Nuclear Chemistry
Mr. Montero
Chemistry
Dr. Michael M. Krop High School
Atomic Symbols
Mass Number (A)
(p+ + n0)
238
92
U
Atomic number (Z)
(number of p+)
Atomic Symbol
Isotopes of Uranium
Uranium - 238
238
92
U
n0 = 146
Uranium - 235
235
92
U
n0 = 143
Uranium - 233
233
92
U
n0 = 141
What is the difference between these three isotopes?
The number of neutrons!
Radioactivity
The spontaneous disintegration of an
unstable atomic nucleus with
accompanying emission of radiation.
Radioactivity
 Only unstable nuclei decay. The majority of
atoms in nature do not decay
 Radioactive nuclei are called Radionuclei.
Atoms containing these nuclei are called
Radioisotopes
 The stability of the nucleus depends on
number of protons and neutrons
Radioactive Emissions α, β, γ
• Alpha particles (α) are the nucleus of Helium (2 protons, 2
neutrons) +2 charge
• Beta particles (β) are high-speed electrons. -1 charge.
• Gamma Rays (γ) are high energy radiation, not particles. Zero
Charge
Separation of α, β, γ emissions
MOVIE
Penetrating ability of α, β, γ radiation
Alpha (α) Emissions
 Helium Nucleus (2 p+ and 2 n0)
4
4
 Notation: 2 He or 2 
Uranium-238 Alpha decay
U 

238
92
Th 
234
90
4
2
He
After alpha decay Uranium changed to Thorium
Alpha (α) Emissions
 What product is formed when radium-226
undergoes alpha decay?
226
88
Ra 
 X
A
Z

4
2
He
2 protons are lost and 2 neutrons so mass number goes down
by 4 and atomic number goes down by 2
226
88
Ra 

222
86
Rn

4
2
He
Beta (β) Emissions
 High Speed electrons (1 e-)
0
0
e
 Notation: 1 or 1 
Iodine-131 Beta decay
I 

131
53
Xe 
131
54
0
1
e
After beta decay Iodine changed to Xenon
Beta (β) Emissions
 What product is formed when Thorium-231
undergoes beta decay?
Th 
 X
231
90
A
Z

0
1
e
No protons are lost and 1 neutron turns into a proton. Atomic
number goes up by one while mass number stays constant
Th 

231
90
Pa 
231
91
0
1
e
Beta (β) Emissions
 If there are no electrons in the
nucleus, where do beta particles
come from?
 Why does the atomic number of an
atom increases by one when there is
beta decay?
Beta (β) Emissions
 The reason is unknown.
 Possible theories involve quarks or particles
which supposedly compose protons and
neutrons
 However, it is known that a neutron
disappears and somehow turns into a proton
1
0
n
 e 
0
1
1
1
p
Positron Emissions
• Same mass 0as electrons but opposite charge
e

1
• Notation:
Carbon-11 Positron decay
C
 B 
11
6
11
5
0
1
e
After positron decay carbon changed to boron
Positron Emissions
• What product is formed when oxygen-15
undergoes positron emission?
O
 X
15
8
A
Z

0
1
e
No neutrons are lost and 1 proton turns into a neutron. Atomic
number goes down by one while mass number stays constant
O
 N
15
8
15
7

0
1
e
Gamma Radiation (γ)
 High energy photons
 Very short wavelength
 Always accompanies other radioactive
emission
 Represents the energy lost when the
remaining nucleons reorganize
Why does radioactivity happen?
 If protons repel each other, how can
they stick together in the nucleus?
 What forces are present in the
nucleus?
STRONG NUCLEAR FORCES
Strong nuclear forces
 When nucleons get close enough a
very strong force binds them together
overcoming electrostatic repulsion
 Neutrons are not repelled so they
help to cement protons together
 As soon as protons are slightly apart
repulsion overcomes strong nuclear
force.
Strong Nuclear Forces
All nucleons attract one another
by Strong Nuclear forces.
Only protons repel one
another
Patterns of Nuclear Stability
 As Z increases the
number of neutrons
must increase to
guarantee stability
 All nuclei with 84 or
more protons are
radioactive and tend
to undergo alpha
decay
Radioactive Series of Uranium
 Uranium-238 does
not become stable
with only one
decay
 This nuclei undergo
a combination of
alpha and beta
decay and finally
reaches stability as
lead (Pb)
Nuclear Transmutation
 Nuclear reactions which are induced (not
spontaneous.)
 Induced by a high speed nucleon striking the
nucleus
 Example: An alpha particle strikes a nitrogen
nucleus to synthesize Oxygen-17
14
7
N

He 
 O 
4
2
17
8
1
1
H
Particle Accelerators
 Charged particles have to be accelerated in
order to overcome the repulsive forces in
the nucleus.
Particle accelerator at Fermi National Lab.
The circumference is 6.3 km.
Using Neutrons
 Neutrons are not charged, therefore they do not
need to be accelerated.
 Cobalt-60 is an important isotope used in radiation
therapy for cancer. It is produced from iron-58
which is bombarded by neutrons
Fe  n 

59
26
Fe 

59
27
Co  n 

60
27
58
26
1
0
59
26
59
27
1
0
Fe
Co 
Co
0
1
e
Transuranium Elements
 Man-made elements heavier than Uranium.
 Many of these only exist for brief moments and then
decay rapidly
 In 1940 elements 93 (Neptunium) and 94 (Plutonium)
were discovered by bombarding Uranium with
neutrons
U  n

238
92
1
0
239
93
Np 

U

239
92
Pu 
239
94
0
1
Np 
239
93
e
0
1
e
Rates of Radioactive Decay
 Different nuclei undergo radioactive
decay at different rates
 Some decay very rapidly (seconds)
others decay very slowly (years)
 Radioactive decay is a first-order
process
First Order Process
 The rate of decay is proportional to
the number of radioactive nuclei (N)
Rate = kN
 Where k is the decay constant
 The more nuclei you have the faster
it decays and vice versa
Half-life
 Time it takes for half of the nuclei to decay.
 Half-life is always constant
First Order process and half-life
MOVIE
Half-life problem
• The half-life of cobalt-60 is 5.3 yr. How much of
a 1000-mg sample of cobalt-60 is left after a
15.9-yr period?
Every half life the initial amount is reduced by half
How many half-lives is 15.9 yrs?
15.9 yr
 3 half lives
5.3 yr
Half-life problem cont’
After three half-lives how much cobalt-60 will
there be if we started with 1000 mg?
1000 mg
500 mg
 500 mg ,
 250 mg,
2
2
250 mg
 125 mg
2
So after 3 half-lives there will be 125 mg of
The previous series is equivalent to:
60Co
1000 mg
mf 
23
Half-Life Calculations
 General formula for Half-Life
problems:
mi
mf  n
2
mf = final mass
mi = initial mass
n = Number of Half-Lives
Another Half-Life Problem
 The half-life of 131I is 0.022 yr. On July 12th
there are 52.3 mg of this radioisotope. How
much 131I was there on June 22nd?
We must find the initial mass mi. Therefore:
mi  m f  2
Number of Half-lives
Answer:
n
1 hl
1 yr
n

 20 d  2.49 hl
0.022 yr 365 d
mi  52.3 mg  2
2.49
 294 mg
Nuclear Binding Energy (Mass Defect)
 Recall: The mass of the carbon-12
nucleus is exactly 12 amu. The mass
of 6 protons and 6 neutrons is 12.096
amu
 How can the mass of 12 subatomic
particles separate be more than their
mass when they combine?
Nuclear Binding Energy (Mass Defect)
The mass and energy of an
object are proportional. If a
system loses mass it loses
energy; if it gains mass it
gains energy
E = mc2
E = Energy
m = mass
c = Speed of light
Nuclear Binding Energy (Mass Defect)
 In every process when energy is released,
mass also changes.
 In chemical reactions the change of mass is
too small to be detected.
 In nuclear reactions the mass change is
very significant.
 Because “c” is so large, even small changes
of mass release huge amounts of energy
Nuclear Binding Energy (Mass Defect)
 Nuclear Binding Energy:
Energy required to separate a nucleus
into its individual nucleons.
 Mass Defect:
Mass difference between a nucleus
and its constituent nucleons
Finding Mass defect and binding
energy for 12C
The mass of one
12C
12C
atom is EXACTLY 12 amu.
has 6 protons and 6 neutrons.
6 x (1.0073) + 6 x (1.0087) = 12.096 u
Mass Defect (Δm) = 12.096 – 12 = 0.096 u
Finding Mass defect and binding energy for
 To find the nuclear binding energy in Joules we must
convert the mass defect to kilograms.
Recall that 1 Joule = 1 kg.m2/s2
Converting to kilograms:
1.6605 10 27 kg
 0.096 u  1.59412 10 28 kg
1u
Finding the Binding Energy “E”
E  mc2  (1.59412 10 28 kg)  (2.99792 108 m / s) 2
E  1.433 10 11 J
12C
Nuclear Binding Energy of Isotopes
Nuclear Fission
 The splitting of a larger nucleus into
smaller ones.
 The splitting of the nucleus produces
a mass change; therefore, energy is
released.
 The fission of a heavy atom like
Uranium is a highly exothermic
process
Nuclear Fission
The splitting of Uranium produces 2 smaller nuclei and 3
neutrons. These neutrons can collide against other atoms of
uranium and make them split as well.
Chain Reaction
 The original fission causes 2 or 3 neutrons. These
neutrons can cause 2 or 3 more fissions and
produce 6 or 9 more neutrons. The process
escalates quickly and violently if not controlled.
Critical Mass
 A chain reaction will not occur unless a certain critical
mass of fuel is present. The critical mass of 235U is about
1 kg.
 If there is less than the critical mass some of the
neutrons may escape and miss the nuclei of uranium and
the reaction fizzles out.
Fission reactors
Cadmium or boron control rods regulate the flux of neutrons
Nuclear Fusion
 The joining of two lighter nuclei to form a
more massive one.
 When two light nuclei fuse into a heavier
one a great amount of energy is released
 Fusion reactions are responsible for energy
produced by the Sun.
 Electrostatic repulsion is the largest
obstacle to overcome when trying to fuse
two light atoms
Nuclear Fusion
 To overcome repulsion an extreme high
temperature environment is necessary.
 The fusion reaction that requires the least
temperature is the fusion of tritium (3H) and
deuterium (2H)
2
1
H H 
 He  n
3
1
4
2
1
0
The temperature necessary for this reaction is 40,000,000 K
Nuclear Fusion
 Such high temperatures have been
achieved using atomic bombs.
 Could we control fusion and harness
its energy?
Plasma Fusion Movie
Uses of Nuclear Energy
 Nuclear Reactors (Power Plants)
 Nuclear Weapons (Fission bomb, HBomb)
 Nuclear Medicine (Drugs,
Radiotherapy, Scanning)
 Food irradiation