# Nuclear Reactions

```Nuclear Reactions
AP Physics B
Montwood High School
R. Casao
Nuclear Reactions

Nuclear reactions are rearrangements of
nuclear components that result from
bombardment by a particle rather than a
spontaneous natural process.
 Nuclear reactions are subject to several
conservation laws:
–
–
–
–
Charge
Momentum
Energy
Number of nucleons

When two nuclei interact, charge
conservation requires that the sum of the
initial atomic numbers must equal the sum
of the final atomic numbers.
 Because of conservation of nucleon
number, the sum of the initial mass numbers
must also equal the sum of the final mass
numbers.
 In general, these are not elastic collisions,
and therefore, the total initial mass does
NOT equal the total final mass.
Reaction Energy

The difference between the masses before and
after the reaction corresponds to the reaction
energy, according to the mass-energy relationship
E = m&middot;c2.
 If initial particles A and B interact to produce final
particles C and D, the reaction energy Q is defined
as:
Q = (MA + MB – MC – MD) &middot;c2
 To balance the electrons, use the neutral atomic
masses.
 Kinetic energy can be used if C and D are in the
ground state: Q = (KC + KD) – (KA – KB)
Reaction Energy

When Q is positive, the total mass decreases
and the total kinetic energy increases.
– Exothermic reaction.

When Q is negative, the total mass
increases and the kinetic energy decreases.
– Endothermic reaction.
– Reaction cannot occur at all unless the initial
kinetic energy is at least equal to Q.
– Threshold energy: the minimum kinetic energy
needed to make the reaction take place.
Example

When Li-7 is bombarded by a proton, two a
particles (He-4) are produced. Determine
the reaction energy.
1
7
4
4
H

Li

He

1
3
2
2 He
MeV
Q  1.007825 u  7.016003 u  (2  4.002603 u)  931 .5
u
Q  17.35 MeV

This is an exothermic reaction because the
final total kinetic energy is 17.35 MeV
greater than the initial total kinetic energy.
Example

Determine the reaction energy for the
nuclear reaction:
4
14
17
1
He

N

O

2
7
8
1H
Q  4.002603 u  14.003074 u  16.999131 u  1.007825 u 
931 .5 MeV
u
Q  1.191 MeV

This is an endothermic reaction.
 The minimum initial kinetic energy
(threshold energy) for this reaction to occur
is 1.191 MeV.

The relationship between the threshold
kinetic energy KEth and the absolute
magnitude of the reaction energy Q can be
determined using conservation of energy
and momentum:
KE th
m

 1    Q
M

– where m = mass of incoming particle and
M = mass of stationary target particle.

Ordinarily, the stationary N-14 nuclei
would be bombarded with a particles from
an accelerator.
The a particles kinetic energy must be
greater than 1.191 MeV.
 When a particle with mass m and kinetic
energy K collides with a stationary particle
with mass M, the total kinetic energy Ktotal
(the energy available to cause reactions) is:

K total
K total
MK

Mm
14.003074 u  1.191 MeV

14.003074 u  4.002603 u
K total  1.532 MeV
The kinetic energy of the a particles must be
at least 1.532 MeV.
 For a charged particle such as a proton or an
a particle to penetrate the nucleus of another
atom and cause a reaction, it must have
enough initial kinetic energy to overcome the
potential-energy barrier caused by the
repulsive electrostatic forces.

k  q1  q 2
U
r

For the reaction of the proton and Li-7, if we treat the
proton and the Li-7 nucleus as spherically symmetric
charges with radii given by R = Ro&middot;A1/3, their centers
will be 3.5 x 10-15 m apart when they touch.
 The repulsive potential energy of the proton and the Li-7
nucleus at this separation r is:
2
N

m
19
19
8.9875 x 10 9

3

1
.
602
x
10
C

1
.
602
x
10
C
2
C
U
3.5x10 15 m
U  2 x 10 13 J  1.2 MeV

Even though the reaction is exothermic, the proton must
have a minimum kinetic energy of 1.2 MeV for the
reaction to occur.
Neutron Absorption

Absorption of neutrons by nuclei forms is
an important class of nuclear reactions.
 Heavy nuclei bombarded by neutrons in a
nuclear reaction undergo a series of neutron
absorptions alternating with beta decays, in
which the mass number A increases by as
much as 25.
 Some of the trans-uranic elements, elements
having an atomic number larger than 92, are
produced in this way.
– These elements have not been found in nature.
Disintegration Energy (a Decay)
 Alpha
decay: parent  daughter + a
– Disintegration energy Q = (mp – md – ma)&middot;c2
– Disintegration energy appears in the form of
kinetic energy in the daughter nucleus and
the alpha particle.
– Alpha decay requires Q &gt; 0; if Q &lt; 0, the
parent nucleus is stable with respect to alpha
decay and alpha decay does not occur.
Disintegration Energy (a Decay)
 Of
the energy available to the daughter
and alpha particle as kinetic energy:
– larger mass has smaller kinetic energy.
– smaller mass has greater kinetic energy.
KE alpha  Q 
mdaughter
mparent
KE daughter  Q 
malpha
mparent
Disintegration Energy (-b Decay)
 Negative
–
beta decay:
parent  daughter + bDisintegration energy Q = (mp – md)&middot;c2
– Example:
14
14
6 C 7
0
N 1e 
v
– C-14 emits a negatively charged beta particle (an
ordinary electron) and the resulting nucleus is N-14.
– Might expect to determine Q using:
Q = (mp – md – mb)&middot;c2

This does not correctly represent the mass of all the
electrons in the parent and daughter atoms.
– mp and md are atomic masses and include the
electron masses.
– The mass for C-14 would be accurate because it
would include the mass of 6 electrons.
– The mass of N-14 would be incorrect because it
would include the mass of 7 electrons.
– In the N-14 formed by the beta decay there are only
the 6 electrons from the parent C-14 atom.
– To compensate for the fact that the daughter atom
contains the mass of 1 more electron than we need,
we do not write the mb term in the energy equation.

Correct equation for disintegration energy
for negative beta decay: Q = (mp – md)&middot;c2
 The nitrogen created in the beta decay of
carbon has only 6 orbiting electrons because
the parent carbon had only six electrons.
 The daughter is nitrogen by definition
because it contains 7 protons.
 The N-14 is positively ionized and will
attract an electron to complete its outer
electron shell.
Disintegration Energy (+b Decay)
 Positive
beta decay:
parent  daughter + b+
– Example: 13
13
0
7 N 6 C 1e
v
– N-13 emits a positively charged beta particle and the
resulting nucleus is C-13.
– The mass for the parent N-13 atom will be correct.
– The mass for the daughter C-13 atom includes the
mass for only the 6 electrons for a normal C-13
atom.

In our daughter C-13 atom, there are 7
electrons from the parent N-13 atom.
 The daughter C-13 is positively ionized.
We have to include an me = mb term in the
disintegration equation to account for this
positive ion.
 The mass of the positron is also mb.
 Disintegration energy:
Q = (mp – md – 2&middot;mb)&middot;c2

The decay of a free proton into a neutron, a
b+, and a neutrino is not possible because
the combined mass of the neutron and beta
particle is greater than the mass of the
proton.
p


1
0
0 n 1e
 ve
However, the same reaction is possible if
the proton is bound within the nucleus
because the required energy is provided by
the motion of the nuclei in the nucleus.
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