Nuclear Chemistry
I. Radioactivity.
A. Review Isotopic Notation.
1. Proton, neutron and electron model.
a. Very approximate and incorrect model. However, it can be used to keep tract of, but not
explain, radioactive decay and nuclear stability.
Mass
Particle
Symbol
1
1H
1
0n
0
-1 e
Proton
Neutron
Electron
Charge
kg
amu (u)
Coulombs
Electron Units
1.67252x10-27 1.007276
1.6022x10-19
+1
1.67496x10-27 1.008665
0
0
9.1095x10-31
0.000549
-1.6022x10-19
-1
2. Protons and neutrons comprise the nucleus = nucleons.
3. Atomic number = Z = Charge on nucleus in electron charge units = number of protons in the
nucleus.
Mass number = A = Total number of nucleons in nucleus = mass in amu rounded to nearest
whole number = number of protons + number of neutrons.
4. Isotopic notations.
A
Z (symbol)
{68 protons
neutrons } in nucleus + 6 electrons outside of nucleus
92 protons
238
92 U {146 neutrons } in nucleus + 92 electrons outside of nucleus.
14
6
Examples:
C
B. Radioactive Decay.
1. Neutrons are important in stabilizing the nucleus. In general, the greater the number of protons in
the nucleus, the greater the number of neutrons required to stabilize it.
a. For the lighter elements, up to Z = 20 (Ca) the stable, nonradioactive, nuclei have neutron to
proton ratios, n/p = 1.
b. As Z increases the n/p ratio increases to a value of approximately 2.5 for
209
83
Bi.
c. All isotopes with Z > 83 are radioactive.
2. Isotopes outside this "belt of stability" will undergo radioactive decay and change there n/p ratio
until a stable isotope is formed.
1
Proton/Neutron Ratio and Stability
2
3. Types of radioactive decay.
a. Alpha particle decay.
4
4
1) Alpha particle is a helium-4 nucleus. Symbol is 2 He or 2 α.
2) In alpha particle decay, Z goes down by 2 and A goes down by 4. Alpha decay is the only
way a radioactive isotope can decrease its mass number.
238
92
3) Example:
4
234
U ----> 2 He + 90 Th
Note in a balanced nuclear equation the sum of the Z's and the sum of the A's on both sides
of the equation must be the same.
4) Very important in the decay of the heavier elements
b. Beta decay.
0
1) Beta particle is an electron that is emitted from the nucleus. Symbol is -1 e or β-.
2) In beta decay, Z goes up by one and A remains unchanged. A neutron is converted into a
1
1
0
proton (0 n ---> 1 H + -1 e ).
14
6
3) Example;
0
14
C -----> -1 e + 7 N
c. Gamma decay.
0
1) Gamma ray is a very high energy photon. Symbol 0 γ or γ .
2) In gamma decay, neither Z nor A changes. Gamma radiation accompanies alpha and beta
decay. It is the way a nucleus can lose energy.
3) Example: 60Co* ---> 60Co + γ
where the asterisk (*) denotes an excited state of the
nucleus.
d. Positron emission.
1) Positron is a particle that is identical to an electron except that it has a positive
0
charge. It is the antimatter equivalent of the electron. Symbol 1 e or β+.
2) In positron emission, Z goes down by one and A remains unchanged. A proton is converted
1
1
0
into a neutron ( 1 H ---> 0 n + 1 e).
22
22
0
3) Example: 11 Na ----> 10 Ne + 1 e
4) Positrons and electron will annihilate one another. The annihilation reaction is:
0
-1
0
e + 1 e ---> 2 γ
e. Electron capture or K capture.
1) In K capture an electron from the lowest energy shell of the atom, the K shell, is captured by
the nucleus.
2) In K capture, Z goes down by one and A remains unchanged. A proton is converted into a
1
0
1
neutron ( 1 H + -1 e ---> 0 n).
3
44
0
44
3) Example: 22 Ti + -1 e ----> 21 Sc
4. The most common forms of radioactive decay found in naturally occurring radioactive isotopes are
alpha, beta, and gamma decay. K capture is also found but is not common.
Positron emission is not found in naturally occurring isotopes.
C. Naturally Occurring Radioactive Isotopes.
1. Heavier atoms with Z>83 are all radioactive. Most of the heavier isotopes are members of one of
three series.
a. The Uranium-radium Series.
The uranium-radium series
4n + 2 series
240
235
230
A
225
220
215
210
205
80
82
84
86
88
90
92
94
Z
1) The only way that the mass number changes is through alpha decay, therefore the mass
numbers of all the isotopes will differ by multiples of four.
238
The heaviest isotope in the series is 92 U. Its mass number (238) is divisible by four with
two left over, that is 238 = 4 (59)+2. Therefore, all daughter isotopes will have mass
numbers that are divisible by four with two left over. This series is also referred as the 4n+2
series.
2) Initial steps and half-lives.
238
234
4
92 U ------> 90 Th + 2 He
234
234
0
90 Th -------> 91 Pa + -1 e
234
234
0
91 Pa -------> 92 U + -1 e
234
230
4
92 U -------> 90 Th + 2 He
4
t1/2 = 4.5x109 yr.
t1/2 = 24.4 da.
t1/2 = 1.14 min.
t1/2 = 2.7x105 yr.
238
206
4
0
3) Net decay: 92 U -----> 82 Pb + 8 2 He + 6 -1 e
The slowest step is the first step with t1/2 = 4.5x109 yr, this is the effective half-life for the
entire series.
b. The uranium-actinium series.
Uranium-actinium series
4n+3 series
240
235
230
A
225
220
215
210
205
80
82
84
86
Z
88
90
92
94
235
1) The heaviest member is 92 U. Note that 235 = 4 (58)+3 . Therefore, all daughters in this
this series will have mass numbers that are divisible by four with three left over. The series is
also called the 4n+3 series.
2) Initial steps and half-lives.
3)
235
231
4
t1/2 = 7.1x108 yr.
92 U -----> 90 Th + 2 He
231
235
0
t1/2 = 24.6 hr.
90 Th -----> 91 Pa + -1 e
235
227
4
t1/2 = 3.2x104 yr.
91 Pa -----> 89 Ac + 2 He
227
227
0
t1/2 = 13.5 yr.
89 Ac -----> 90 Th + -1 e
235
207
4
0
Τhe net decay is 92 U -----> 82 Pb + 7 2 He + 4 -1 e
The slowest step is the first step with a t1/2 = 7.1x108 yr. This
series.
5
is the effective half-life of the
c. The Thorium series.
Thorium series
4n series
235
230
A
225
220
215
210
205
80
82
84
86
Z
88
90
92
232
1) The heaviest member is 90 Th. Note that 232 = 4 (58). Therefore, all the daughter
isotopes will have mass numbers that are divisible by four with zero left over. The series
is also called the 4n series.
2) Initial steps and half-lives.
232
90
228
88
228
89
228
90
3)
228
4
Th ------> 88 Ra + 2 He
228
0
228
0
Ra ------> 89 Ac + -1 e
Ac ------> 90 Th + -1 e
224
4
Th -------> 88 Ac + 2 He
232
208
Net decay: 90 Th ------> 82 Pb
t1/2 = 1.39x1010 yr.
t1/2 = 6.7 yr.
t1/2 = 6.1 hr.
t1/2 = 1.9 yr.
4
0
+ 6 2 He + 4 -1 e
The slowest step is the first with a t1/2 = 1.39x1010 yr. The is the effective half-life of the
series.
d. For completeness there should be another series whose mass number would be divisible by
four with one left over. The 4n+1 series does not exist in nature but it has been synthesized
in nuclear reactors. This series is called the neptunium series.
6
Neptunium series
4n+1 series
245
240
235
A
230
225
220
215
210
205
80
82
84
86
88
Z
90
92
94
96
1) Initial steps and half-lives.
237
93 Np
233
91 Pa
233
92 U
229
90 Th
233
4
t1/2 = 2.25x106 yr.
-------> 91 Pa + 2 He
233
0
-------> 92 U + -1 e
t1/2 = 27.4 da.
229
4
-------> 90 Th + 2 He
t1/2 = 1.62x105 yr.
225
4
-------> 88 Ra + 2 He
t1/2 = 7.0x103 yr.
237
209
4
0
2) The net decay: 93 Np ------> 83 Bi + 7 2 He + 4 -1 e
The slowest step is the first step with a t1/2 = 2.25x106 yr.
This is much faster than the
steps in the other series. It has long since disappeared from the earth's crust.
2. Other naturally occurring radioactive isotopes.
a. There are some radioactive elements found in the earth's crust that are not members of one of
the heavy atom series. Some have half-lives that make them useful in radiochemical dating.
b.
14
6
C is a naturally occurring radioactive isotope that decays by beta emission with a half-life of
5730 yrs. It is being continually produced in the upper atmosphere by the interaction of
14
cosmic rays with 7 N.
The production and decay reactions are;
Production
decay
14
7 N
14
6 C
1
14
1
+ 0 n ------> 6 C + 1 H
14
0
-----> 7 N + -1 e
7
3
3
3
0
c. 1 H (tritium) 1 H ----> 2 He + -1 e
40
t1/2 = 12.3 yr.
40
d. 19 K. 19 K undergoes two modes of decay,
40
19
40
19
40
0
K -----> 20 Ca + -1 e
0
40
K + -1 e ---> 18 Ar
(89%)
(11%)
40
19
K decay is 1.28x109 yr. Although K capture is not the main mode of decay,
40Ar
40
the 18 Ar isotope is fairly unique and the 40K ratio is used in dating minerals.
The half-life for
D. Artificial or Induced Radioactivity.
1. Bombardment reactions.
a. In 1934 Frédéric and Iréne Joliot-Curie reported that B, Mg, and Al could be made radioactive
when bombarded with alpha particles. This was the first instance of the synthesis of new
isotopes. The nuclear reactions were as follows:
10
5
27
13
24
12
4
13
1
4
30
1
27
1
B + 2 He ----> 7 N + 0 n
Al + 2 He ----> 15 P + 0 n
4
Mg + 2 He ----> 14 Si + 0 n
All of these "artificial" radioactive isotopes decay by positron emission with half lives of
10 min. for 13N, 2.5 min. for 30P, and 4.2 s. for 27Si.
b. These nuclear reactions can be abbreviated as:
10
5
24
12
13
27
30
B(α,n) 7 N [this is called a alpha-in-neutron-out reaction]; 13 Al(α,n)15 P; and
27
Mg(α,n)14 Si.
c. A large number of artificial radioactive isotopes have been made by bombarding stable nuclei
with light charged particles (alpha particles, deuterium, protium, boron, etc.)
1) Such reactions are called nuclear transformation reactions and the process is called
nucleosynthesis.
2) Since the transformation reactions require the merging of two nuclei, they have extremely
high activation energies, and are usually carried out using particle accelerators, such as,
cyclotrons.
8
2. Transuranium elements.
a. Uranium is the heaviest naturally occurring element. All elements with Z's greater than 92 have
been produced by nucleosynthesis. Some examples are shown in the following Table.
Selected Transuranium Elements
Z
Symbol
93
Np
94
Pu
95
Am
96
Cm
99
Es
101
Md
103
Lr
104
Rf
105
Db
106
Sg
109
Mt
Synthetic Method
238
92
239
93
239
94
239
94
238
92
253
99
252
98
249
98
249
98
249
98
209
83
Half-life
1
+0
239
0
U
n ----> 93 Np + -1 e
239
0
Np -----> 94 Pu + -1 e
1
240
0
Pu + 0 n ---> 95 Am + -1 e
4
242
1
Pu + 2 He ---> 96 Cm + 0 n
1
253
0
U + 15 0 n ---> 99 Es + 7 -1 e
4
256
1
Es + 2 He ---> 101 Md + 0 n
10
257
1
Cf + 5 B ---> 103 Lr + 5 0 n
12
257
1
Cf + 6 C ---> 104 Unq + 4 0 n
15
260
1
Cf + 7 N ---> 105 Unp + 4 0 n
18
263
1
Cf + 8 O ---> 106 Unh + 4 0 n
58
266
1
Bi + 26 Fe ---> 109 Une + 0 n
2.35 da.
86.4 yr.
458 yr.
4.5 hr.
20 da.
1.5 hr.
8 s.
4.5 s.
---<10-2s.
-----
b. Symbols and nomenclature of the transuranium elements.
1) Until a name can be agreed on, IUPAC has systemized the "naming" of the newly discovered
elements. They are just called by their atomic number. Therefore, element with the atomic
number of 112 is just called element-112.
2) A more formal way of expressing the atomic number and naming is to use a three letter symbol
that gives its atomic number. The roots are as follows:
digit
0
name
nil
digit
4
Name
quad
digit
8
1
un
5
pent
9
2
bi
6
hex
3
tri
7
sept
Name
oct
enn
3) Examples:
element 104 = Unq = unnilquadium (the elemental ium ending is appended last).
element 112 = Uub = ununbium
9
II. Half-Life and the Rates of Decay.
A. Review.
1. Radioactive decay is a first order process. Therefore, the half-life is independent of the amount.
2. Consider a radioactive isotope whose half-life = t1/2. After a time, t = n t1/2, has elapsed, the
1
fraction remaining f = 2
( )n .
3. The useful equations are:
t
number of half-lives elapsed = n = t1/2
( )n
1
fraction remaining = f = 2
amount remaining = (f) (original amount)
14
B. 6 C. Radiocarbon Dating.
1 As discussed above, 14C is being produced in the upper atmosphere and is decaying with a half14
life of 5740 yrs. Since 6 C is being produced and consumed at fairly constant rates, a small
steady state concentration of
14
6
C exists is nature, so that
14
6C
the 12
6C
1
ratio in nature is about 1012
14
2. The 6 C can react with O2 in the atmosphere to form 14CO2 . The CO2 can be taken up by
plants and thus enters the food chain. Therefore, all living organisms are somewhat radioactive
14
14
due to the 6 C. The amount of 6 C in living organisms is enough to give about 15.3 counts per
min. per gram of carbon on a Geiger Counter. All living organisms are continually
14
exchanging CO2, and thus maintain a constant level of 6 C. They should give 15.3 counts per
14
min. per g of C. When the organism dies, the 6 C is trapped, and will decay with a half-life of
equal to 5730 yrs; the counts would slowly decrease from the 15.3 counts per min. per g.
of C found for living organisms. This can be used to date old carbonaceous samples.
3. Example. The Lescaux Cave near Montignac, France contains some remarkable drawings made
by prehistoric man. A sample of charcoal from campsite in the cave gave 2.34 counts per min.
per g. of C. Given that freshly made charcoal gives 15.3 counts per min. per g of C,
calculate the age of the campsite.
1) Since radioactive decay is kinetically a first order process, at a time
1 n
t = (n) (t1/2), the fraction remaining, f = (2 ) .
2.34 counts
14
1 n
2) Fraction of 6 C remaining = 15.3 counts = 0.153 = (2 )
10
∴ (0.5)n = 0.153 or n log(0.5) = log(0.153)
log(0.153) -0.8153
∴ n = log(0.5) = -0.3010 = 2.71
∴ t = (2.71) (5730 yrs.) = 15,520 yrs.
C. Other dating methods.
1. The radium-uranium series begins with 238U and ends with 206Pb. The slowest step in the series is:
238
92
234
4
U -----> 90 Th + 2 He with a t1/2 = 4.5x1010 yr. This is basically the half-life for the series.
238
238
That is if you start with 1 mole of 92 U in 4.5x109 yrs, you will have 0.5 mole of 92 U and 0.5
238U
206
mole of 82 Pb. Therefore, the ratio of 206Pb in a uranium containing mineral can be use to date the
mineral.
2. Example: Suppose that a rock sample was analyzed and found to contain 1.25% 238U and 0.62%
206Pb. Calculate the age of the rock.
A 100 g sample of the mineral would contain 1.25 g 238U and 0.62 g 206Pb. Assuming that all the
206Pb
can from the decay of 238U, then
1.25 g
mole of 238U = 238 g/mol = 5.25x10-3 mol
0.62 g
mole of 206Pb = 206 g/mol = 3.01x10-3 mol
Total moles of 238U originally present = 8.26x10-3 mol
5.25x10-3
1 n
∴ the fraction remaining, f = 7.26x10-3 =0.636 = (2 )
log0.636
-0.1968
n = log0.5 = -0.3010 = 0.645
age = n x t1/2 = (0.645)(4.5x109yr) = 2.9x109 yrs.
235U
232Th
40K
87Rb
3. In the same way, the 207Pb ratio, the 208Pb ratio, the 40Ar ratio, and the 87Sr ratio
(t1/2 = 5x1011 yr) have also been used in dating rocks.
D. Rates of disintegration and the biological effects of radiation.
1. The fundamental unit of radiation is the Curie (Ci).
a. 1 Ci = 3.70x1010 disintegrations per second. This is equivalent to the radiation produced by 1 g
of radium.
b. 1 Becquerel (Bq) = 1 disintegration per second (1 Ci = 3.70x1010 Bq.).
2. Absorbed doses of radiation.
a. rad (radiation absorbed dose) = amount of radiation that is equivalent to 1x10-5 J/g of
irradiated material.
11
b. The biological effectiveness of the radiation depends on the type of radiation and the type of
tissue being irradiated. The biological effect of the radiation can be given by a parameter called
RBE (relative biological effectiveness).
1) α particles have extremely high energies and are about 20 times as effective (damaging) than
are β and γ particles, which are about equal in there biological effectiveness. Therefore, the
RBE of β and γ radiation is set equal to 1, the RBE of α particles is 20.
2) Some tissues are more susceptible to radiation damage than are others. Whole body radiation
is given a RBE = 1, some tissues, such as bone marrow have much greater RBE's.
3) The product of 1 rad x 1 RBE = 1 rem (roentgen equivalent man)
c. Biological effects of radiation doses. (These are continually being revised See end of notes for
more information))
Effect
Dose(rem)
No observable effects
0 - 25
Decrease in white blood cell count
25 - 50
Nausea, no deaths
100 - 200
Death to 50% of irradiated subjects
450
100% death rate
>800
d. One other factor that is important in determining the biological danger of radiation is its
penetrating power.
1) α particles, while extremely ionizing are not very penetrating. The can be stopped by
clothing or a sheet of paper and do not penetrate past the dead outer layer of skin. Therefore,
this type of radiation from an outside source is not too dangerous. However, if an α emitter
is ingested or inhaled, it is very deadly.
2) β particles are more penetrating, they can pass through a piece of paper or light clothing but
are stopped by a piece of wood. They can penetrate the dead outer layer of skin, but are
stopped within the skin. They can cause skin damage so that it appears as if the skin were
rburned. Ingested or inhaled, β emitters are very damaging.
3) γ rays are very penetrating, it requires about 3 cm of lead to stop γ rays. They will
completely penetrate the body causing extensive cell damage.
II. Nuclear Stability.
A. Binding Energy.
1. Any nucleus is more stable than is its composite protons and neutrons. Therefore, the process:
Z+N
Z protons + N neutrons ------> Z
is exothermic.
12
X (where Z+N = A = mass number )
a. The binding energy is the energy liberated in this process.
binding energy
b. The binding energy per nucleon =
.
A
The most stable nuclei will have the highest values of the binding energy per nucleon.
c. According to Einstein's special theory of relativity, when an amount of mass of M kg is
destroyed, the amount of energy, E, liberated, in J, is given by:
E = Mc2
where c is the speed of light ( 3.0x108 m/s ). In ordinary chemical reactions the energy changes
are so small that mass differences between reactants and products are beyond the detection
range of our most sensitive balances, and the law of conservation of mass holds. However, for
nuclear processes, such as the formation of nuclei, the mass changes are measurable.
d. The mass defect = mass lost when nuclei are formed from their composite protons and
neutrons.
7
2. Example: The mass of a 3 Li nucleus is 7.016005 amu. Given that the mass of a proton is
1.007276 amu and that of a neutron is 1.008665 amu, calculate the mass defect, the binding
7
energy, and the binding energy per nucleon of 3 Li.
a. mass defect.
1
1
7
The process is: 3 1 H + 4 0 n ------> 3 Li
mass of protons = 3 (1.007276 amu) = 3.021828 amu
mass of neutrons = 4 (1.008665 amu) = 4.034660 amu
Total mass of protons & neutrons = 7.056488 amu
7
- mass of 3 Li nucleus = -7.016005 amu
mass defect = 0.040483 amu
7
When the 3 Li is formed from its protons and neutrons, this much mass is converted into energy .
b. Binding energy.
0.040483 amu
The mass defect in kg = 6.023x1026 amu/kg = 6.721x10-29 kg
The binding energy in J = (6.721x10-19 kg) ( 2.998x108 m/s)2 = 6.04x10-12 J
7
This is the energy liberated when one 3 Li nucleus is formed, it is equivalent to 3.64x1012 J/mol.
A more convenient unit of energy is the MeV (Million electron Volt).
An electron volt is the energy change when 1 electron (charge = 1.602x10-19 C) is transferred
across a potential energy difference of 1 volt (1 J/C). Therefore,
1 eV = 1.602x01-19 J
1 MeV = 1.602x10-13 J
13
6.04x10-12 J
The binding energy in MeV = 1.602x10-13 J/MeV = 37.7 MeV
7
c. The binding energy per nucleon of 3 Li =
37.7 MeV
= 5.38 MeV per nucleon.
7
A convenient conversion factor to remember is that the destruction of 1 amu of mass will yield
933.2 MeV. Therefore, the binding energy in MeV can be obtained by multiplying the mass defect
in amu by 933.2.
3. The binding energy per nucleon has been calculated for most nuclei. A plot of binding energy per
nucleon vs. the mass number (A) gives a display of the relative stabilities of the nuclei. Such a plot is
shown below.
56
4. The most stable nuclei have mass numbers in the 50-60 range. The most stable nucleus is 26 Fe with a
binding energy per nucleon of 8.6 MeV.
4
a. Note that 2 He, the alpha particle, has an unusually high binding energy per nucleon.
b. The stability of the lighter nuclei (A's less than about 40) increase rapidly with increasing mass
number, while changes are much smaller for the heavier isotopes.
B. Fission and Fusion.
1. Fission = the splitting apart of a heavy nucleus into lighter ones.
a. Fission will be exothermic for isotopes with A's greater than 56.
235
239
b. There are certain isotopes, such as 92 U and 94 Pu, that can undergo sustained fission
chain reactions.
14
235
236
When a 92 U captures a neutron, the resulting 92 U that is formed will split apart in
two lighter nuclei, one in the mass range of ≈ 100 and another in the mass range of
≈140. An example of this fission process is:
235
92
1
236
90
143
1
U + 0 n ---> [ 92 U] ----> 38 Sr + 54 Xe + 3 0 n
1) Note that 3 neutrons are also produced in the fission reaction.
235
2) If enough 92 U is located in the immediate vicinity, the probability of one of the product
235
neutrons encountering another 92 U will be large enough so that a chain reaction is set
up. Since there are more neutrons produced than consumed, the reaction will be a
branching chain reaction and the rate would begin to increase without bounds leading to
a nuclear explosion.
3) The critical mass = the minimum amount of fissionable material necessary to generate a
235
239
self-sustained chain reaction. In an atomic bomb a critical mass of 92 U or 94 Pu is
suddenly brought together. The actual amount will depend on the configuration and the
shape of the sample. The bombs dropped on Hiroshima and Nagasaki most probably
235
239
consisted of a few kilograms of 92 U ( used on Hiroshima) or 94 Pu (used on Nagasaki)
that were suddenly compressed into a small sphere; the explosions were equivalent to about
20 kilotons of TNT (1 kiloton of TNT will liberate 4 billion kJ of energy).
2. Fusion = merging of two nuclei into a heavier one.
a. Fusion reactions are exothermic for the lighter isotopes. That is, for fusion reactions in which
isotopes with A's < 56 are produced.
b. Since the binding energy vs. A curve rises much more steeply for A's < 56 than it drops for
A's > 56, more energy is obtained per gram of nuclear fuel for fusion reactions than for fission
reactions.
1) Since fusion reactions have extremely large activation energies, very high temperatures, in
excess of several million degrees, are required to initiate fusion reactions.
2) The energy of the sun and stars is furnished by the fusion of hydrogen into helium, and
ultimately, other elements.
15
At temperatures less than 20 million degrees the conversion can be envisioned to take place
through the following steps.
1
1
2
1
3
2
1
2
1
3
3
4
0
H + 1 H ----> 1 H + 1 e
H + 1 H ----> 2 He + γ
1
He + 2 H ----> 2 He + 2 1 H
1
4
0
The net reaction is the conversion of 4 1 H into a 2 He (and 2 1 e's). At much higher
temperatures further fusion processes occur producing carbon and heavier elements.
c. Thermonuclear bombs derive most of their energy from a fusion reaction. A modern 20
megaton superbomb is a three stage fission-fusion-fission device. Its first stage (the detonator)
consists of a few kilograms of 239Pu. The fission bomb detonator is surrounded by about 150
kg of LiD that can undergo fusion (6Li2D ----> 2 4He); this is the second stage. The LiD is
surrounded by about 1000 pounds of 238U, that can undergo fission with fast neutrons, that acts
as the third stage.
The total weight of a 20 megaton bomb is about 1.5 tons; its explosive power is roughly three
times the total of all the bombs used in the Second World War.
C. Nuclear Reactors - Controlled Fission Reactions.
1. When a critical mass of 235U, or some other fissionable isotope, is brought together, a branching
chain reaction will take place and an explosion will result.
a. Chain branching occurs because in the fission reaction a neutron is required to induce fission
and three neutron are produced in the reaction.
235
92
1
236
90
143
1
U + 0 n ---> [ 92 U] ----> 38 Sr + 54 Xe + 3 0 n
If some of the product neutrons are absorbed before they strike other 235U nuclei, then
the rate of the chain reaction can be controlled and a sustained chain reaction can be
achieved.
b. In a Nuclear Reactor a sustained chain reaction takes place and a constant amount of
energy is produced. ( See the accompanying diagram.)
2. Components.
a. Fuel Rods- Contain uranium, usually in the form of uranium oxide (U3O8).
1) Natural uranium is only about 0.7% 235U, which is too low to support a chain
reaction.
2) The uranium used must be enriched in 235U. One way in which this has been
accomplished is through the diffusion, or centrification, of UF6(g).
16
F is unusual in that it is one of the few naturally occurring substances in which only
one isotope is found in nature. Therefore, molecules of UF6 will have slightly
different masses depending on the isotope of uranium that is incorporated in the
Steam to
Turbine
Heat
Exchanger
Shield
Moderator
Control
Rods
Fuel Rods
(U 3 O 8)
CORE
Schematic of Nuclear Reac tor
compound. A molecule of 238UF6 will be heavier than a molecule of 235UF6 by 3
amu. In accord with Graham's Law of Diffusion, the 235UF6 will diffuse fasted than
a molecule of 238UF6 by an amount given by the expression,
17
RateU-235
RateU-238 =
M238UF6
=
M235UF6
352
349 = 1.004
This is not much of a rate enhancement, but given long enough diffusion paths, the
leading front of UF6(g) will be enriched in 235U.
3) Other methods, such as the use of centrifuges and mass spectrometers can also be
used to enrich uranium. The enriched uranium is 3 - 4% uranium-235.
b. Control Rods. Rods of Cd and B, elements with large neutron capture radii, that are
inserted between the fuel rods to absorb neutrons and slowdown the chain reaction.
When the control rods are fully inserted, the reactor is shut down.
c. Moderators. These are substances that decrease the kinetic energy of the neutrons.
Because the fission reaction is very exothermic, the emitted neutrons have very high
kinetic energies. These fast neutrons are not easily captured by 235U nuclei. To
improve the efficiency of neutron capture a moderator is used to slowdown the
neutrons.
1) The moderator is a substance that surrounds the fuel rods that slow the neutrons and
also carry off the heat generated by the reactor. The most common moderator is
water.
2) Light Water Reactors. Use naturally occurring water, that is composed mainly
of 1H2O. Light water reactors require the use of enriched uranium. Most U. S.
reactors are light water reactors.
3) Heavy Water Reactors. Heavy water is D2O (D = deuterium = 2H). D2O is not
as efficient in slowing neutrons as is light water. Therefore, the faster neutrons will
travel greater distances and a 235U will have a higher probability of being struck.
Essentially all the uranium-235 is reacted.
Do not need enriched uranium to run heavy water reactors.
3. Breeder Reactors. Reactors that produce more fissionable fuel than they consume.
239
a. Uranium-238 can absorb fast neutrons to give 94 Pu, which is fissionable.
238
92
1
239
239
0
U + 0 n ------> 92 U ---------> 93 Np + -1 e
239
93
Np ------->
239
94
Pu +
0
-1
e
(t1/2 = 23.4 min)
(t1/2 = 2.35 da)
The half-life of plutonium-239 is 24,400 yr.
b. Uranium-233 is also a fissionable isotope that can be produced from 232Th in a breeder
reactor. The equations are,
232
90
1
233
0
Th + 0 n ---------> 92 U + 2 -1 e
18
c. In a breeder reactor the core is surrounded by a blanket of uranium-238,
so that some of the emitted neutrons will form plutonium-239. If, on the average, for
every fission process more than one neutron is absorbed by a uranium-238, then the
amount of plutonium-239 produced will be greater than the amount of nuclear fuel
consumed.
The doubling time = time required for the fuel to double. For fast breeders doubling
times are 7 - 10 yrs.
d. There are a number of problems associated with breeder reactors, the neutrons used are
fast neutrons so that a moderator, such as water, is not used. Therefore other
substances, such as liquid sodium, are used as coolants. These are very corrosive.
19
Table of exposure levels and symptoms
Dose-equivalents are presently stated in sieverts (Sv). 1 Sv = 100 REM
0.05–0.2 Sv (5–20 REM)
No symptoms. Possible potential for cancer and mutation of genetic material. A few researchers
contend that low dose radiation may be beneficial.
50 mSv is the yearly federal limit for radiation workers in the United States.
In the UK the yearly limit for a classified radiation worker is 20 mSv.
In Canada, the single-year maximum is 50 mSv, but the maximum 5-year dose is only 100 mSv.
Company limits are usually stricter so as not to violate federal limits.
0.2–0.5 Sv (20–50 REM)
No noticeable symptoms. Red blood cell count decreases temporarily.
0.5–1 Sv (50–100 REM)
Mild radiation sickness with headache and increased risk of infection due to disruption of
immunity cells. Temporary male sterility is possible.
1–2 Sv (100–200 REM)
Light radiation poisoning, 10% fatality after 30 days (LD 10/30). Typical symptoms include
mild to moderate nausea (50% probability at 2 Sv), with occasional vomiting, beginning 3 to 6
hours after irradiation and lasting for up to one day. This is followed by a 10 to 14 day latent
phase, after which light symptoms like general illness, and fatigue appear (50% probability at 2
Sv). The immune system is depressed, with convalescence extended and increased risk of
infection. Temporary male sterility is common. Spontaneous abortion or stillbirth will occur in
pregnant women.
2–3 Sv (200–300 REM)
Severe radiation poisoning, 35% fatality after 30 days (LD 35/30). Nausea is common (100% at 3 Sv),
with 50% risk of vomiting at 2.8 Sv. Symptoms onset at 1 to 6 hours after irradiation and last for 1 to 2
days. After that, there is a 7 to 14 day latent phase, after which the following symptoms appear: loss
of hair all over the body (50% probability at 3 Sv), fatigue and general illness. There is a massive loss of
leukocytes (white blood cells), greatly increasing the risk of infection. Permanent female sterility is
possible. Convalescence takes one to several months.
20
3–4 Sv (300–400 REM)
Severe radiation poisoning, 50% fatality after 30 days (LD 50/30). Other symptoms are similar
to the 2–3 Sv dose, with uncontrollable bleeding in the mouth, under the skin and in the kidneys
(50% probability at 4 Sv) after the latent phase.
4–6 Sv (400–600 REM)
Acute radiation poisoning, 60% fatality after 30 days (LD 60/30). Fatality increases from 60% at
4.5 Sv to 90% at 6 Sv (unless there is intense medical care). Symptoms start half an hour to two
hours after irradiation and last for up to 2 days. After that, there is a 7 to 14 day latent phase,
after which generally the same symptoms appear as with 3-4 Sv irradiation, with increased
intensity. Female sterility is common at this point. Convalescence takes several months to a
year. The primary causes of death (in general 2 to 12 weeks after irradiation) are infections and
internal bleeding.
6–10 Sv (600–1,000 REM)
Acute radiation poisoning, near 100% fatality after 14 days (LD 100/14). Survival depends on
intense medical care. Bone marrow is nearly or completely destroyed, so a bone marrow
transplant is required. Gastric and intestinal tissue are severely damaged. Symptoms start 15 to
30 minutes after irradiation and last for up to 2 days. Subsequently, there is a 5 to 10 day latent
phase, after which the person dies of infection or internal bleeding. Recovery would take several
years and probably would never be complete.
Devair Alves Ferreira received a dose of approximately 7.0 Sv (700 REM) during the Goiânia
accident and lived partially due to his fractionated exposure.
10–50 Sv (1,000–5,000 REM)
Acute radiation poisoning, 100% fatality after 7 days (LD 100/7). An exposure this high leads to
spontaneous symptoms after 5 to 30 minutes. After powerful fatigue and immediate nausea caused by
direct activation of chemical receptors in the brain by the irradiation, there is a period of several days
of comparative well-being, called the latent (or "walking ghost") phase. After that, cell death in the
gastric and intestinal tissue, causing massive diarrhea, intestinal bleeding and loss of water, leads to
water-electrolyte imbalance. Death sets in with delirium and coma due to breakdown of circulation.
Death is currently inevitable; the only treatment that can be offered is pain therapy. Louis Slotin was
exposed to approximately 21 Sv in a criticality accident on 21 May 1946, and died nine days later on
30 May.
21
50–80 Sv (5,000–8,000 REM)
Immediate disorientation and coma in seconds or minutes. Death occurs after a few hours by
total collapse of nervous system.
More than 80 Sv (>8,000 REM)
U.S. military forces expect immediate death. A worker receiving 100 Sv (10,000 REM) in an
accident at Wood River, Rhode Island, USA on 24 July 1964 survived for 49 hours after
exposure, and an operator receiving 120 Sv (12,000 REM) to his upper body in an accident at
Los Alamos, New Mexico, USA on 30 December 1958 survived for 36 hours.
22
Problems
1. Complete and balance the following nuclear equations.
234
0
214
a. 91 Pa ------> -1 e +
12
12
14
c. 5 B -------> 6 C +
238
4
4
1
d. 7 N + 2 He ------> 1 H +
0
250
e. 92 U + 2 He -------> -1 e +
8
210
b. 84 Po ------> 82 Pb +
11
1
f. 98 Cf + 5 B ------> 40 n +
8
g. 5 B -----> 4 Be +
h.
30
15
30
P -------> 14 Si +
11
2. Calculate the binding energy per nucleon, in eV and kJ/mol, of the 5 B nucleus, given that the
11
mass of a 5 B nucleus is 11.00656 amu.
3. complete the notations for the following processes.
a.
24
26
2
4
2
1
Mg(1 H,2 He)_____
b. 12 Mg(1 H,1 H)_____
40
4
1
c. 18 Ar(2 He,1 H)______
12
2
1
d. 6 C(1 H,0 n)____
130
2
55
1
59
1
4
g. 27 Co(0 n, 2 He)______
1
e. 52 Te(1 H,20 n)_____
0
f. 25 Mn(0 n,0 γ)_______
4. At the centers of very heavy stars, two 16O atoms can fuse to produce an atom of 32S. Calculate
the energy released in kJ/mol. (mass 32S = 31.9721 amu, 16O = 15.9949 amu )
5. Through a series of decay steps in the 4n+2 series, 238U is converted to 214Pb. How many
alpha particles and beta particles are lost in this process?
6. The binding energy of 2H is 1.7x10-16kJ per nucleon and that of 4He is 1.1x10-15kJ per
nucleon. Calculate the energy released when 2 moles (4.0g) of 2H is fused to form of 4He.
Given that the heat of combustion of CH4 is 890 kJ/mol, how many grams of methane would
have to be burned in order to produce this much energy?
7. Given that the binding energy per nucleon of 12C = 1.233x10-15 kJ, calculate the mass of the
12C
in amu. From this mass show that the mass of the 12C isotope is 12.000 amu.
8. If a 235U, after the absorption of a slow neutron, undergoes fusion to give 142Ba and 93Kr, write
the balanced nuclear reaction.
9. Given that 13B undergoes beta decay to give 13C and the isotopic mass of 13B = 13.0178 amu
and that of 13C = 13.0034 amu, calculate the energy change for the radioactive decay process.
(Note that these masses are not nuclear masses)
10. Assume that you have prepared 100 g of a compound containing a radioactive element whose
half-life is 20 min. How many grams of the radioactive compound would remain 2.0 hrs after
preparation?
23
11. 50.0 grams of a radioactive isotope was found to decay at a rate such that only 6.75 g of the
isotope remains after 30 hrs. What is the half-life of the isotope?
12. A wooden artifact from an Egyptian tomb was dated using 14C. Suppose that the artifact gave
8.3 counts per min per g of C. Given that fresh wood gives 15.3 counts per min per g of C and
the half-life of 14C is 5730 yrs, calculate the age of the artifact.
13. Potassium-40 undergoes K capture to form 40Ar with a half-life of 1.3x109 yr. Suppose that
analysis of a moon rock gave an potassium-40/argon-40 molar ratio of 0.816, calculate the age
of the rock.
14. The age of the earth is about 2 billion years. Using the half-lives given in this chapter, calculate
the maximum ratio of 206Pb to 238U you would expect in the oldest possible rock sample.
24