QUIZ CHAPTER #19
ANSWER KEY
Student’s Name
1. A sodium nucleus, 23
11Na, is bombarded with alpha particles to form a radioactive nuclide. This
product nuclide can decay by several routes. Which of the following sets of products does not
represent a potential route of decay of the product nuclide?
27
13Al
a)
𝟐𝟏
𝟏𝟎𝐍𝐞
+ 𝟐𝟏𝐇
23
11Na
b)
26
Mg
12
+ 11H
When 27
13Al undergoes nuclear decay, the sum of the mass numbers
c)
23
11Na
+ 42He
of the products must be 27, the sum of the atomic numbers of the
d)
26
13Al
+ 10n
products must be 13.
+ 42He →
(product nuclide)
2. A radioactive isotope decays by  emission followed by two  emissions. What is the overall
change in the mass number and atomic number of the original isotope?
a) The mass number decreases by 4 and the atomic number is unchanged.
b) The mass number increases by 4 and the atomic number increases by 2.
c) The mass number increases by 4 and the atomic number increases by 4.
d) The mass number decreases by 2 and the atomic number decreases by 2.
208
3. In the radioactive decay series of plutonium-244, 244
94Pu, to form lead-208, 82Pb, how many
alpha particles and how many beta particles are emitted per plutonium atom?
a) 6 alpha particles and 6 beta particles
b) 3 alpha particles and 3 beta particles
c) 9 alpha particles and 6 beta particles
d) 12 alpha particles and 8 beta particles
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4. Consider a certain type of nucleus that has a rate constant of 2.5610-2 min-1. Calculate the
time required for the sample to decay to one-fourth of its initial value.
a) 27.1 min
b) 54.1 min
c) 33.8 min
d) 2.56 min
𝑙𝑛
𝑁𝑡
= −𝑘𝑡
𝑁0
1
𝑁𝑡
1
1
𝑡 = − ∙ 𝑙𝑛
=−
∙
𝑙𝑛
= 54.1 𝑚𝑖𝑛
𝑘
𝑁0
2.56 × 10−2 𝑚𝑖𝑛−1
4
5. What is the energy change for the following nuclear bombardment reaction?
40
43
19K(𝛼, 𝑛) 21Sc
Particle
Mass (amu)
1
0n
1.008665
4
2He
4.00260
40
19K
39.96400
43
21Sc
42.96115
c = 3.00108 m/s, 1 amu = 1.6605410–27 kg, 1 MeV = 1.60210–13 J
a) 7.47103 MeV
40
19K
1
+ 42He → 43
21Sc + 0n
b) 2.80103 MeV
40
4
1
∆m = m( 43
21Sc) + m( 0n) − m( 19K) − m( 2He)
c) 8.20104 MeV
∆m = 42.96115 + 1.008665 − 39.96400 − 4.00260
d) 3.00 MeV
∆m = 0.0032𝟏5 amu
1.66054 × 10−27 kg
∆m = 0.0032𝟏5 amu ×
= 5.3𝟑8 × 10−30 kg
1 amu
m 2
∆E = ∆m × c 2 = 5.3𝟑8 × 10−30 kg × (3.00 × 108 ) = 4.8𝟎4 × 10−13 J
s
1 MeV
∆E = 4.8𝟎4 × 10−13 J ×
= 3.00 MeV
1.602 × 10−13 J
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