Chemist’s Daily
Chapter 22-1
1. Write balanced nuclear equations for the following:
11
6
214
83
237
93
C
Bi
decays by positron emission:
11
6
C
decays by emission of a beta particle
0
→
1
214
83
Np
2.
195
79
39
19
237
produces an alpha particle
93
Np
Bi
→
e
→
4
2
11
+
0
-1
5
e
He
Supply the missing particle:
Au
K
0
+
-1
→
38
19
e
K
→
+
195
78
1
0
Pt
neutron
Electron capture
Neutron emission
B
214
+
+
233
91
84
Pa
Po
Chemist’s Daily
Chapter 22-2
1.
Technetium-99m is used to form pictures of internal organs and is often used to assess
heart damage.
emission.
The "m" indicates an excited nuclear state that decays by gamma
The rate constant for decay of technetium-99m is known to be 1.16 x 10
-1
What is the half-life?
t1/2 = ln 2 / k = ln 2 / 1.16 x 10
2.
-1
h
-1
=
5.96 hrs
The half-life of molybdenum-99 is 67.0 hours.
molybdenum-99 is left after 335 hours?
t1/2 = ln 2 / k
k = ln 2 / t1/2 = ln 2 / 67 hrs = 0.0103 h
ln (No / N) = kt
-1
h .
How much of a 1.00 mg sample of
-1
= (0.0103 h ) (335 h) =
-1
ln (No / N) = 3.47
3.47
(No / N) = 32.0
N = No / 32.0
= 1.00 mg / 32.0 = 0.0312 mg
3. The remnants of an ancient fire in a cave in Africa showed a carbon-14 decay rate of
3.1 counts per minute per gram of carbon.
Assuming that the decay rate of C-14 in
freshly cut wood is 13.6 counts per minute per gram of carbon, calculate the age of the
remnants.
The half-life of C-14 is 5715 years.
Initial activity = Ao = 13.6 counts per min per gram
Current activity = A = 3.1 counts per min per gram
t1/2 = ln 2 / k
= ln 2 / 5715 years = 1.21 x 10
ln (Ao / A) = kt
=
-4
ln (13.6 / 3.1) = (1.21 x 10
t = 12, 200 years old
yr
-1
-4
yr ) t
-1