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: A z X ; e.g. 14 6 C 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 92 U
234 90 Th + 4 2 He 2
Radioactivity
There are three
– –
types of radiation which we consider:
Radiation is the loss of 4 2 He from the nucleus,
Radiation is the loss of an electron from the nucleus
–
(electrons represented as either 0 -1 e or 0 -1 β)
Radiation is the loss of high-energy photon from the nucleus.
Example of α emission : Example of β emission :
238 92 U 234 90 Th + 131 53 I 131 54 Xe + 0 1 e 4 2 He
Example of γ emission :
43 Tc 99 43 Tc + 0 0 hν
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Radioactivity -
Separating the types of radiation (-) (++) α β γ _ 4 2 He nucleus electron high energy photons Charge 2+ 1 Mass(g) 6.64x10
-24 9.11x10
-28 0 0 Rel. mass 7,300 1 0 Rel. penetration 1 100 10,000 4
Radioactivity
Complete the following nuclear reactions :
32 16 S + 0 1 n 1 1 p + ____ 7 4
Be
+ 0 1
e
_____
235 92 U + 0 1 n 135 54 Xe + 2 0 1 n + ____ 2 1
H
+ 2 1
H
2 3
He
+
____
98 42 Mo + 2 1 H 0 1 n + ____
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Patterns of Nuclear Stability
Neutron-to-Proton Ratio
Neutron/proton ratio increases as atoms become larger Above 83 Bi, all nuclei are unstable and belt of stability ends.
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Radioactive Series
238
U Series
238 U 92 α α 226 Ra 88 234 90 Th α β 222 Rn 86 234 91 Pa α β 234 U α 92 218 Po 84 α 230 214 Pb 82 90 Th 214 83 Bi β β 210 Po 84 α 214 Po 84 206 82 Pb α 210 82 Pb β 210 83 Bi
Stable 7
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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 N 0 is the initial number of nuclei and N
t
number of nuclei at time t, then is the
ln
N o N t
kt
with half-life t 1/2 =0.693/k 10
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 12 C to 14 C has been constant over time.
For us to detect 14 C, the object must be less than 50,000 years old.
The half-life of 14 C is 5,730 years.
Its abundance is <1% (the most common isotope of carbon is C-12) 12
Rates of Radioactive Decay
14 C is created in upper atmosphere by bombardment of nitrogen with cosmic neutrons:
14 7 N + 0 1 n 14 6 C + 1 1 p
14 C itself decays to stable 14 N by β emission:
14 6 C 14 7 N + 0 1 e
with a half-life of t 1/2 = 5715 yr The amount of 14 C in the environment is constant. 13
How does 14 C dating work?
•
When an organism dies, it no longer takes in carbon compounds but its 14 C continues to decay.
•
5715 years (or one half-life) after the death of the organism (or 1 half-life), the relative amount of 14 C is half that found in living matter.
•
11,430 years after the death of the organism (two half-lives), the relative amount of 14 C is ¼ that found in living matter.
First order rate law:
ln [ 14 C] 0 [ 14 C] t kt
and t 1/2 =.693/k Since [ 14 C] is proportional to radiation emitted (in counts/min or cpm), the law become:
cpm(initia ln cpm(curren l) t) kt
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Rates of Radioactive Decay 14 C Dating
21.37 Artifact has 14 C 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: t 1/2 = 5715 year.
k=.693/t 1/2 = .693/5715 yr = 1.21x10
-4 yr -1 Use first order expression
ln N o N t kt ln 32.5
(1.21x10
4 yr 1 )t 24 .
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2200 yr = t 15
40 K – 40 Ar Dating
40 K is a strange beast – with a half life of 1.3 x 10 9 yr, it has two simultaneous modes of decay.
88.8% decays by electron-emission to give Ca-40:
40 19 K 40 20 Ca + 0 1 e
11.2% decays by electron-capture (of one of the orbital electrons) to give Ar-40:
40 19 K + 0 1 e 40 18 Ar
40 K 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 40 K/ 40 Ar mass ratio of 3/1.
How old is the mineral?
Answer: For the purposes of calculation, let’s assume a femtogram total mass of 40 K and 40 Ar. 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.
ln N o N t kt k = 0.693/t 1/2 ln 2.98
(5.33 x 10 -10 0.75
t = 2.58 x 10 9 yr
yr -1 ) t
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Half-lives
Isotope Half-life Type of decay U-238 4.5 x 10 9 U-235 7.0 x 10 8 yr Alpha yr Alpha Th-232 1.4 x 10 10 yr Alpha K-40 1.3 x 10 9 yr Beta-capture or -emission C-14 5715 yr Beta Rn-222 3.825 days Alpha Tc-99 210,000 yr Beta 18
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 = mc
2 If a system
loses
mass it
loses
energy (exothermic).
If a system
gains
Since c 2 mass it
gains
energy (endothermic).
is a large number (8.99
10 16 m 2 /s 2 ) 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 : 238 92 U
Molar masses: 238.0003 g
234 90 Th + 4 2 He 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 238 92 U:
E
mc
2 (
m
)
c
2 3 .
00 10 8 m/s 2 0 .
0046 g 1 kg 1000 g 4 .
1 10 11 kg m 2 4 .
1 10 11 J s 2
This is a very large number!!
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• •
Nuclear Fission
Splitting of heavy nuclei is exothermic for large mass numbers.
Consider a neutron bombarding a 235 U nucleus:
0 1 n + 235 U 92 142 Ba 56 + 91 Kr 36 + 3 0 1 n 137 52 Te + 97 40 Zr + 2 0 1 n
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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 + 1 1 H 2 1 H + 0 1 e 2 1 H + 1 1 H 2 3 He 2 3 He + 1 1 H 4 2 He + 0 1 e
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 1 H + 3 1 H
4 2 He + 1 0 n 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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