21
Nuclear Chemistry
Visualizing Concepts
21.1
Analyze. Given the name and mass number of a nuclide, decide if it lies within the belt
of stability. If not, suggest a process that moves it toward the belt.
Plan. Calculate the number of protons and neutrons in each nuclide. Locate this point
on Figure 21.2. If the point is above the belt, β-decay increases protons and decreases
neutrons, decreasing the neutron-to-proton ratio. If the point is below the belt, either
positron emission or neutron capture decreases protons and increases neutrons, increasing the neutron-to-proton ratio.
Solve.
21.4
(a)
24
(b)
32
(c)
1 08
(d)
2 16
Ne: 10 p, 14 n, just above the belt of stability. Reduce the neutron-to proton ratio
via β-decay.
Cl: 17 p, 15 n, just below the belt of stability. Increase the neutron-to-proton ratio via positron emission or orbital electron capture.
Sn: 50 p, 58 n, just below the belt of stability. Increase the neutron-to-proton ratio via positron emission or orbital electron capture.
Po: 84 p, 132 n, just beyond the belt of stability. Nuclei with
atomic numbers ≥ 84 tend to decay via alpha emission, which decreases both protons and neutrons.
Analyze/Plan. Write the balanced equation for the decay. Nuclear decay is a first-order
process; use appropriate relationships for first-order processes to determine t 1 /2, k and
88
remaining Mo after 12 minutes.
Solve.
(a)
t 1 /2 is the time required for half of the original nuclide to decay. Relative to the
88
graph, this is the time when the amount of Mo is reduced from 1.0 to 0.5. This
time is 7 minutes.
(b)
For a first-order process, t 1 /2 = 0.693/k or k = 0.693/t 1 /2.
k = 0.693/7 min = 0.0990 = 0.1 min
(c)
From the graph, the fraction of
–1
88
Mo remaining after 12 min is 0.3/1.0 = 0.3
Check. In(N t /N o ) = –kt = –(0.099)(12) = –1.188; N t /N o = e
(d)
613
– 1.188
= 0.30.
21 Nuclear Chemistry
Solutions to Exercises
Radioactivity
21.7
Analyze/Plan. Given various nuclide descriptions, determine the number of protons and
neutrons in each nuclide. The left superscript is the mass number, protons plus neutrons. If there is a left subscript, it is the atomic number, the number of protons. Protons
can always be determined from chemical symbol; all isotopes of the same element have
the same number of protons. A number following the element name, as in part (c) is the
mass number.
Solve.
p = protons, n = neutrons, e = electrons; number of protons = atomic number;
number of neutrons = mass number – atomic number
(a)
21.9
(b)
21.13
21.15
Hg: 80p, 121n
(c)
39
K: 19p, 20n
Analyze/Plan. See definitions in Section 21.1. In each case, the left superscript is mass
number, the left subscript is related to atomic number.
Solve.
(a)
21.11
2 01
(b)
(c)
Analyze/Plan. Follow the logic in Sample Exercises 21.1 and 21.2. Pay attention to definitions of decay particles and conservation of mass and charge.
Solve.
(a)
(b)
(c)
(d)
Analyze/Plan. Using definitions of the decay processes and conservation of mass number and atomic number, work backwards to the reactants in the nuclear reactions.
Solve.
(a)
(b)
(c)
(d)
Analyze/Plan. Given the starting and ending nuclides in a nuclear decay sequence, we
are asked to determine the number of alpha and beta emissions. Use the total change in
A and Z, along with definitions of alpha and beta decay, to answer the question. Solve.
The total mass number change is (235–207) = 28. Since each α particle emission decreases the mass number by four, whereas emission of a β particle does not correspond
to a mass change, there are 7 α particle emissions. The change in atomic number in the
series is 10. Each α particle results in an atomic number lower by two. The 7 α particle
emissions alone would cause a decrease of 14 in atomic number. Each β particle emission raises the atomic number by one. To obtain the observed lowering of 10 in the
series, there must be 4 β emissions.
Nuclear Stability
21.17
Analyze/Plan. Follow the logic in Sample Exercise 21.3, paying attention to the guidelines for neutron-to-proton ratio.
Solve.
(a)
- low neutron/proton ratio, positron emission (for low atomic numbers, positron emission is more common than orbital electron capture)
614
21 Nuclear Chemistry
(b)
(c)
(d)
21.19
Solutions to Exercises
- high neutron/proton ratio, beta emission
- slightly high neutron/proton ratio, beta emission
- high neutron/proton ratio, beta emission
Analyze/Plan. Use the criteria listed in Table 21.3.
(a)
Stable:
odd proton, even neutron more abundant than odd proton, odd
neutron; 20 neutrons is a magic number.
(b)
Stable:
odd proton, even neutron more abundant than odd proton, odd
neutron; 126 neutrons is a magic number.
(c)
Stable:
even proton, even neutron more likely to be stable than even proton, odd neutron;
21.21
21.23
Solve.
has high neutron/proton ratio
Analyze/Plan. For each nuclide, determine the number of protons and neutrons and decide if they are magic numbers.
Solve.
(a)
, both
(b)
(c)
, both
(d)
has a magic number of protons, but not neutrons
has neither
(e)
, both
Consider the stability of the two emitted particles. The alpha particle,
, has a magic number of both protons and neutrons. The proton bas no magic numbers, and is an
odd proton – even neutron particle. The alpha is a very stable emitted particle, which
makes alpha emission a favorable process. The proton is not a stable emitted particle
and its formation does not encourage proton emission as a process.
Nuclear Transmutations
21.25
Protons and alpha particles are positively charged and must be moving very fast to
overcome electrostatic forces which would repel them from the target nucleus. Neutrons are electrically neutral and not repelled by the nucleus.
21.27
Analyze/Plan. Determine A and Z for the missing particle by conservation principles.
Find the appropriate symbol for the particle.
Solve.
(a)
(b)
(c)
(d)
(e)
21.29
Analyze/Plan. Follow the logic in Sample Exercise 21.5, paying attention to conservation
of A and Z.
Solve.
(a)
(b)
(c)
615
21 Nuclear Chemistry
Solutions to Exercises
Rates of Radioactive Decay
21.31
21.33
(a)
True. k = 0.693/t1/2. The decay rate constant, k, and half-life, t1/2 are inversely
related.
(b)
False. If X is not radioactive, it does not spontaneously decay and its half-life is
essentially infinity.
(c)
True. Changes in the amount of A would be measurable over the 40-year time
frame, while changes in the amount of X would be very small and difficult to detect.
Analyze/Plan. The half-life is 12.3 yr. Use t1/2 to calculate k and the mass remaining after 100 yr.
Solve. After 12.3 yr, one half-life, there are 1/2(56.2) = 28.1 mg remaining.
k = 0.693/ t 1 /2 = 0.693/12.3 yr = 0.056341 = 0.0563 yr
=
21.35
kt; lnNt – lnNo =
1
kt; lnNt + lnNo
lnN100 =
0.056341 yr 1(100 yr) + ln 56.2 mg
lnN100 =
5.6341 + 4.0289 =
1.60522; N100 = 0.201 mg
Analyze/Plan. We are given half-life of cobalt-60, and replacement time when the activity of the sample is 75% of the initial value. Consider the rate law for (first-order)
nuclear decay: ln(Nt/No) = −kt. Solve.
(a)
–1
k = 0.693 / t 1 /2 = 0.693/5.26 yr = 0.1317 = 0.132 yr ; Nt/No = 0.75
2.18 yr = 26.2 mo = 797 d. The source would have been replaced sometime in the
summer of 2008, probably in August.
21.37
(b)
Even at 75% of initial activity, decommissioned medical samples are quite radioactive. These so-called “sources” should be encased in a material such as lead
that readily absorbs gamma-rays. Since the half-life is relatively short, they can
be housed in a safe, remote storage facility until they are no longer active, and
then treated as traditional metal waste. This can be (and usually is) the same facility that stores spent fuel from nuclear reactors.
(a)
Analyze/Plan.
2 26
2 26
1 α particle is produced for each
Ra that decays. Calculate the mass of
Ra
remaining after 1.0 min, calculate by subtraction the mass that has decayed, and
use Avogadro’s number to get the number of
–1
Calculate k in min .
616
particles.
Solve.
21 Nuclear Chemistry
Solutions to Exercises
(don’t round here!)
[
is a number very close to 1. In this calculation, it is conveneint to ex–9
press the number as (1 – 4.1 × 10 )].
N t = 10.0 × 10
10.0 × 10
–3
–3
–9
g (1.00 – 4.1 × 10 ) The amount that decays is N o – N t :
g – [10.0 × 10
–3
–9
(1.00 – 4.1 × 10 )] = 10.0 × 10
–3
–9
g (4.1 × 10 )
= ~4.1 × 10
– 11
g Ra
(In terms of sig figs, [N o – N t ] is a very small number, found by subtracting two
numbers known only to two sig figs and one decimal place. At best, we can express the result as an order of magnitude.)
= 1.1 × 10
(b)
21.39
11
= ~ 10
11
α particles emitted in 5.0 min
Plan. The result from (a) is disintegrations/5.0 min. Change this to dis/s and ap10
ply the definition 1 Ci = 3.7 × 10 dis/s.
Analyze/Plan. Calculate k in yr
N t = 9.7/min/g. Solve.
–1
and solve Equation [21.19] for t. N o = 16.3/min/g,
k = 0.693/t 1 /2 = 0.693/5715 yr = 1.213 × 10
21.41
–4
= 1.21 × 10
–4
yr
–1
Analyze/Plan. Follow the procedure outlined in Sample Exercise 21.7. If the mass of
40
Ar is 4.2 times that of 40K, the original mass of 40K must have been 4.2 + 1 = 5.2 times
the amount of 40K present now. Solve.
9
k = 0.693/1.27 × 10 yr = 5.457 × 10
– 10
= 5.46 × 10
617
– 10
yr
–1
21 Nuclear Chemistry
Solutions to Exercises
Energy Changes
21.43
Analyze/Plan. Given an energy change, find the corresponding change in mass. Use Eq2
uation [21.22], E = mc .
Solve.
2
2
2
ΔE = c Δm; Δm = ΔE/c ; 1 J = kg - m /s
21.45
2
Analyze/Plan. Follow the logic in Sample Exercise 21.9. Given the mass of an 27Al atom,
subtract the mass of 13 electrons to get the mass of an 27Al nucleus. Calculate the mass
difference between the 27Al nucleus and the separate nucleons, convert this to energy
using Equation [21.22]. Use the molar mass of 27Al to calculate the energy required for
100 g of 27Al. Solve.
The mass of an electron is 5.485799 × 10−4 amu (inside back cover of the text). The mass
of a 27Al nucleus is then 26.9815386 amu −13(5.485799 × 10−4 amu ) = 26.9744071 amu.
Δm = 13(1.0072765 amu) + 14(1.0086649 amu) − 26.9744071 amu = 0.2414960 amu.
ΔE = (2.9979246 × 108 m/s)2 × 0.2414960 amu ×
= 3.604129 × 10−11 = 3.604129 × 10−11 J/27Al nucleus required
If the mass change for a single
mole of 27Al is 0.2414960 g.
27
Al nucleus is 0.2414960 amu, the mass change for 1
13
10
= 8.044234 × 10 J = 8.044234 × 10 kJ/ 1 0 0 g
21.47
27
Al
Analyze/Plan. In each case, calculate the mass defect (Δm), total nuclear binding energy
and then binding energy per nucleon.
Solve.
(a)
Δm = 6(1.0072765) + 6(1.0086649) – 11.996708 = 0.0989404 = 0.098940 amu
= 1.476604 × 10
– 11
= 1.4766 × 10
– 11
J
binding energy/nucleon = 1.476604 × 10
(b)
– 11
– 12
J/nucleon
Δm = 17(1.0072765) + 20(1.0086649) – 36.956576 = 0.3404225 = 0.340423 amu
= 5.080525 × 10
– 11
= 5.08053 × 10
– 11
J
binding energy/ nucleon = 5.080525 × 10
(c)
J/12 = 1.2305 × 10
– 11
J / 37 = 1.37312 × 10
– 12
J/nucleon
Calculate the nuclear mass by subtracting the electron mass from the atomic
618
21 Nuclear Chemistry
mass.
Solutions to Exercises
102.905504 amu – 45(5.485799 × 10
–4
amu) = 102.880818 amu
Δm = 45(1.0072765) + 58(1.0086649) – 102.880818 = 0.9491888/0.949189 amu
= 1.416586 × 10
– 10
= 1.41659 × 10
– 10
J
binding energy/nucleon = 1.416586 × 10
21.49
– 10
J/103 = 1.37533 × 10
– 12
J/nucleon
Analyze/Plan. Use Equation [21.22] to calculate the mass equivalence of the solar radiation.
Solve.
(a)
(b)
Analyze/Plan. Calculate the mass change in the given nuclear reaction, then a
2 35
conversion factor for g U to mass equivalent.
Solve.
Δm = 140.8833 + 91.9021 + 2(1.0086649) – 234.9935 = –0.19077 = –0.1908 amu
Converting from atoms to moles and amu to grams, it requires 1.000 mol or
235.0 g
2 35
U to produce energy equivalent to a change in mass of 0.1908 g.
5
2
5
0.10% of 1.714 × 10 kg is 1.714 × 10 kg = 1.714 × 10 g
(This is about 230 tons of
21.51
2 35
U per day.)
We can use Figure 21.13 to see that the binding energy per nucleon (which gives rise to
the mass defect) is greatest for nuclei of mass numbers around 50. Thus (a)
should possess the greatest mass defect per nucleon.
Effects and Uses of Radioisotopes
21.53
(a)
NaI is a good source of iodine, because it is a strong electrolyte and completely
dissociated into ions in aqueous solution. The I−(aq) are mobile and immediately
available for bio-uptake. They do not need to be digested or processed in the
body before uptake can occur. Also, iodine is a large percentage of the total mass
of NaI.
(b)
After ingestion, I−(aq) must enter the bloodstream, travel to the thyroid and then
be absorbed. This requires a finite amount of time. A Geiger counter placed near
the thyroid immediately after ingestion will register background, then gradually
increase in signal until the concentration of I−(aq) in the thyroid reaches a maximum. Then, over time, iodine-131 decays, and the signal decreases.
(c)
Analyze/Plan. The half-life of iodine-131 is 8.02 days. Use t1/2 to calculate the de-
619
21 Nuclear Chemistry
Solutions to Exercises
cay rate constant, k. Then solve Equation [21.19] for t. No = 0.12 (12% of ingested
iodine absorbed); Nt = 0.0001 (0.01% of the original ingested amount). Solve.
k = 0.693 t1/2 = 0.693/8.02 d = 0.086409 = 0.0864 d−1
ln(Nt/No) = −kt; t = −ln(Nt/No)/k
= 82.05 = 82 d
Check. Nt is given to 1 sig fig, so 8 × 101 (70 – 90) days is a more correct representation of the time frame for decay.
21.55
21.57
(a)
In a chain reaction, one neutron initiates a nuclear transformation that produces
more than one neutron. The product neutrons initiate more transformations, so
that the reaction is self-sustaining.
(b)
Critical mass is the mass of fissionable material required to sustain a chain reaction so that only one product neutron is effective at initiating a new
transformation.
(a)
(b)
21.59
21.61
(a)
The extremely high temperature is required to overcome the electrostatic charge
repulsions between the nuclei so that they come together to react.
(b)
The sun is not solid. No element or compound is solid at temperatures of
1,000,000 to 10,000,000 K.
Analyze/Plan. Hydroxyl radical is electrically neutral but has an unpaired electron,
−
•OH. Hydroxide is an anion, OH . Solve.
Hydrogen abstraction: RCOOH + •OH → RCOO• + H2O
RCOOH + OH− → RCOO− + H2O
Deprotonation:
Hydroxyl radical is more toxic to living systems, because it produces other radicals
when it reacts with molecules in the organism. This often starts a disruptive chain of
reactions, each producing a different free radical.
–
Hydroxide ion, OH , on the other hand, will be readily neutralized in the buffered cell
environment. Its most common reaction is ubiquitous and innocuous:
+
–
–
H + OH → H 2 O. The acid-base reactions of OH are usually much less disruptive to
the organism than the chain of redox reactions initiated by •OH radical.
21.63
Analyze/Plan. Use definitions of the various radiation units and conversion factors to
calculate the specified quantities. Pay particular attention to units.
Solve.
(a)
1 Ci = 3.7 × 10
10
disintegrations(dis)/s; 1 Bq = 1 dis/s
620
21 Nuclear Chemistry
Solutions to Exercises
–2
(b)
1 rad = 1 × 10 J/kg; 1 Gy = 1 J/kg = 100 rad. From part (a), the activity of the
8
source is 5.3 × 10 dis/s.
(c)
rem = rad (RBE); Sv = Gy (RBE) , where 1 Sv = 100 rem
2
3
3
mrem = 6.14 × 10 mrad (9.5) = 5.83 × 10 = 5.8 × 10 mrem (or 1.5 rem)
Sv = 6.14 × 10
–3
Gy (9.5) = 5.83 × 10
–2
= 5.8 × 10
–2
Sv
Additional Exercises
21.65
This corresponds to a reduction in mass number of (3 × 4 =) 12 and a reduction in
atomic number of (3 × 2 – 2) = 4. The stable nucleus is
(This is part of the se-
quence in Figure 21.5.)
21.69
(a)
(b)
(c)
(d)
21.73
The
bond of the acid and the
bond of the alcohol break in this reaction.
18
18
Initially, O is present in the
group of the alcohol. In order for O to end up
in the ester, the
bond of the alcohol must break. This requires that the
bond in the acid also breaks. The unlabeled O from the acid ends up in the H 2 O product.
21.75
Because of the relationship ΔE = Δmc , the mass defect (Δm) is directly related to the
binding energy (ΔE) of the nucleus.
2
7
Be: 4p, 3n; 4(1.0072765) + 3(1.0086649) = 7.05510 amu
Total mass defect = 7.0551 – 7.0147 = 0.0404 amu
0.0404 amu/7 nucleons = 5.77 × 10
–3
amu/nucleon
621
21 Nuclear Chemistry
9
Solutions to Exercises
Be: 4p, 5n; 4(1.0072765) + 5(1.0086649) = 9.07243 amu
Total mass defect = 9.0724 – 9.0100 = 0.06243 = 0.0624 amu
0.0624 amu/9 nucleons = 6.937 × 10
6.937 × 10
–3
–3
= 6.94 × 10
amu/nucleon × 1.4925 × 10
– 10
–3
amu/nucleon
J/amu = 1.035 × 10
– 12
= 1.04 × 10
– 12
J/nucleon
10
Be: 4p, 6n; 4(1.0072765) + 6(1.0086649) = 10.0811 amu
Total mass defect = 10.0811 – 10.0113 = 0.0698 amu
0.0698 amu/10 nucleons = 6.98 × 10
6.98 × 10
–3
–3
amu/nucleon
amu/nucleon × 1.4925 × 10
– 10
J/amu = 1.042 × 10
9
The binding energies/nucleon for Be and
higher.
– 12
= 1.04 × 10
10
Be are very similar; that for
– 12
J/nucleon
10
Be is slightly
Integrative Exercises
21.81
Calculate the amount of energy produced by the nuclear fusion reaction, the enthalpy
of combustion, ΔH°, of C 8 H 1 8 , and then the mass of C 8 H 1 8 required.
= 0.027583 = 0.02758 amu
= 4.11654 × 10
= 6.1495 × 10
11
– 12
= 4.117 × 10
8
– 12
1
J/4 H nuclei
1
J = 6.1 × 10 kJ produced by the fusion of 1.0 g H.
C8H18(l) + 25/2O2(g) → 8CO2(g) + 9H2O(g)
ΔH° = 8(−393.5 kJ) + 9(–241.82 kJ) – (–250.1 kJ) = −5074.3 kJ
14,000 kg C 8 H 1 8 (l) would have to be burned to produce the same amount of energy as fu1
sion of 1.0 g H.
622