Chapter 20
Nuclear Chemistry
Glenn T. Seaborg
1912-1999.*
Transuranium
elements.
Pierre and Marie Curie.
1859-1906,* 1867-1934.**
Discovered radium;
defined “radioactivity.”
1
Basics: Radioactivity
Nuclear Equations
• Nucleons: particles in the nucleus:
p+: proton
n: neutron.
•
•
•
•
Mass number A: the number of p+ + n.
Atomic number Z: the number of p+.
Symbol: AzX; e.g. 146C is “carbon-14”
Isotopes: have the same number of p+ and different
numbers of n (and therefore, different mass)
• In nuclear equations, the total number of nucleons is
conserved:
238 U  234 Th + 4 He
92
90
2
2
Radioactivity
There are three types of radiation which we consider:
– -Radiation is the loss of 42He from the nucleus,
– -Radiation is the loss of an electron from the nucleus
(electrons represented as either 0-1e or 0-1β)
– -Radiation is the loss of high-energy photon from the
nucleus.
Example of α emission: 238 U  234 Th + 4 He
92
Example of β emission:
131
53
2
I Xe + e
Example of γ emission: 99m Tc
43
90
131
54
0
1
 Tc + hν
99
43
0
0
3
Radioactivity - Separating the types of radiation
(-)
(++)
α
4
2He
nucleus
Charge
2+
Mass(g)
6.64x10-24
Rel. mass
7,300
Rel. penetration
1
β
electron
19.11x10-28
1
100
γ _
high energy photons
0
0
0
10,000
4
Radioactivity
Complete the following nuclear reactions:
32
16
7
4
S+ n  p + ____
1
0
1
1
Be + e  _____
235
92
2
1
0
1
U + n  Xe + 2 n + ____
1
0
135
54
1
0
H+ H He + ____
98
42
2
1
3
2
Mo + H n + ____
2
1
1
0
5
Patterns of Nuclear Stability
Neutron-to-Proton Ratio
Neutron/proton ratio increases
as atoms become larger
Above 83Bi, all nuclei are unstable
and belt of stability ends.
6
Radioactive Series
238U
238
92
Series
U
 Th 
 Pa 
 U 
 Th
α
β
234
90
β
234
91
α
234
92
230
90

 Ra 
 Rn 
 Po 
 Pb
α
226
88
214
83


β
α
222
86
α
α
218
84
214
82
Bi 
 Po 
 Pb 
 Bi
β
214
84
α
210
82
β
210
83

 Po 
 Pb
β
210
84
α
206
82
Stable
7
8
Nuclear Transmutations
Using Charged Particles - cyclotron
9
Rates of Radioactive Decay
Calculations Based on Half-Life
• Radioactive decay is a first order process:
Rate = kN
• In radioactive decay the constant, k, is called the
decay constant, and N is number of nuclei.
• The rate of decay is called activity (disintegrations per
unit time).
• If N0 is the initial number of nuclei and Nt is the
number of nuclei at time t, then
No
ln
 kt
Nt
with half-life t1/2=0.693/k
10
Rates of Radioactive Decay
5.0
2.5
1.25
11
Rates of Radioactive Decay
Dating
• Carbon-14 is a radioactive isotope of carbon and is
used to determine the ages of organic compounds.
• We assume the ratio of 12C to 14C has been constant
over time.
• For us to detect 14C, the object must be less than
50,000 years old.
• The half-life of 14C is 5,730 years.
• Its abundance is <1% (the most common isotope of
carbon is C-12)
12
Rates of Radioactive Decay
14C
is created in upper atmosphere by bombardment
of nitrogen with cosmic neutrons:
14
7
14C
N + n  C+ p
1
0
14
6
1
1
itself decays to stable 14N by β emission:
14
6
C N + e
14
7
0
1
with a half-life of t1/2= 5715 yr
The amount of 14C in the environment is constant.
13
How does 14C dating work?
•When an organism dies, it no longer takes in carbon
compounds but its 14C continues to decay.
•5715 years (or one half-life) after the death of the organism (or
1 half-life), the relative amount of 14C is half that found in living
matter.
•11,430 years after the death of the organism (two half-lives),
the relative amount of 14C is ¼ that found in living matter.
First order rate law:
[14 C]0
ln 14
 kt
[ C]t
and t1/2 =.693/k
Since [14C] is proportional to radiation emitted (in counts/min or
cpm), the law become:
cpm(initia l)
ln
 kt
cpm(curren t)
14
Rates of Radioactive Decay-14C Dating
21.37 Artifact has 14C activity of 24.9 counts/m, compared
to current count of 32.5 counts/m for a standard. What is
the age of the artifact? Given: t1/2 = 5715 year.
k=.693/t1/2 = .693/5715 yr = 1.21x10-4 yr-1
Use first order expression
No
ln
 kt
Nt
 32.5 
4
-1
ln 
  (1.21x10 yr )t
 24.9 
2200 yr = t
15
40K
– 40Ar Dating
is a strange beast – with a half life of 1.3 x 109 yr,
it has two simultaneous modes of decay.
88.8% decays by electron-emission to give Ca-40:
40K
40
19
K
40
20
0
+
Ca 1 e
11.2% decays by electron-capture (of one of the
orbital electrons) to give Ar-40:
40
19
K + e Ar
0
1
40
18
40K
constitutes 0.01% of the natural abundance of
potassium in the earth’s crust.
16
Problem: a mineral is found with a 40K/40Ar mass ratio of 3/1.
How old is the mineral?
Answer: For the purposes of calculation, let’s assume a femtogram
total mass of 40K and 40Ar. The respective masses of K-40 and
Ar-40 will be 0.75 fg K-40 and 0.25 fg Ar-40. Since the mode of
decay giving Ar-40 is 11.2%, the mass of K-40 which decayed is
0.25/0.112 = 2.23 fg. Hence, the original mass of K-40 was
0.75 + 2.23 = 2.98 fg.
No
k = 0.693/t
ln
 kt
Nt
2.98
 (5.33 x 10-10 yr-1) t
ln
0.75
1/2
t = 2.58 x 109 yr
17
Half-lives
Isotope
Half-life
U-238
U-235
Th-232
K-40
C-14
Rn-222
Tc-99
4.5 x 109 yr
7.0 x 108 yr
1.4 x 1010 yr
1.3 x 109 yr
5715 yr
3.825 days
210,000 yr
Type of decay
Alpha
Alpha
Alpha
Beta-capture or -emission
Beta
Alpha
Beta
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Detection of Radioactivity
• Matter is ionized by radiation.
• Geiger counter determines the amount of ionization by
detecting an electric current.
• A thin window is penetrated by the radiation and causes the
ionization of Ar gas.
• The ionized gas carried a charge and so current is produced.
• The current pulse generated when the radiation enters is
amplified and counted.
19
Energy Changes in Nuclear Reactions
• Einstein showed that mass and energy are
proportional:
E = mc2
• If a system loses mass it loses energy (exothermic).
• If a system gains mass it gains energy (endothermic).
• Since c2 is a large number (8.99  1016 m2/s2) small
changes in mass cause large changes in energy.
• Mass and energy changed in nuclear reactions are
much greater than in chemical reactions.
20
Energy Changes in Nuclear Reactions
Consider reaction: 23892U  23490Th + 42He
– Molar masses:
238.0003 g  233.9942 g + 4.0015 g.
– The change in mass during reaction is
233.9942 g + 4.0015 g - 238.0003 g = -0.0046 g. =m
– The process is exothermic because the system has lost mass.
– To calculate the energy change per mole of 23892U:
E  mc 2  (m)c 2
 1 kg 
 
 3.00 10 m/s  0.0046 g 
 1000 g 
2
kg
m
11
 4.11011


4
.
1

10
J
2
s
This is a very large number!!

8

2
21
Nuclear Fission
• Splitting of heavy nuclei is exothermic for large mass
numbers.
• Consider a neutron bombarding a 235U nucleus:
1
0
n+
235
92
U
142
56
Ba + Kr +3 n
91
36
1
0
Te + Zr + 2 n
137
52
97
40
1
0
22
Nuclear Fusion
• Light nuclei can fuse to form heavier nuclei.
• Most reactions in the Sun are fusion.
Fusion processes occurring in sun:
1
1
H + H H + e
1
1
2
1
0
1
2
1
H + H  He
3
2
He +11 H  42 He + 01 e
1
1
3
2
Sun currently is about 75% H and 25% He.
Another 15b yrs or so to go before sun “burns” up
all its H.
23
Nuclear Fusion
• Fusion of tritium and deuterium requires about
40,000,000K:
2 H + 3 H  4 He + 1 n
1
1
2
0
• These temperatures can be achieved in a nuclear
bomb or a tokamak.
• A tokamak is a magnetic bottle: strong magnetic fields
contained a high temperature plasma so the plasma
does not come into contact with the walls. (No known
material can survive the temperatures for fusion.)
• To date, about 3,000,000 K has been achieved in a
tokamak.
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