Nuclear Reactions
Elementary Particles
The only atomic particles that play a part in
nuclear reactions are the protons and the neutrons;
electrons do not play a part in nuclear reactions.
We use the following notation to describe the
specific element we are considering in a nuclear
reaction.
X represents the element, Z represents the atomic
number (number of protons), and A represents the
atomic mass number (the number of protons and
neutrons) of the element.
A
Z
X
Isotopes
For the nuclides below, determine the number of neutrons and
protons.
A  # p  #n
Z #p
87
41
Nb
48
24
Cr
60
29
Cu
44
18
Ar
147
69
Tm
Atomic Radii
Use the formula below in order to find the atomic radii of the
following Nuclide.

r  1.2  10 m  A


15
28
12
Mg

1
3




Particle Masses
Nuclear masses are specified in unified atomic mass units or amu
The atomic mass of a neutron is 1.008665 u.
The atomic mass of a proton is 1.007276 u.
A neutral hydrogen atom 11H , which has an electron and a proton but no neutron,
has a mass of 1.007825 u.
We will use the mass of a neutral hydrogen in the place of the mass of a proton.
The mass of a 24 He nucleus is known to be 4.002602 u.
Compare this mass to the masses of the appropriate number of protons and
neutrons combined.
2m0 p  2m 1n  2(1.007825u)  2(1.008665u)
1
0
What did you discover?  4.03298u
In the question above, you found that the actual mass was less than the mass of its
constituent parts.
What do you think happened to the “missing mass?”
Binding Energy
The difference in masses discovered on the previous slide is known
as the total binding energy of the nucleus.
This energy represents the amount of energy that must be put into
the nucleus in order to break apart its protons and neutrons.
For a given nucleon Az X, the total binding energy, Eb, is

Eb  931.5MeV  Zm 1H   ( A  Z )mn  m A X
1
Z

Find the amount of energy put into the following nuclide in order to
break it apart.
24
12
Mg
Nuclides
This notation is very important because it allows us to represent different
isotopes of an element X.
For a given atom, carbon for instance, nuclei are found that contain
different numbers of neutrons even though they contain the same amount of
protons.
Here are the symbols of some different isotopes of carbon.
They are all Carbon because they all have 6 protons; however, they have
different numbers of neutrons.
These “different” carbons are known as Isotopes of each other.
The carbon to the left is known as “Carbon 12” because it has six neutrons
and six protons.
The carbon to the right is known as “Carbon 16” and has six protons and
ten neutrons.
A
Z
X
12
6
C
13
6
C
14
6
C
15
6
C
16
6
C
Alpha Decay
Alpha decay is one of several types of nuclear decay.
In alpha decay a parent nucleus is broken apart to yield a daughter
nucleus and an alpha particle.
The general equation for alpha decay is as follows.
An alpha particle is a neutral helium atom.
A
Z
X
A4
Z 2
Y  He
4
2
Q-Value
In alpha decay, the masses of the daughter nucleus and alpha particle combined are
less than the mass of the parent nucleus.
The “missing mass” is converted into the kinetic energy of the alpha particle and
the daughter nucleus.
This “missing mass” or released energy is known as the disintegration energy.
It is also known as the Q-value for the particular parent nucleus.
The equation used to determine the Q-value is


Q  m A X  m A 4Y  m 4 He c 2
Z 2
Z
2
Find the daughter nuclide due to alpha decay and the Q-values or the nuclides
below.
209
83
Bi
Beta Minus Decay
In Beta decay a parent nucleus is broken apart to yield a
daughter nucleus and a Beta particle.
A
A Beta particle is either an electron (e-) or a positron (e+). Z
If an electron (also known as a -) is emitted during the
decay process, then the decay process is known as a “Beta
minus decay.”
The general equation for beta minus decay is as follows.
The underlying reaction is the conversion of a neutron into
a proton, electron, and an anti-neutrino.
A
Z
X
A
Z 1
Y
X  Z A1Y     
Neutron Proton
n  p    


Beta Minus Decay
•Calculate the KE for Beta Minus decay for
A
Z
X  Y   

A
Z 1
32
15
32
15
P
n  p    
32
P  16
S  e 
m  M p  (M s  e )
0
KEmax
M  mass of the nucleus
m  (M p 15e )  (M s  e 15e )
m  (M p  15e )  (M s  16e )
m  mp  ms
Q=[mAZ - mAZ+1 ]c2
Beta Plus Decay
If a Positron (also known as a +) is emitted
during the decay process, then the decay process is
known as a “Beta plus decay.”
The general equation for beta plus decay is as
follows.
The underlying reaction is the conversion of a
proton into a neutron, positron, and a neutrino.
A
Z
X
A
Z 1
Y
Neutron
Proton
A
Z
X  Z A1Y     
p  n   



Beta Plus Decay
•Calculate the KE for Beta Minus decay for
A
Z
X  Y   

A
Z 1
22
11
22
11
Na
n  p    
22
Na  10
Ne  e 
m  M Na  (M Ne  e )
0
KEmax
M  mass of the nucleus
m  (M Na 11e )  (M Ne 11e  e )
m  (M Na  11e )  (M Ne  10e  2e )
m  mNa  mNe  2e
Q=[mAZ - mAZ-1 -2me ]c2
Electron Capture
In electron capture, an orbital electron is captured by
the nucleus, combines with a proton, and forms a
neutron and a neutrino.
The general equation for electron capture is as
follows.
The underlying equation for electron capture is the
conversion of a proton and an electron into a neutron
and a neutrino.
A
Z
X
A
Z 1
Y
A
Z
Neutron Proton
X  Y 
A
Z 1
p     n 

