Binding Energy
Nuclear Physics Lesson 13
Learning Objectives

To define binding energy.

To define mass defect.

To know which are the most stable nuclei.

Explain why energy is released in nuclear fission
and nuclear fusion.
The Atomic Mass Unit

The atomic mass unit (u) is far more
convenient to use with nuclear masses.

It uses carbon-12 as a reference and is defined
as:
Exactly 1/12th the mass of a carbon 12 atom.

1 atomic mass unit (u) = 1.661 × 10-27 kg.
The table shows particle masses in
atomic mass units.

Note that the numbers are expressed to a large number
of significant figures as the changes are quite subtle.
Particle
Mass (u)
Electron
Neutron
Proton
Hydrogen atom (1p+ + 1e-)
0.000549
1.008665
1.007276
1.007825
Helium atom (2p+ + 2n + 2e-)
4.002603
α particle (2p+ + 2n)
4.001505
Be Careful!

Remember to distinguish between the atomic
mass and the nuclear mass.
The atomic mass is the mass of an atom complete
with its electrons.
 The nuclear mass is the mass of the nucleus
alone. To get the nuclear mass we need to take away
the mass of the electrons.

Binding Energy

If you want to remove a nucleon from the
nucleus of an atom, then you have to do work
to overcome the strong nuclear force.

Definition from specification book:-

The binding energy of a nucleus is the work that
must be done to separate a nucleus into its
constituent neutrons and protons.

The removed nucleons gain potential energy.
Binding Energy II

If we run that process in reverse, then the
binding energy can also be defined as the energy
released when a nucleus is assembled from its
constituent nucleons.

This means that the mass of the nucleus is less
than the mass of the separate nucleons because
energy has been released…
Mass Defect


Definition from the specification book:-
The mass defect Δm of a nucleus is defined as
the difference between the mass of the
separated nucleons and the mass of the nucleus.
Calculating Mass Defect

For a nucleus of an isotope ZAX, composed of Z
protons and (A-Z) neutrons with mass MNUC,
the mass defect, Δm, is given by:-
m  ZmP  ( A  Z )mN  M NUC
Question 1

What is the mass defect of a helium atom?
Answer 1

What is the mass defect of a helium atom?
Particle
Mass (u)
Number
Total (u)
Proton
Neutron
1.007276
1.008665
2
2
2.014552
2.017330
Electron
0.000549
2
0.001098
Total
4.032980

The atomic mass of a helium atom is 4.002603 u

Therefore mass defect, Δm=0.030377 u
Calculating Binding Energy

The mass defect Δm exists because energy is
released when the constituent nucleons bind
together to form a nucleus.

The energy released is equal to the binding
energy of the nucleus:-
binding energy of a nucleus  mc

2
Note that Δm must be in kg to get energy in J.
Question 2

What is the binding energy of the helium atom whose
mass defect is 0.030377 u? Express your answer in
MeV.

m = 0.030377 u × 1.661 × 10-27 kg = 5.046 × 10-27 kg


E = mc2 = 5.046 × 10-27 kg × (3 × 108)2 = 4.541 × 1012 J
E = 4.541 × 10-12 J = 28.38 × 106 eV= 28.38 MeV

Note: 1 u = 931.3 MeV
Binding Energy Per Nucleon

If we know the binding energy in a nucleus, and the
number of nucleons, we can work out the binding
energy per nucleon, which is the work done needed to
remove each nucleon.

The higher the binding energy per nucleon, the more
stable is the nucleus. For helium (4He) the binding
energy per nucleon is:

Binding energy per nucleon = 28.38 MeV/4 = 7.1 MeV
Question 3

What is the mass defect in atomic mass units (u)
and in kilograms for the lithium nucleus which
has 7 nucleons, and a proton number of
3? What is the binding energy in J and
eV? What is the binding energy per nucleon in
eV? The nuclear mass = 7.014353 u.
Answer 3






Add them together to get 7.056488 u
Now take away the nuclear mass from the number above to get
the mass deficit. 7.056488 u - 7.014353 u = 0.042135 u
Now convert to kilograms: 1 u = 1.661 ´ 10-27 kg 0.042135 u ×
1.661 ´ 10-27 kg = 6.9986235 × 10-29 kg
Now use E = mc2 to work out the binding energy: E =
6.9986235 × 10-29 kg × (3 × 108 m/s)2 = 6.3 × 10-12 J
In electron volts, this is 6.3 × 10-12 J ÷ 1.6 × 10-19 eV/J = 3.9
× 107 eV = 39 MeV.
There are 7 nucleons so the binding energy per nucleon = 3.9 ×
107 eV ÷ 7 = 5.6 × 106 eV
Question 4

What is the mass defect in atomic mass units (u)
and in kilograms for the copper nucleus which
has 63 nucleons, and a proton number of
29? What is the binding energy in J and
eV? What is the binding energy per nucleon in
eV? The nuclear mass = 62.91367 u.
Answer 4









Number of protons = 29; number of neutrons = 63 – 29 = 34
Mass of protons = 29 ´ 1.007276 = 29.211004 u
Mass of neutrons = 34 ´ 1.008665 = 34.29461 u
Total mass = 29.211004 u + 34.29461 u = 63.505614 u
Mass defect = 63.505614 u – 62.91367 u = 0.591944 u
Mass defect in kg = 0.591944 ´ 1.661 ´ 10-27 = 9.83218 ´ 10-28
kg
Binding energy = mc2 = 9.83218 ´ 10-28 kg ´ (3 ´ 108 m/s)2 =
8.85 ´ 10-11 J
Binding energy in eV = 8.85 ´ 10-11 J ¸ 1.6 ´ 10-19 J/eV = 5.53 ´
108 eV
Binding energy per nucleon = 5.53 ´ 108 eV ¸ 63 = 8.78 ´ 106
eV
Binding Energy per Nucleon

We can plot a graph of binding energy per
nucleon against nucleon number.
From this graph we can see that




The maximum value is about binding energy of 8.7
MeV per nucleon and iron is the most stable nuclide.
Helium has a particularly high value of binding energy
per nucleon, much higher than the light isotopes of
hydrogen.
There is a trend for nuclides of nucleon numbers in
multiples of 4 to be particularly stable (i.e. have a high
binding energy).
The largest nuclides tend to be less stable, with slightly
lower binding energies per nucleon.
The Most Stable Nucleus




Iron has the highest binding energy per nucleon so is
the most stable nucleus.
If we look at large nuclei (greater than iron), we find
that the further to the right (greater nucleon number)
the less stable the nuclei.
This is because the binding energy per nucleon is
getting less.
The explanation for this observation lies in that the
strong nuclear force that binds the nucleus together has
a very limited range, and there is a limit to the number
of nucleons that can be crammed into a particular
space.
Nuclear Fission

A large unstable nucleus splits into two
fragments which are more stable than the
original nucleus.

The binding energy per nucleon increases in this
process and energy is released.

The change in binding energy per nucleon is
about 0.5 MeV in a fission reaction.
Nuclear Fusion

Small nuclei fuse together to form a larger
nucleus.

The product nucleus has a higher binding per
nucleon as long as A is no greater than ~50.

The change in binding energy per nucleon can
be more than 10 times greater in a fusion
reaction than a fission reaction.
Summary

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

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Atomic Mass Unit: 1/12th the mass of a carbon atom
Mass defect: Difference between the mass of
nucleons separately and together within a nucleus.
Difference between the two sides of a nuclear
interaction equation. Energy worked out by E =
mc2.
Binding Energy: Energy equivalent of the mass defect
in a nucleus. Binding energy per nucleon increases in
more stable nuclei.
Fission Splitting of a nucleus. Rarely
spontaneous. Occurs after the nucleus has been tickled
with a neutron
Fusion Joining together of two light nuclei to make a
heavier nucleus.