Nuclear Chemistry
Chapter 18
Hein and Arena
Version 2.0
12th Edition
Eugene Passer
Chemistry Department
Bronx Community
1 College
© John Wiley and Sons, Inc
Chapter Outline
18.1 Discovery of Radioactivity
18.2 Natural Radioactivity
18.8 Nuclear Fission
18.9 Nuclear Power
18.3 Alpha Particles, Beta
Particles and Gamma Rays
18.10 The Atomic Bomb
18.4 Radioactive Disintegration
Series
18.11 Nuclear Fusion
18.5 Transmutation of Elements
18.6 Artificial Radioactivity
18.7 Measurement of
Radioactivity
18.12 Mass-Energy Relationships
in Nuclear Reactions
18.13 Transuranium Elements2
18.1
Discovery of
Radioactivity
3
Roentgen
• In 1895 Wilhelm Konrad Roentgen
discovered X-rays.
• Roentgen observed that a vacuum
discharge tube enclosed in a thin,
black cardboard box had caused a
nearby piece of paper coated with
the salt barium platinocyanide to glow
with phosphorescence.
4
Roentgen
• From this and other experiments he
concluded that certain rays, which he
called X-rays, were emitted from the
discharge tube, and penetrated the box
and caused the salt to glow.
5
Becquerel
• Shortly after Roentgen’s discovery,
Antoine Henri Becquerel attempted to
show a relationship between X-rays
and the phosphorescence of uranium
salts.
• Becquerel wrapped a photographic
plate in black paper, sprinkled a sample
of a uranium salt on it, and exposed it
to sunlight.
6
Becquerel
• When Becquerel attempted to repeat
the experiment the sunlight was
intermittent.
• He took the photographic plate
wrapped in black paper with the
uranium sample on it and placed the
whole setup in a drawer.
7
Becquerel
• Several days later he developed the
film and was amazed to find an intense
image of the uranium salt on the plate.
• He repeated the experiment in total
darkness with the same result.
8
Becquerel
• This proved that the uranium salt emitted
rays that affected the photographic plate
and that these rays were not a result of
phosphorescence due to exposure to
sunlight.
• Two years later, in 1896, Marie Curie
coined the name radioactivity.
Radioactivity is the spontaneous emission of
Elements
havingrays
this from
property
are radioactive.
particles and/or
the nucleus
of an
atom.
9
Rutherford
• In 1899 Rutherford began to investigate
the nature of the rays emitted by uranium.
• He found two particles in the rays. He
called them alpha and beta particles.
• Rutherford’s nuclear atom description led
scientists to attribute the phenomenon of
radioactivity to reactions taking place in
the nuclei of atoms.
10
Villiard
• The gamma ray, a third type of emission
from radioactive material, was discovered
by Paul Villiard in 1900.
11
Definitions
12
mass number the total number of nucleons
nucleon a proton or a neutron
in the nucleus.
13
isotope atoms of the same element with
nuclide any isotope of any atom
different masses
14
Isotopic Notation
15
16
6 protons + 6 neutrons
12
C
6
6 protons
A nuclide of carbon
17
8 protons + 8 neutrons
16
O
8
8 protons
A nuclide of oxygen
18
8 protons + 9 neutrons
17
O
8
8 protons
A nuclide of oxygen
19
8 protons + 10 neutrons
18
O
8
8 protons
A nuclide of oxygen
20
21
18.2
Natural Radioactivity
22
• Radioactive elements continuously
undergo radioactive decay or
disintegration to form different
elements.
• Radioactivity is a property of an
atom’s nucleus. It is not affected
by temperature, pressure, chemical
change or physical state.
23
radioactive decay the process by which a
radioactive element emits particles or rays
and is transformed into another element.
24
• Each radioactive nuclide disintegrates
at a specific and constant rate, which is
expressed in units of half-life.
• The half-life (t1/2) is the time required
for one-half of a specific amount of a
radioactive nuclide to disintegrate.
1.0 g
226
88
Rn
t

1/2
1620 yrs
0.5 g
t
Rn  0.25 g
226
88
1/2
1620 yrs
226
88
Rn
25
131
131
53
The half-life of I is 8 days. How much
32-g sample remains after five half-lives?
half-lives
131
131
53
I from a
435210
number of days 32
24
40
16
8
amount
remaining
16
32
2418 g
Trace a horizontal line from this point on the plotted
Take a perpendicular line from any multiple of 8 days
line to the y-axis and read the corresponding grams
26
on 131
the x-axis to the line on the graph.
131
I
18.1 of
53 I.
• Nuclides are said to be either stable
(nonradioactive) or unstable (radioactive).
• Elements that have an atomic number
greater than 83 are naturally radioactive.
• Some of the naturally occurring nuclides of
elements 81, 82 and 83 are radioactive and
some are stable.
27
• Only a few naturally occurring
elements that have atomic numbers
less than 81 are radioactive.
• No stable isotopes of element 43
(technetium) or of element 61
(promethium) are known.
28
• Radioactivity is believed to be a result
of an unstable ratio of neutrons to
protons in the nucleus.
• Stable nuclides of elements up to about
atomic number 20 generally have
about a 1:1 neutron-to-proton ratio.
29
• In elements above atomic number 20,
the neutron-to-proton ratio in the stable
nuclides gradually increases to about
1.5:1 in element number 83 (bismuth).
• When the neutron to proton ratio is too
high or too low, alpha, beta, or other
particles are emitted to achieve a more
stable nucleus.
30
31
18.3
Alpha Particles,
Beta Particles and
Gamma Rays
32
Marie Curie, in a classic experiment, proved
that alpha and beta particles are oppositely
charged.
three types of radiation are
Beta rays
areare
strongly
deflected
to
radiation
passes
between
Gamma
rays
not
detected
by deflected
a the
photographic
thethe
positive
pole.
poles
of
an
electromagnet
by
magnet.
plate
Alpha rays are less strongly
radioactive source
deflected to the negativeapole.
was placed inside a
lead block
18.1
33
Alpha Particles
34
The symbols of an alpha particle are
An alpha particle is a helium nucleus.
4
2
He

It consists of two protons and two neutrons.
It has a charge of +2.
It has a mass of 4 amu.
35
Loss of an alpha particle from the nucleus
results in
loss of 4 in the mass number (A)
loss of 2 in the atomic number (Z)
36
Formation of thorium from the radioactive
decay of uranium can be written as
mass number decreases by 4
238
92
U 
234
90
Th + α
atomic number decreases by 2
or
238
92
U 
234
90
4
2
Th + He
37
To have a balanced nuclear equation
• the sum of the mass numbers (superscripts)
on both sides of the equation must be equal.
•
the sum of the atomic numbers (subscripts)
on both sides of the equation must be equal.
sum of mass numbers = 238
238
92
U 
234
90
4
2
Th + He
sum of atomic numbers = 92
38
Beta Particles
39
The symbols of the beta particle are
The beta particle is identical in mass and
charge to an electron.
0
-1
e

Its charge is -1.
40
A proton and a beta particle are formed by
the decomposition of a neutron.
1
0
n  p+ e
1
0
0
-1
The beta particle leaves the nucleus and the
proton remains in the nucleus.
41
Loss of a beta particle from the nucleus
results in
– no change in the mass number
– an increase of 1 in the atomic number
234
90
234
91
Th 
Pa 
Pa + β
234
91
234
92
0
-1
Pa + e
42
Gamma Rays
43
The symbol of a gamma ray is
A gamma ray is a high energy photon.

It is emitted by radioactive nuclei.
It has no electrical charge.
It has no measurable mass.
44
Loss of a gamma ray from the nucleus
results in
– no change in the mass number
– no change in atomic number
45
Write an equation for the loss of an alpha particle from
194Pt
the nuclide 194
78 Pt.
46
Write an equation for the loss of an alpha particle from
194Pt
the nuclide 194
78 Pt.
Mass number
(sum of protons and
neutrons in the nucleus)
78 protons + 116 neutrons
194
78
Atomic number
(number of protons
in the nucleus)
Pt
78 protons
A nuclide of platinum
47
Write an equation for the loss of an alpha particle from
194Pt
the nuclide 194
78 Pt.
44
2
Loss of an alpha particle, He
He, results in a decrease
of 4 in the mass number and a decrease of 2 in the
atomic number.
A = mass number
A-4 or 194 – 4 = 190
Mass of new nuclide:
Z = atomic number
Atomic number of new nuclide: Z-2 or
78 – 2 = 76
Element number 76 is Os, osmium.
The equation is
194
78
Pt 
190
76
4
2
Os + He
48
What nuclide is formed when
particle from its nucleus?
228
194
88
Ra loses a beta
49
What nuclide is formed when
particle from its nucleus?
Mass number
(sum of protons and
neutrons in the nucleus)
228
194
88
Ra loses a beta
88 protons + 140 neutrons
228
Ra
88
Atomic number
(number of protons
in the nucleus)
88 protons
A nuclide of radium
50
What nuclide is formed when
particle from its nucleus?
228
194
88
Ra loses a beta
194Ra nucleus means
The loss of a beta particle from a 228
88
a gain of 1 in the atomic number with no essential
change in mass.
A = mass number
A-0 or 228 – 0 = 228
Mass of new nuclide:
Z = atomic number
Atomic number of new nuclide: Z-(-1) or 88 + 1 = 89
The equation is
228
88
Ra 
228
89
Ac + 0-1e
51
Penetrating Power
of Radiation
52
The ability of radioactive rays to pass through
various objects is in proportion to the speed at
which they leave the nucleus.
sheet 5-cm
of lead block –
Thin sheet of Thin
will reduce, but not
paper – stops  aluminum – stops
completely stop 
 and  particles.
particles.
radiation
53
18.2
54
18.4
Radioactive
Disintegration Series
55
The naturally occurring radioactive
elements with a higher atomic number than
lead fall into three orderly disintegration
series. Each series proceeds from one
element to the next with the loss of either
an alpha or a beta particle, finally ending in
a nonradioactive nuclide.
56
– The uranium series starts with 238
92U and
ends with 206
82Pb.
– The thorium series starts with 232
90Th and
ends with 208
82Pb.
– The actinium series starts with 235
92U and
ends with 207
82Pb.
57
A fourth series begins with the synthetic
element plutonium.
– The neptunium series begins with 241
94Pu
and ends with 238
83Bi.
58
238
The uranium disintegration series. 238
92 U decays by a
series of alpha () and beta () emissions to the stable
206
nuclide 208
Pb
82 Pb.
59
18.3
18.5
Transmutation
of Elements
60
Transmutation
is
the
conversion
of
one
Transmutation occurs spontaneously in
element
into
another
by
either
natural
or
natural radioactive disintegrations.
artificial means.
226
88
Ra 
222
86
210
82
Pb 
210
83
Rn + α
0
-1
Bi + e
61
Alchemists
for transmutation
hundreds ofoccurred
years
The
first artificial
attempted
to transmute
mercury
and lead
in
1919 when
Rutherford
succeeded
in
into gold by
artificial
means.
They were
producing
oxygen
from
nitrogen.
never successful.
14
7
N + He 
4
2
17
8
1
1
O+ H
62
Some of these transmutations are:
7
3
209
83
Li + H  2 He
1
1
Li + H 
2
1
4
2
210
84
1
0
Po + n
63
18.6
Artificial
Radioactivity
64
Irene and Frederick Curie observed that
the bombardment of aluminum-27 with
alpha particles resulted in the emission of
neutrons and positrons.
27
13
Al + He  neutrons and positrons
4
2
65
This suggested
When
alpha particle
that bombardment
neutron emission
is halted
and
neutron emission
positron
emissionstops,
were but
a result
positron
of emission
separate
continues.
reactions.
66
Further investigation on their part showed that
when aluminum-27 is bombarded with alpha
particles phosphorous-30 is produced.
27
13
Al + He 
4
2
30
15
1
0
P+ n
67
Phosphorous-30 is radioactive, has a 2.5
minute half-life, and decays to silicon30 with the emission of a positron.
30
15
P 
30
14
0
+1
Si + e
68
The radioactivity
Joliot-Curies of
received
nuclides
theproduced
Nobel Prize
in this
in
manner
chemistryis in
known
1935as for
artificial
the discovery
or induced
of
radioactivity.
artificial, or induced, radioactivity.
30
15
P 
30
14
0
+1
Si + e
69
18.7
Measurement
of Radioactivity
70
Radiation from a radioactive source can
be measured by a variety of instruments.
– Geiger counters
– film badges
– scintillation counters
71
The curie is the unit indicating the rate of
decay of a radioactive substance.
One curie (Ci) = 3.7 x 1010 disintegrations per
second.
Because
This
verya high
curieradiation
is so large
level
theismillicurie
the amount
(one
of
radiation emitted
thousandth
of a curie)
by 1and
gram
theof
microcurie
radium in(one
one
second. of a curie) are more commonly used.
millionth
72
73
18.8
Nuclear Fission
74
As
the atomfission,
splits, ait heavy
releases
energysplits
and
In nuclear
nuclide
two
other intermediate
neutrons, each
of
into or
twothree
or more
sized
which
can cause
fission.
fragments
when another
struck nuclear
in a particular
way by a neutron
1
0
U+ n 
139
56
94
36
1
0
U+ n 
144
54
90
38
1
0
235
92
235
92
1
0
Ba + Kr + 3 n
Xe + Sr + 2 n
75
Characteristics of Nuclear Fission
1. Upon absorption of a neutron, a heavy
nuclide splits into one or more smaller
nuclides (fission products).
2. The mass of the nuclides ranges from
abut 70-160 amu.
3. Two or more neutrons are produced
from the fission of each atom.
76
Characteristics of Nuclear Fission
4. Large quantities of energy are produced
as a result of the conversion of a small
amount of mass into energy.
5. Many nuclides produced are radioactive
and continue to decay until they reach a
stable nucleus.
77
Fission of
235
235
92
U
Each time fission occurs three neutrons
and two nuclei are produced.
78
18.5
•
•
In a chain reaction the products
cause the reaction to continue or
magnify.
For a chain reaction to continue,
enough fissionable material must be
present so that each atomic fission
causes, on average, at least one
additional fission.
79
• The minimum quantity of an element
needed to support a self-sustaining
chain reaction is called the critical
mass.
• Since energy is released in each atomic
fission, chain reactions provide a
steady supply of energy.
80
Fission and chain
reaction of 235U.
81
18.6
18.9
Nuclear Power
82
• A nuclear power plant is a thermal
power plant in which heat is produced
by a nuclear reactor instead of by
combustion of fossil fuel. The major
components of a nuclear reactor are:b
– an arrangement of nuclear fuel, called the
reactor core.
– a control system, which regulates the rate
of fission and thereby the rate of heat
generation.
83
84
• In the United States breeder reactors are
used to generate nuclear power.
• Breeder reactors use U3O8 that is
enriched with scarce, fissionable U-235.
• In a breeder reactor, excess neutrons
convert nonfissionable isotopes, such as
U-238 or Th-232, to fissionable isotopes,
Pu-239 or U-233.
238
92
U+ n 
1
0
U 
239
92

Np 
239
93

239
94
Pu
85
18.10
The Atomic Bomb
86
• The atomic bomb is a fission bomb.
• “Wild” or uncontrolled fission occurs in
an atom bomb, whereas in a nuclear
reactor the fission is carefully controlled.
• A minimum critical mass of fissionable
material is required for a bomb.
87
• When a quantity of fissionable material
smaller than the critical mass is used, too
many neutrons escape and a chain
reaction does not occur.
• The fissionable material of an atomic
bomb is stored as two or more subcritical
masses and are then brought together to
achieve a nuclear detonation.
88
• Uranium-235 and plutonium-239 are
the nuclides used to construct an
atomic bomb.
• 99.3% of uranium is nonfissionable
uranium-238. Uranium-238 can be
transmuted to fissionable plutonium239.
238
92
U+ n 
1
0
U 
239
92

Np 
239
93

239
94
Pu
89
18.11
Nuclear Fusion
90
Nuclear fusion is the process of uniting
two light elements to form one heavier
element.
91
The masses of the two nuclei that fuse
into a single nucleus are greater than the
mass of the nucleus formed by their
fusion.
The difference
3
2
4
1
n + energy
in + 0mass
is
1 H + 1H  2He
tritium deuterium
released
as
3
1
4
energy.
H
+
H

He
+ energy
1
1
2
3.0150 1.0079
amu
amu
4.0229
amu
4.0026
amu
4.0229 amu – 4.0026 amu = 0.0203 amu
92
• Fusion reactions require temperatures
on the order of tens of millions of
degrees for initiation.
• Such temperatures are present in the
Sun but have been produced only
momentarily on earth.
93
Fusion power will be far superior to fission
power because
– Virtually infinite amounts of energy are
possible from fusion power.
– While uranium supplies are limited, deuterium
supplies are abundant.
– It is estimated that the deuterium present in a
cubic mile of seawater used as fusion fuel can
provide more energy than the petroleum
reserves of the entire world.
94
Fusion power will be far superior to fission
power because
– Fusion power is much “cleaner” than fission
power.
– Fusion reactions (unlike uranium and
plutonium fission reactions) do not produce
large amounts of long-lived and dangerously
radioactive isotopes.
95
18.12
Mass-Energy Relationship in
Nuclear Reactions
96
• In fission reactions, about 0.1% of the
mass of the reactants is converted into
energy.
• In fusion reactions as much as 0.5% of
This is 4,000,000 times greater
the mass of the reactants may be
The energy
than
the energy
equivalent
releasedoffrom
this
changed into
energy.
mass
the
combustion
is 1.62 x 1012
ofJ. 1 mol of
carbon.
4
Li + H  2He + He + energy
7
1
3
1
7.016 g 1.008 g
8.024 g
4.003 g
4
2
4.003 g
8.006 g
8.024 g – 8.006 g = 0.018 g
97
• The mass of a nucleus is less than the
sum of the masses of the protons and
neutrons that makes up that nucleus.
• The difference between the mass of the
protons and the neutrons in a nucleus is
known as the mass defect.
• The energy equivalent to this difference
in mass is known as the nuclear binding
energy.
98
• The nuclear binding energy is the
amount of energy that would be required
to break a nucleus into its individual
protons and neutrons.
• The higher the binding energy, the more
stable the nucleus.
99
• Neutrons and protons attain more
stable arrangements through nuclear
fission or fusion reactions.
– when uranium undergoes fission the
products have less mass (greater binding
energy) than the reactants.
– when hydrogen and lithium fuse to form
helium, the helium has less mass (greater
binding energy) than the hydrogen and
lithium.
100
Calculate the mass defect and the nuclear binding
energy for an  particle (helium nucleus).
Data: proton mass = 1.0073 g
neutron mass = 1.0087 g/mol
 mass = 4.0015 g/mol
1.0 g = 9.0 x 1013J when converted into energy
Solution: First calculate the sum of the individual
parts of an  particle:
2 protons:
2 x 1.0073 g = 2.0146 g/mol
2 neutrons:
2 x 1.0087 g = 2.0174 g/mol
4.0320 g/mol
101
Calculate the mass defect and the nuclear binding
energy for an  particle (helium nucleus).
Mass the difference between the mass of the 
defect: particle and its component parts.
mass of component particles = 4.0320 g/mol
mass of  particle = 4.0015 g/mol
mass defect = 0.0305 g/mol
Nuclear binding energy:
convert mass defect to its energy equivalent:
g 

13 J 
12 J
= 2.7 x 10
 0.305
  9.0 x 10

mol  
g

mol
102
18.13
Transuranium
Elements
103
They
are synthetic
radioactive
transuranium
elements
followelements;
uranium
none
them occur
naturally.
in theof periodic
table
and have atomic
numbers greater than 92.
104
• The first transuranium element,
number 93, was discovered in 1939 by
Edwin McMillan who named it
neptunium for the planet Neptune.
• In 1941, element 94, plutonium, was
identified as a beta-decay product of
neptunium.
238
93
239
93
Np 
238
94
0
-1
Np 
239
94
0
-1
Pu + e
Pu + e
Plutonium is one of the most important fissionable elements
105
known today.
106