Exam 3 additional review session: Mon 6:30-8, 201 Reic
Chem 106
Thurs 28 April 2011
Chapt 23:
1) First-order decay kinetics
2) Radiocarbon dating
3) Nuclear fission
4) Review
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Decay kinetics same as 1st order chemical reactions, except
Molarity [ ] is not used. Decay is a property of each nucleus, and it
is totally independent of the local chemistry.
Call “C” = amount of radioactive atoms remaining UN-decayed.
If you have less, then
the rate of decay is
less too.
Replace “ΔC” with
“dC”, and integrate.
Put two ln terms
together, and raise
both sides to power
of e.
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C

 rate of disappearance of unstable atoms  kC
t
ln C  kt  ln Co
In exponential form :
C  C0e kt
2
Half-life () of a radioactive isotope
C  Co e  kt
Co
if C 
2
then we set t   , the half - life
Co
 Co e  k
2
1
 e  k
2
2  e k
ln 2  k
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ln 2

k
3
Practice problem
The decay constant for the spontaneous decay of
iodine-131 is 9.98 x 10-7 s-1. How much of a 0.50
g sample of this isotope remains after 3.0 days?
a.
b.
c.
d.
e.
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0.11 g
0.13 g
0.19 g
0.28 g
0.39 g
4
Practice problem
The decay constant for the spontaneous decay of
iodine-131 is 9.98 x 10-7 s-1. How much of a 0.50
g sample of this isotope remains after 3.0 days?
a.
b.
c.
d.
e.
0.11 g
0.13 g
0.19 g
0.28 g
0.39 g
Or, calculate half-life, and use it…
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If the half-life of an isotope = 20 min,
approximately what % will be present after 1 hour?
1.
2.
3.
4.
88%
33%
20%
13%
83%
13%
%
13
20
%
33
%
88
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%
3%
0%
6
(4)
After 20 minutes, 50% remaining, 50% gone
After 40 minutes, half of 50% = 25% remaining, 75% gone
After 60 minutes, half of 25% = 12.5% remaining, 88.5% gone
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Radiocarbon dating
Some common carbon isotopes
11
6
12
6
13
6
14
6
20 min
(stable)
(stable)
5780 y
-
98.9%
1.1%
C
Half-life
% natural
abundance
C
C
C
1 x 10-10 %
14
6
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147 N  10
C
8
  atoms 
 01n  11p  10e
14
7
14
7
1


N 0n 
N
14
6
14
7
N

 146 C  11p
 CO2
C  O2 
14
Carbon fixation by
plants and algae
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http://www.physics.arizona.edu/ams/index.htm
9
About 1/1012 carbon atoms on Earth is the 14C isotope.
Determined by the cosmic ray intensity reaching Earth. The atmosphere
itself has provided an essentially constant amount of N during the past
50,000 years.
How radioactive is 1-g of “new” natural carbon?
Calculate radioactivity of 1.00-g of pure 14C, then take a %.
1 molC 6.02 x10 23 atoms C

1.00 g C 
14.0 g C
1 mol C
 4.30 x1022 atoms C
After 5780 years, you would have ½ that, or 2.15 x 1022 atoms
And an equal number of disintegrations would have occurred.
2.15x10 22 disintegrations 1 y 1 da
1h



 7.08 x1012 dpm
5780 years
365 da 24 h 60 min
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7.08 x10 dpm 
12
1
@ 7 dpm
12
10
10
1 Curie = 1 Ci = 2.2 x 1012 disintegrations per min (dpm)
Pure C-14 carbon has activity of 3 Ci/g (hot!)
C-14 labelled organic compounds are normally
diluted 100- to 1000-fold with C-12 to reduce the
hazard and enable more accurate counting.
1 microcurie = 1 μCi = 2.2 x 106 disintegrations per min
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14C
produces a beta ray that is detectable by Geiger Counter
A beta is called “ionizing radiation” . When it hits Ar atoms inside the Geiger
counter, it knocks off electrons that are detected as a current pulse.
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Radiocarbon dating is limited to about 50,000 y (or ~ nine half-lives)
1-g of “new” C sample counted for 10 days = 10,080 counts
1-g of 52,020-year old C counted for 10 days = 10,080/29 = 20 counts

Background counts are removed using dual
detectors. -rays cause simultaneous counts in both
channels, which are discounted.
Beta particles are detected only on one side or the
other; these are counted.
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Nuclear Fission
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Nuclear Fission
Fission chain occurs in three steps:
1. Initiation. Reaction of a single atom starts
the chain (e.g., 235U + neutron)
2. Propagation.
236U
fission releases neutrons
that initiate other fissions
3. Termination. Capture of neutrons by “nonfissile” atoms (cadmium or boron).
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23.6 Homework: Nuclear Fission
No (no -1e0)
No (no 4He2)
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23.6 Homework: Nuclear Fission
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Basic reactor design
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Total mass = 9000 lb., release energy equiv to 40,000,000 lb TNT
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