Chemistry 1000
Problem Set #11: Chapter 23
Answers to Practice Problems
1.
<text omitted>
(a)
There is only one stable odd-odd nucleus with a mass greater than 20 µ. It is 158La.
Name the other four elements that have a stable isotope with both an odd number of
protons and an odd number of neutrons. Give the symbol (including mass and atomic
numbers) for each of these four stable isotopes.
2
6
(b)
H (hydrogen) = 1 proton + 1 neutron
10
Li (lithium) = 3 protons + 3 neutrons
14
B (boron) = 5 protons + 5 neutrons
N (nitrogen) = 7 protons + 7 neutrons
Give the symbol (including mass and atomic numbers) for one stable isotope falling into
each of the other three categories.
There are many possible answers to this question. The answers below are just examples.
e.g.
even/even:
4
He = 2 protons + 2 neutrons (or 12C, 16O, 20Ne, 40Ca, etc.)
even/odd:
9
Be = 4 protons + 5 neutrons (only stable isotope of Be)
odd/even:
19
F = 9 protons + 10 neutrons (only stable isotope of F)
2.
The following isotopes have stable nuclei: 4He, 40Ca, 93Nb, 112Sn, 159Tb and 204Pb.
i
Use the above information to predict whether the isotopes listed below would be
stable or unstable.
ii
For each unstable isotope, predict which mode(s) of decay the nucleus would be
most likely to undergo.
(a)
95
Mo
(b)
8
(c)
47
(d)
B
K
37
K
42 protons
53 neutrons
N/Z = 1.26
stable (N/Z should be ~1.27 since, of the stable isotopes listed,
closest in mass to 95Mo)
93
Nb is
5 protons
3 neutrons
N/Z = 0.6
not stable (N/Z is less than 1)
electron capture or positron emission (too many protons)
(actually undergoes electron capture)
19 protons
28 neutrons
N/Z = 1.47
not stable (N/Z should be ~1 since, of the stable isotopes listed,
closest in mass to 40Ca)
β-emission (too many neutrons)
19 protons
18 neutrons
N/Z = 0.95
not stable (N/Z is less than 1)
electron capture or positron emission (too many protons)
(actually undergoes electron capture)
47
K is
3.
(a)
Write a balanced equation for each of the following nuclear reactions.
electron capture by 26Al
26
13
(b)
Al
+
0
-1
26
e
12
β-emission by 208Au
208
208
Au
79
(c)
Hg
80
a nuclear transmutation reaction in which
and one neutron
253
99
(d)
Mg
Es
4
+
2
+
253
0
β
-1
4
Es reacts with He to produce a large nucleus
256
He
101
Md
+
1
0
n
fission of 236U to give 138Xe and 95Sr
236
92
138
U
54
+
95
38
Sr
+
1
3 n
0
4.
18
(a)
Calculate the nuclear binding energy per nucleon for each of these isotopes of fluorine.
18
A = 18
F
F has a mass of 18.000938µ.
Z=9
19
Xe
F has a mass of 18.99840321µ.
N=9
∆m = [Z mp + Z mp + N mn] – MF-18
= [(9)(1.0072765 g/mol) + (9)(1.0086649 g/mol) + (9)(0.00054688 g/mol)]
- 18.000938 g/mol
∆m = 0.1474565 g/mol
∆E = ∆mc2
= (0.1474565 g/mol)(2.9979 x 108 m/s)2 ×
1 kg ×
1000 g
∆E = 1.3253 × 1010 kJ/mol
Eb = ∆E = 1.3253 × 1010 kJ/mol = 7.3625 × 108 kJ/mol
A
18
Similarly,
19
F
A = 19
Z=9
∆m = 0.158656 g/mol
∆E = 1.4259 × 1010 kJ/mol
Eb = 7.5048 × 108 kJ/mol
N = 10
1J
× 1 kJ .
2 2
1 kgm /s
1000 J
(b)
Which of these two isotopes is the only stable isotope of fluorine? Give two good
reasons for your choice.
19
F
1.
2.
5.
19
F has a higher nuclear binding energy than 18F, so it is a more stable nucleus.
The mass of fluorine listed on the periodic table is 18.998 g/mol. This matches
the mass of 19F.
Note that this is question 37 from Kotz, Treichel and Weaver.1
A sample of wood from a Thracian chariot found in an excavation in Bulgaria has a 14C
activity of 11.2 dpm/g. Estimate the age of the chariot and the year in which it was made.
(t½ for 14C is 5.73 × 103 years, and the activity of 14C in living material is 14.0 dpm/g).
ln(A/A0) = -kt
where k = ln(2)
t½
ln(A/A0) = -t × ln(2)
t½
t = -t½ × ln(A/A0) = -(5.73 × 103 y) × ln[(11.2 dpm·g-1)/(14.0 dpm·g-1)] = 1.84 × 103 y
ln(2)
ln(2)
The chariot is approx. 1845 years old, so it was made sometime around the year 160.
6.
Jerri prepared a sample of 93Tc and measured its activity to be 46.1 dps. After 35
minutes, she measures the activity of the sample again, and it has decreased to 39.8 dps.
(a)
93
Tc decays to give 93Mo. What type of radiation is emitted by this reaction?
There are two reasonable answers to this question:
(b)
1.
positron (93Mo has the same mass number but the atomic number is one less) and
gamma (emitted by all nuclear reactions), or
2.
gamma radiation only, if reaction occurs by electron capture (this is what actually
happens)
Assuming that the amount radiation produced by 93Mo is negligible, calculate the half life
of 93Tc.
k = ln(2)
t½
where ln(A/A0) = -kt
therefore k = -ln(A/A0)
t
-ln(A/A0) = ln(2)
t
t½
t½ = - t × ln(2) =
-(35 min) × ln(2)
= 1.7 × 102 min = 2.8 hours
ln(A/A0)
ln[(39.8 dps)/(46.1 dps)]
1
Kotz, J.C., Treichel, P.M. and Weaver, G.C. (2006) Chemistry and Chemical Reactivity 6th edition, Thomson
Brooks/Cole, p.1143.