Nuclear Reactions
Natural Transmutation
1 term on reactant side
Original isotope
2 terms on product side
Emitted Particle
New Isotope
Happens all by itself (spontaneous)
Not affected by anything in environment
Natural Transmutation
16N
7

1 term on
reactant side
0e
-1
+
16O
8
2 terms on
product side
Artificial Transmutation
• Cause it to happen by smashing
particles into one another
• 2 terms on reactant side
• Original Isotope
• Particle that hits it
–neutron, proton, or -particle
• Product side: usually 2 terms
Artificial Transmutation
13
27Al
+
4He
2
Original isotope
or target nucleus

30P + 1n
15
0
“Bullet”
-what hits isotope
Artificial Transmutation
27Al
13
14N
7
+ 4He  30P + 1n
2
+ 42He 
15
17O
8
75As
+ 4He 
37Cl
17
+ 1n  38
Cl
17
33
2
0
0
+ 11H
78Br
35
+ 1n
0
All of these
equations have
2 reactants!
Bombarding with Protons or 
Protons and -particles have positive charge and
mass
• do some damage when hit target nucleus
• must be accelerated to high speeds to overcome
repulsive forces between nucleus & particle (both are
+)
What is an accelerator?
• vacuum chamber (usually a long pipe)
– surrounded by vacuum pumps, magnets, radiofrequency cavities, high voltage instruments and
electronic circuits
• inside the pipe particles are accelerated to
very high speeds then smashed into each
other
Fission Reaction

Splitting heavy nucleus into 2 lighter nuclei

Requires a critical mass of fissionable isotope
Controlled – nuclear reactor
Uncontrolled – bomb
Fission
Reactant side: 2 terms
• 1 heavy isotope (examples: U-235 or Pu-239)
• Bombarding particle – usually a neutron
• Product side: at least 2 terms
• 2 medium-weight isotopes
• 1 or more neutrons
• Huge amount of energy is released
• Fission = Division
Fission
235U
+ 1n 
91Kr
36
+ 142Ba + 31n + energy
235U
92
+ 1n 
72Zn
30
1n + energy
+ 160
Sm
+
4
62
92
0
0
56
0
0
More than 200 different product isotopes
identified from fission of U-235
A small amount of mass is converted to
energy according to E = mc2
Fission Chain Reaction
Fusion
• Reactant side has 2 small nuclei:
– H + H;
H + He;
He + He
• Product side:
– 1 nucleus (still small) and maybe a particle
• Source of sun’s energy
• 2 nuclei unite
2H
1
+ 3H  4He + 1n + energy
1
2
0
CERN
27 kilometer ring
•Particles travel just below
speed of light
•In 10 hrs: particles make 400
million revolutions of the ring
FermiLab
4 miles in circumference!
Balancing Nuclear
Equations
Nuclear Equations - tasks
• Identify type (4 types)
• Balance to find 1 unknown term
Natural Transmutation – ID
• 1 term on reactant side
– starting isotope
• 2 terms on product side
– ending isotope and emitted particle
• Type of particle emitted characteristic
of isotope – Table N
Nuclear Equations
• To balance: use conservation of both
atomic number & mass number
• Mass number = left superscript
• Atomic Number = left subscript
Balancing Nuclear Equations
16N
7

0e
-1
+
16O
8
Conservation of mass number:
16 = 0 + 16
Conservation of atomic number:
7 = -1 + 8
Writing Equations
• Write the equation for the decay of
Thorium-232
• Use Table N to find the decay mode: α
• Write the initial equation:
232Th  4He + X
90
2
figure out
what element
it turned into
Write an equation for the α
decay of Am-241
241
95
Am  4He + YX
2
What’s X?
Z
so Y = 228
232 = 4 + Y
232Th
90

4He
2
+
Y
X
Z
Conservation of Mass Number:
sum of mass numbers on left side must =
sum of mass numbers on right side
232Th
90
 42He + 228
X
Z
90 = 2 + Z
so Z = 88
Conservation of Atomic Number:
sum of atomic numbers on left side must =
sum of atomic numbers on right side
232Th
 4He + 228X
2
90
88
Use the PT to find X:
232Th
90
 4He + 228Ra
2
88
X = Ra
Alpha (α) decay:
233U

92
232Th
90
78
+ 4He
2
 228Ra + 4He
90
175Pt
229Th
88

171Os
76
2
+ 4He
2
How does the mass number or atomic
number change in α,β or γ decay?
• go to Table N:
– find isotope that decays by alpha or β decay
– write the equation
– see how the mass number (or atomic number)
changes
•
226 Ra
88

4 
2
• X is Rn-222
+ X so X has to be
222 X
86
– mass number decreases by 4; atomic number
decreases by 2
Write an equation for the 
decay of Am-241
241 = 4 + Y
241 Am
so Y = 237
 4He + YX
95
2
Z
95 = 2 + Z
What’s X?
so Z = 93
X = Np
Radioactive Decay Series
• Sometimes 1 transmutation isn’t enough
to achieve stability
• Some radioisotopes go through several
changes before they achieve stability
(and are no longer radioactive)
β-
β+
18F
9
14C
6
 147 N +
 18
O
+
8
0e
+1
0e
-1
How does the mass number or atomic
number change in  or  decay?
• Go to Table N; find an isotope that decays by α,
 or , write the equation; see how the mass
number (or atomic number) changes
• 226Ra  4 + X so X has to be 222X
88
2
• X is Ra-222
– mass number decreases by 4
– atomic number decreases by 2
86