Chem 401 Discussion Workshop #10
Name________________________
Nuclear Chemistry
1 MeV = 1.602 x 10-13 J
Speed of light: c = 2.998 x 108 m/s
t 1/2 =
Rate = kNt
0.693
k
ln
Nt
= − kt
N0
After decay amount = (original amount)(1/2)n
E = mc2 (mass in kg, speed of light in m/s)
Nt
= e − kt
N0
before decay amount = (end amount)(2)n
ΔE = (Δm)c2
I. Complete and balance the following nuclear reactions then classify each reaction as
alpha emission, beta emission, gamma emission, positron emission, electron capture, fission or fusion.
Reaction
Classification
218
84
Po → 42 He + _______ 214
82 Pb
Alpha emission
254
102
No → 42 He + _______ 250
100 Fm
Alpha emission
Tc + _______ 0 γ
Gamma emission
0
+1
e + _______ 207
83 Bi
Positron emission
0
−1
e + _______ 228
89 Ac
Beta emission
Tc →
99
43
207
84
Po →
228
88
Ra →
99m
43
2
1
0
H + 31H → 42 He + _______ 01 n
U
_______ →
4
2
He +
234
90
Alpha emission
Kr
_______ →
0
-1
e+
87
37
Rb
Beta emission
Fe →
55
25
Mn
Electron capture
238
92
87
36
0
-1
e
13
7
N _______ → 1e +
_______ +
55
26
0
235
92
14
6
Fusion
U + 01n → 2 01 n +
C→
0
−1
13
6
Th
C
137
52
Te _______ 97
40 Zr
e + _______ 147 N
Positron emission
Fission
Beta emission
D10-1
Chem 401 Discussion Workshop #10
Name________________________
II Problem Solving
1. The carbon-14 activity of a fiber from an Egyptian mummy shroud is 7.50 disintegrations
per minute per gram of carbon. How old is the fiber? The half-life of C-14 is 5730 years
and the activity of fresh C-14 is 15.2 disintegrations per minute per gram of carbon.
(Ans: 5.84 x 103 years)
k = 0.693 = 0.693 = 1.21 x 10-4/yr
t1/2
5730y
ln Nt = -kt
No
t = ln (Nt/No) = ln (7.50/15.2) =
-0.706
= 5.84 x 103 years
-4
-4
-k
-1.21 x 10 /y -1.21 x 10 /y
2. Radon in a tube is often used in cervical cancer therapy. The half-life of a radioactive
radon is 3.8 days. If 11 micrograms of radon are sealed in a tube, how many micrograms
remain after 21 days? (Ans: 0.24 μg)
OR
n = 21 = 5.53 half-lives
3.8
Nt/No = e-kt
Nt = No(e-kt)
After left = (original amount)(1/2)n
5.53
After left = (11μg)(1/2)
k = 0.693 = 0.693 = 0.182/d
t1/2
3.8d
-(0.182/d)(21d)
= 0.24μg
Nt = (11μg) e
-3.82
= (11 μg) e
Nt = (11 ug)(0.219) = 0.24μg
3. How long will it take before the 11μg of radon in the previous problem is reduced to
1.0 x 10-6 micrograms. (Ans: 89 days)
OR
amount left= (original amount)(1/2)n
-6
amt left
original amt
-8
k = 0.693 = 0.693 = 0.182/d
t1/2
3.8d
n
= 1.0 x 10 μg = (0.5)
11μg
ln Nt / No = -kt
n
9.09 x 10 = (0.5)
t = ln (Nt/No) = ln (1.0 x 10-6 μg /11 μg)
-k
-0.182/d
-8
ln 9.09 x 10 = (n) ln0.5
n = ln9.09 x 10-8 = -1.62 = 23.4
ln0.5
-0.693
t==
t = (23.4)(3.8) = 89 days
t = = 89 days
D10-2
ln9.09 x 10-8 =
-16.2
-0.182/d
-0.182/d
Chem 401 Discussion Workshop #10
Name________________________
4. A Se-75 source is producing 300 rem at a distance of 2.0 m. What distance is needed to
decrease the intensity of exposure to below 25 rem? (Ans: 6.9 m)
I A d 2B
=
I B d 2A
d=
d 2B =
I A 2 300rem
dA =
(2m) 2 = (12)(4) = 48m 2
25rem
IB
48m 2 =6.9m
5. A sample of Pr-147 is prepared and put in a scintillation counter. The initial counting rate
is 200/min. After 36 minutes, the rate is 25 counts/min. What is the half-life of Pr-147?
(Ans: 12 min)
ln Nt = -kt
No
k = ln(Nt/No) = ln(25/200) = -2.08 = 0.0578/min
-t
-36min
-36min
t1/2 = 0.693 =
0.693
= 12min
k
0.0578/min
6. Calculate the binding energy per nucleon in MeV of an atom of Pb-208 which has a mass
of 208.060 amu. The masses of the proton, neutron, and electron are 1.007277 amu,
1.008665 amu, and 0.0005486 amu respectively. (Ans: 7.50 MeV/nucleon)
protons
neutrons
electrons
82 (1.007277) = 82.596714
126 (1.008665) = 127.09179
82 (0.0005486) =
0.044985
209.733489 amu/atom or g/mol
mass defect = 209.733489
-208.060
1.673489 g/mol
ΔE = (Δm)c2 = 1.6735g x 1 kg x (2.998 x 108m/s)2= 1.504 x 1014J
mol
1000g
mol
1.504 x 1014J x
1 Mev
x
1 mol
= 1.559 x 103J
1 mol
1.602 x 10-13J 6.022 x 1023atoms
atom
1.559 x 103J x
atom
1 atom
= 7.496 MeV/nucleon
208 nucleons
D10-3