 + 206 (product side); x = 206
Subscripts: y (reactant side) = –1 + 81 (product side); y = 80
 + 109
47 Ag
Superscripts: 109 (reactant side) = 0 + x (product side); x = 109
Subscripts: 46 (reactant side) = –1 + y (product side); y = 47
+
a.
206
80
Hg 
0
-1
b.
109
46
Pd 
0
-1
c.
245
96
Cm  24 +
241
94
Pu
Superscripts: x (reactant side) = 4 + 241 (product side); x = 245
Subscripts: y (reactant side) = 2 + 94 (product side); y = 96
d.
249
100
Fm  24 +
245
98
Cf
Superscripts: 249 (reactant side) = 4 + x (product side); x = 245
Subscripts: 100 (reactant side) = 2 + y (product side); y = 98
206
81
1
1
a.
left = 3
8
2
3 half-lives x
18 hrs
= 54 hrs
1 half-life
b.
1
1
left = 5
32
2
5 half-lives x
18 hrs
= 90 hrs
1 half-life
c.
1
1
left = 6
64
2
6 half-lives x
18 hrs
= 108 hrs
1 half-life
d.
1
1
left = 7
128
2
7 half-lives x
18 hrs
= 126 hrs
1 half-life
11.69 In a bombardment reaction, the bombarding particle is written on the reactant side of the
nuclear equation. Write and balance the mass numbers and the atomic numbers to identify the
missing parts of the equations.
11.70
Pu + 24 
a.
243
94
b.
246
96 Cm
c.
27
13
Al + 24 
30
15
d.
23
11
Na + 21 H 
21
10
142
60
246
96
+ 126 C 
Nd + 01 n 
143
61
Cm + 01 n
254
102No
Superscripts: x + 4 (react.) = 246 +1 (prod.); x = 243
Subscripts: y + 2 (react.) = 96 + 0 (prod.); y = 94
+ 410 n Superscripts: 246 + x (react.) = 254 +4(1) (prod.); x = 12
Subscripts: 96 + y (react.) = 102 + 4(0) (prod.); y = 6
P + 01 n
Superscripts: 27 + 4 (react.) = x + 1 (prod.); x = 30
Subscripts: 13 + 2 (react.) = y + 0 (prod.); y = 15
Ne + 24
Superscripts: 23 + 2 (react.) = 21 + x (prod.); x = 4
Subscripts: 11 + 1(react.) = 10 + y (prod.); y = 2
Pm + -10 
11.71 Table 11.2 lists the most stable nuclides of transuranium elements. Fourteen of these elements
have nuclides with half-lives less than 1.0 day.
11.72 four neutrons
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426
Problem-Set Solutions Chapter 11
11.73 When we consider the decay series E  F  G  H, we can see that the half-lives of E and F
are both very short as compared with the total time (50 days). Therefore, we can say that there
are no (or a negligible amount of) E or F atoms left at the end of 50 days. The half-life of G is
not short (12.5 days), so we can use the half-life decay equation to calculate the fraction of
G atoms remaining at the end of 50 days (50 days/12.5 days = 4 half-lives):
1
2n
=
1
2
4
=
1
16
Multiply the fraction of G atoms remaining by the total atoms (1/16 x 1000 = 63 G atoms).
Subtract the number of G atoms from the number of total atoms to find the number of H atoms
(1000 – 63 = 937 H atoms).
At the end of 50 days there are the following numbers of atoms: E = 0 (negligible amount),
F = 0 (negligible amount), G = 63 atoms, H = 937 atoms
11.74
228
Ac, 228Th, 224Ac
11.75 Statement c. is incorrect. A gamma ray is a form of high-energy radiation without mass or
charge.
11.76 b
11.77 The daughter nuclide for the alpha decay of polonium-212 is lead-208.
212
84
Po  24  +
208
82
Pb
11.78 c
11.79 After 3 half-lives have elapsed, the amount of radioactive sample which is not decayed is 1/8
the original amount.
1
2
n
=
1
2
3
=
1
8
11.80 c
11.81 The general production process for synthetic elements involves both a transmutation reaction
and a bombardment reaction.
11.82 a
11.83 Statement d. is incorrect. A piece of aluminum foil will stop alpha particles but not gamma
radiation.
11.84 a
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