Transcript General Chemistry

Chapter 20
Nuclear Chemistry
Glenn T. Seaborg
1912-1999.*
Transuranium
elements.
Pierre and Marie Curie.
1859-1906,* 1867-1934.**
Discovered radium;
defined “radioactivity.”
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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
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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
90
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Example of β emission: 131 I131 Xe+ 0 e
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Example of γ emission: 99m Tc
43
1
54
 Tc + hν
99
43
0
0
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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
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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
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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.
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Radioactive Series
238U
238
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Series
U
 T h
 Pa
 U 
 T h
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
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8
Nuclear Transmutations
Using Charged Particles - cyclotron
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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
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Rates of Radioactive Decay
5.0
2.5
1.25
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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)
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Rates of Radioactive Decay
14C
is created in upper atmosphere by bombardment
of nitrogen with cosmic neutrons:
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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.
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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(initial)
ln
 kt
cpm(current)
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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
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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
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40K
constitutes 0.01% of the natural abundance of
potassium in the earth’s crust.
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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
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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.
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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.
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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  m c2  (m)c 2
 1 kg 
 
 3.00 10 m/s  0.0046g 
 1000g 
2
kg
m
11
 4.11011


4
.
1

10
J
2
s
This is a very large number!!

8

2
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Nuclear Fission
• Splitting of heavy nuclei is exothermic for large mass
numbers.
• Consider a neutron bombarding a 235U nucleus:
1
0
n+ U
235
92
142
56
Ba + Kr +3 n
91
36
1
0
Te+ Zr +2 n
137
52
97
40
1
0
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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.
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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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