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Ch. 21 Kimball Spring 2019
Discussion Worksheet - Chapter 21
Nuclear Chemistry KEY
Dr Cabirac Lecture
1. The table below is a summary of different modes of nuclear decay. Fill in the changes in atomic
number (Z), number of neutrons (N), and mass number (A) in each case. Use “+” for increase,
2. Complete the following nuclear equations and identify X in each case:
A. X = 𝑁𝑎
11
23
B. X = 𝑝
1
1
C. X = 𝑛
0
1
D. X = 𝐶𝑎
20
40
E. 𝐾 → 𝐴𝑟 + 𝑋
18
38
19
38 X = 𝛽
+1
0
F. 𝐹𝑒 + 𝑋 → 𝑀𝑛
25
55
26
55 X = 𝑒
−1
0
G. 𝑈
92
235 + 𝑛 → 𝐵𝑟
35
87 + 3 𝑛
0
1 + 𝑋
0
1X = 𝐿𝑎
57
145
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Ch. 21 Kimball Spring 2019
3. Calculate the nuclear binding energy (inJ) and the binding energy per nucleon of 35Cl (34.95952
amu). (mass of proton = 1.007825amu and mass of neutron = 1.008668amu)
Find Δm first: Δm = -0.32948 amu
Now find ΔE: ΔE = Δmc2
𝛥𝐸=−𝟒.𝟖𝒙𝟏𝟎−𝟏𝟏𝑱
This is the binding energy for one 35Cl atom
A nucleon is simply a nuclear particle (so a proton or a neutron), so 35Cl has 35 nucleons. Finding the
binding energy per nucleon is:
−𝟒.𝟖𝒙𝟏𝟎−𝟏𝟏 𝑱𝑪𝒍
𝟏𝟕
𝟑𝟓 (𝟏 𝑪𝒍
𝟏𝟕
𝟑𝟓
𝟑𝟓 𝒏𝒖𝒄𝒍𝒆𝒐𝒏𝒔)=−𝟏.𝟒𝒙𝟏𝟎−𝟏𝟐𝑱
4. Fill in the blanks in the following radioactive decay series:
5. In the below, predict which one you would expect to be radioactive:
The principal factor that determines nucleus stability is the neutron to proton ratio (n/p)
a. 𝑁𝑒
10
20 𝑜𝑟 𝑁𝑒
10
17
𝑁𝑒
10
20 𝑛
𝑝=10
10 =1 ℎ𝑖𝑔ℎ𝑙𝑦 𝑠𝑡𝑎𝑏𝑙𝑒 𝑵𝒆
𝟏𝟎
𝟏𝟕 𝒏
𝒑=𝟕
𝟏𝟎 =𝟎.𝟕<𝟏 𝒖𝒏𝒔𝒕𝒂𝒃𝒍𝒆,𝒓𝒂𝒅𝒊𝒐𝒂𝒄𝒕𝒊𝒗𝒆
will emit a positron to convert a proton to a neutron
b. 𝑀𝑜 𝑜𝑟 𝑇𝑐
43
92
42
95
𝑀𝑜
42
95 𝑛
𝑝=53
42 =1.26 𝑻𝒄
𝟒𝟑
𝟗𝟐 𝒏
𝒑=𝟒𝟗
𝟒𝟑 =𝟏.𝟏𝟑
The n/p ratio for 𝑀𝑜
42
95 deviates more from 1 than for 𝑻𝒄
𝟒𝟑
𝟗𝟐 so 𝑀𝑜
42
95 is expected to be radioactive.
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Ch. 21 Kimball Spring 2019
c. 𝐵𝑖 𝑜𝑟 𝐶𝑚
96
242
83
209
𝐵𝑖
83
209 𝑛
𝑝=126
83 =1.51 𝑪𝒎
𝟗𝟔
𝟐𝟒𝟐 𝒏
𝒑=𝟏𝟒𝟔
𝟗𝟔 =𝟏.𝟓𝟐
The n/p ratio for both isotopes are greater than 1.5, so both are expected to be radioactive.
6. A radioactive substance undergoes decay as shown below. Calculate the first-order decay constant
and the half-life of the reaction.
Chose any 2 experiments and use first order rate law to find k (rate constant)
𝑙𝑛𝑁𝑡
𝑁0=−𝑘𝑡 k = 0.251 d-1
(mass can replace # particles)
Now calculate half-life:
𝒕𝟏/𝟐 =𝟐.𝟕𝟔 𝒅𝒂𝒚𝒔
7. The radioactive decay of Tl-206 has a half-life of 4.20 min. Starting with 5.00 x 1022 atoms of Tl-206,
calculate the number of atoms left after 42.0 min.
First find half-life to find rate constant k:
𝑘=𝟎.𝟏𝟔𝟓 𝒎𝒊𝒏−𝟏
Now use the first order integrated rate law to find atoms left after 42.0 min:
𝑙𝑛𝑁𝑡
𝑁0=−𝑘𝑡 Nt = 4.89 x 1019 atoms
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Ch. 21 Kimball Spring 2019
8. Consider the decay series: A → B → C → D
where A, B, and C are radioactive isotopes with half-lives if 4.50 s, 15.0 days, and 1.00 s, respectively,
and D is nonradioactive. Starting with 1.00 mole of A and none of B, C, or D, calculate the number of
moles of A, B, C, and D left after 30 days.
[A] after thirty days is 0 moles.
0.25 mol of B remaining.
[C] after 30 days is 0 moles.
D produced, 0.75 mol
9. Write balanced nuclear equations for the following reactions and identify X:
a) 𝑆𝑒
34
80 + 𝐻
1
2 → 𝑆𝑒
34
81 + 𝑝
1
1
b) 𝐵𝑒
4
9+ 𝐻
1
2 → 𝐿𝑖
3
9+ 2 𝑝
1
1
c) 𝐵
5
10 + 𝑛
0
1 → 𝐿𝑖
3
7 + 𝛼
2
4
10. The diagram below shows part of the thorium decay series. Write a nuclear equation for each step of
the decay. Use the AZX symbol for each isotope.
For the first two steps, the atomic number increases by one and the number of neutrons decreases by one, so this
indicates beta decay:
𝑇ℎ
90
234 →𝑃𝑎
91
234 + 𝛽
−1
0
𝑃𝑎
91
234 → 𝑈
92
234 + 𝛽
−1
0
In the next three steps the atomic number and the number of neutrons both decrease by 2, so alpha decay
𝑈
92
234 → 𝑇ℎ
90
230 + 𝛼
2
4
𝑇ℎ
90
230 →𝑅𝑎
88
226 + 𝛼
2
4
𝑅𝑎
88
226 →𝑅𝑛
86
222 + 𝛼
2
4
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Ch. 21 Kimball Spring 2019
11. In 2006, an ex-KGB agent was murdered in London and the cause of death was found to be poisoning
with the radioactive isotope 210Po, which was added to his food and drink.
a) 210Po is prepared by bombarding 209Bi with neutrons. Write and equation for this reaction.
𝐵𝑖
83
209 + 𝑛
0
1→ 𝐵𝑖
83
210
𝐵𝑖
83
210 → 𝑃𝑜
94
210 + 𝛽
−1
0
b) Who discovered the element polonium? Marie Curie
c) The half-life of 210Po is 138 days. It decays with the emission of an 𝛂 particle. Write an
equation for the decay process.
𝑃𝑜
84
210 → 𝑃𝑏
82
206 + 𝛼
2
4
d) Calculate the energy of an emitted 𝛂 particle. Assume both the parent and the daughter nuclei
to have zero kinetic energy. The atomic masses are 206Po (209.98285 amu), 206Pb (205.97444
amu), and 𝛂 (4.00150 amu).
Δm = mPb206 + m𝜶 - mPb210
= 205.97444 amu + 400150 amu - 209.98285 amu = -0.00691 amu
−0.00691 𝑎𝑚𝑢 ( 1 𝑔
6.022 𝑥 1023𝑎𝑚𝑢)( 1 𝑘𝑔
1000 𝑔)= −1.14746 𝑥 10−29 kg
ΔE = Δmc2
= (-1.14746 x 10-29 kg)(3.0 x 108 m/s)2
= -1.03 x 10-12 kg·m2/s2
So 1.03 x 10-12 J of kinetic energy is left over for the alpha particle
e) Ingestion of 1 𝜇g of 210Po can be fatal. What is the total energy released by the decay of this
quantity of 210Po?
We will assume all the 210Po decays
1 𝜇𝑔 ( 1 𝑔
1 𝑥 106𝜇𝑔)(1 𝑚𝑜𝑙
210 𝑔)(6.022 𝑥 1023𝑎𝑡𝑜𝑚𝑠
1 𝑚𝑜𝑙 )=2.87 𝑥 1015 𝑎𝑡𝑜𝑚𝑠
2.87 𝑥 1015 𝑎𝑡𝑜𝑚𝑠 ( 1 𝛼
1 𝑎𝑡𝑜𝑚)(1.03 𝑥 10−12𝐽
1 𝛼 )=2.961 𝑥 103𝐽=𝟐.𝟗𝟔 𝒌𝑱
This energy loss will take many years because the half-life is long (138 days)
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