Uranium-235 undergoes fission, releasing 0.1% of its mass as energy. A 100 MW power plant would require 96 grams of Uranium-235 to undergo fission per day. Coal provides 32.6 MJ/kg upon burning, so a similar power plant would use 265 metric tons of coal per day. Nuclear fusion in the Sun converts four protons into a helium nucleus, releasing 24.7 MeV of energy. Radioactive decay changes the atomic number and mass number in predictable ways. The half-life of an isotope is the time for half of a sample to decay.

1. NUCLEAR REACTIONS - FISSION AND
FUSION
Problem 1
When U undergoes fission, about 0.1 % of the
235
92
original mass is released as energy.
a) How much energy is released when 1 kg of
U undergoes fission?
235
92
b) How much U must undergo fission per day
235
92
in a nuclear reactor that provides energy to a
100 MW electric power plant?
c) When coal is burned, about 32.6 MJ/kg of
heat is released. How many kilograms of
coal would be consumed per day by a
conventional coal-powered 100 MW electric
power plant?
Solution
a) E = mc2 = (0.001 kg)(3 x 108 m/s)2 = 9 * 1013
J
b) Energy = power x time, so
E = Pt = (108 W) (3600 s/h)(24 h/d) = 8.64 x
1012 J/day
1

2. Hence the mass of 235
92 U required is
12
8.64 *10 J / day
13
= 9.6 *10 −2 kg / day = 96 g / day
9 * 10 J / kg
c) The energy liberated per kg of coal burned is
3.26 x 107 J. Hence the mass of coal required is
12
8.64 *10 J / day
7
= 2.65 *10 kg / day ,
5
which is 265 metric
3.26 *10 J / kg
tonnes.
The core of the Sun is considered to extend
from the center to about 0.2 solar radii. It has a
density of up to 150,000 kg/m³ (150 times the
density of water on Earth) and a temperature
of close to 13,600,000 kelvins (by contrast, the
surface of the Sun is around 5,800 kelvins).
Problem 2
In the sun and other stars the principal process
that liberates energy is the process of conversion
of hydrogen to helium in a series of nuclear
fusion reactions where positrons (positively
charged electrons) are emitted.
(a) write the equation for the overall
process in which 4 protons form a helium
nucleus (alpha particle) .
2

3. (b) how much energy is liberated in such a
process?
Given the masses of H , He and the electrons are
1
1
4
2
1.007825, 4.002603, and 0.000549 u.
Solution
(a) the positrons must be given off so that
charge will be conserved. Hence the overall
process is
+ +
1
1H + H + H + H → He + e + e
1
1
1
1
1
1
4
2
(b) since helium atoms has only 2 electrons
around its nucleus (2 p+ = 2 e-,ie. x p+ = x e-,
right?), 2 electrons as well as 2 positrons are lost
when each helium atom is formed. The mass
change is therefore
∆m = 4mH + 0mn – (mHe + 4me) = (4) (1.007825
u) – [4.002603 u + (4) (0.000549 u)]
= 0.026501 u
Hence, the amount of energy liberated is
(0.026501 u) (931 MeV/u) = 24.7 MeV.
3

4. RADIOACTIVITY
Unstable nuclei undergo radioactive decay to
become more stable nuclei. There are 5 types of
radioactive decay (decay events):
Decay event What causes
this decay type
gamma ray Gamma ray Nuclear has
emission excessive
reduces energy energy
of nucleus
alpha decay Alpha particle Nucleus is too
emission large
reduces size of
nucleus
beta decay Emission of Nucleus has
electron by too many
nuclear neutrons
neutron relative to
changes it to a number of
proton protons
electron Capture of Nucleus has
capture electron by too many
nuclear proton protons
4

5. changes it to a relative to
neutron number of
neutrons
positron Emission of Nucleus has
emission positron by too many
nuclear proton protons
changes it to relative to
proton number of
neutrons
Problem 3
What happens to the atomic number and mass
number of a nucleus that
(a) emits an electron,
(b) undergoes electron capture,
(c) emits an alpha particle.
Solution
(a) Z increases by 1, A is unchanged.
(b) Z decreases by 1, A is unchanged.
(c) Z decreases by 2, A decreases by 4.
5

6. Problem 4
How many successive α-decays occur in the
decay of the thorium isotope Th into the lead
228
90
isotope Pb ?
212
82
Solution
Each α-decay means a reduction of 2 in atomic
number and 4 in mass number. Here Z decreases
by 8 and A by 16, which means that 4 α-
particles are emitted.
HALF LIFE, t 1/2 = ln 2 / λ
The time required for half of the atoms in any
given quantity of a radioactive isotope to decay
is the half-life of that isotope. Each particular
isotope has its own half-life. For example, the
half-life of 238U is 4.5 billion years. That is, in
4.5 billion years, half of the 238U on Earth will
have decayed into other elements. In another 4.5
billion years, half of the remaining 238U will
have decayed. One fourth of the original
material will remain on Earth after 9 billion
years. The half-life of 14C is 5730 years, thus it
is useful for dating archaeological material.
6

7. Nuclear half-lives range from tiny fractions of a
second to many, many times the age of the
universe. (http://www.lbl.gov/abc/Basic.html).
Problem 5
Tritium is the hydrogen isotope H whose nucleus
3
1
contains two neutrons and one proton. Tritium is
beta-radioactive and emits an electron.
(a) What does tritium become after
undergoing beta decay?
(b) The half-life of tritium is 12.5 year.
How much of the 1-g sample will remain
undecayed after 25 years?
Solution
(a) in the beta decay of nucleus one of the
neutrons become a proton. Since the atomic
number A= 2 corresponds to helium, the
beta decay of tritium is
3
1 H → 23 He + e −
and the new atom is 3
2 He .
(b) 25 years = 2 x t1/2 , and so ½ x ½ x 1 g = ¼
g of tritium remains undecayed.
7

8. Problem 6
The half-life of the sodium isotope Na against
24
11
beta decay is 15 h. how long does it take for ⅞th
of a sample of this isotope to decay?
After ⅞th has decayed, ⅛th is left,and
1 1 1 1
= × × which is 3 half-lives. Hence the answer
8 2 2 2
is (3)(15) h = 45 h.
Problem 7
The carbon isotope carbon-14 ( C ) called
14
6
radiocarbon is beta-radioactive with a half-life
of 5600 years. Radiocarbon is produced in the
earth’s atmosphere by the action of cosmic rays
on nitrogen atoms, and the carbon dioxide of the
atmosphere contains a small proportion of
radiocarbon as a result. All plants and animals
therefore contain a certain amount of
radiocarbon along the stable isotope of C . When 12
6
a living thing dies, it stops taking in radiocarbon
and the radiocarbon it already contains decays
steadily. By measuring the ratio between C and 12
6
C of the remains of an animal or plant and
14
6
8

9. comparing it with the ratio of these isotopes in
living organisms, the time that has passed since
the demise of the animal or the plant can be
found.
(a) How old is a piece of wood from an
ancient dwelling if its relative radiocarbon
content is ¼th that of a modern specimen?
(b) If it is 1/16th that of a modern specimen?
Solution
(a) since ¼ = ½ x ½ , the specimen is two half-
lives old, which is 11,200 yr old.
(b) since 116 = ½ x ½ x ½ x ½ , the specimen is 4
half-lives old, which is 22,400 yrs old.
9

10. Problem 8
The rate at which a sample of radioactive
substance decays is called its activity. A unit of
activity is the curie(Ci) where 1 curie =3.7 *
1010 decays/s. if a luminous watch dial contains
5 µCi of the radium isotope Ra , how many
226
88
decays per second occur in it? (This isotope is –
alpha-radioactive).
Solution
Since 1 µCi is 10-6 Ci, the activity of the watch
dial is
( 3.7 ×10 decays / s ) (5 x 10-6 Ci) = 1.85 x 105 decays/s.
10
Ci
10