Transcript Nuclear Reactions

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 to happen: smash 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
27Al
13
+
original isotope
or
target nucleus
4He
2

30P
15
+ 1n
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 these equations
have 2 reactants!
Bombarding with protons or 

Protons & -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 long pipe)
• surrounded by vacuum pumps, magnets, radiofrequency cavities, high voltage instruments &
electronic circuits

inside pipe particles are accelerated to very
high speeds then smashed into each other
Fission Reaction

Splitting heavy nucleus into 2 lighter nuclei
Requires critical mass of fissionable isotope
• controlled: nuclear reactor
• uncontrolled: bomb
Fission

Reactant side: 2 terms



Product side: at least 2 terms




1 heavy isotope (examples: U-235 or Pu-239)
bombarding particle – usually a neutron
2 medium-weight isotopes
1 or more neutrons
huge amount energy released
Fission = Division
Fission Chain Reaction
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
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
+ 13H  42He + 01n + energy
CERN
27 kilometer ring
•Particles travel just below
speed of light
•In 10 hrs: particles make 400
million revolutions of 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 & 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 equation for decay of Thorium-232
use Table N to find decay mode: α
write initial equation:
232Th  4He + X
90
2
figure out
what element
it turned into
What’s under the hat?
Little cats X, Y, & Z!
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 mass numbers on left side must
=
sum mass numbers on right side
232Th
90
 4He + 228X
2
90 = 2 + Z
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 +
90
228X
2
88
X = Ra
use PT to find X:
232Th
90
 4He + 228Ra
2
88
Alpha (α) decay:
233U

92
232Th
90
175Pt
78
229Th
90
+ 4He
2
 228Ra +
88

171Os
76
+
4He
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 α or β decay
• write equation
• see how mass number (or atomic number) changes

226 Ra
88

X is Rn-222

4 
2
+X
so X has to be
• mass number decreases by 4
• atomic number decreases by 2
222 X
86
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 achieve stability (no
longer radioactive)
β+
β-
14C
18F
 18
O
+
8
9
6
 147 N +
0e
+1
0e
-1
How does the mass number or
atomic number change in  or 
decay?

go to Table N; find isotope that decays by α, 
or ; write equation; see how mass number (or
atomic number) changes
226Ra  4 + X so X has to be 222X

X is Ra-222

88
2
• mass number decreases by 4
• atomic number decreases by 2
86