
The Mass Defect of the Nucleus and Nuclear
Binding Energy  


Two ideas that we have studied previously come into play as we discuss the
binding energy of a nucleus. These are the rest energy of an object (Section
28.6) and mass (Section 4.2). In Einstein’s theory of special relativity, energy
and mass are equivalent; in fact, the rest energy
and the mass
m are related via
(Equation 28.5), where
c is the speed of light in a vacuum.
Therefore, a change
in the rest energy of the system is equivalent to a
change
in the
mass of the system, according to
. We see, then, that the binding energy used in
Figure 313 to disassemble the nucleus appears as extra mass of the separated
and stationary nucleons. In other words, the sum of the individual masses of the
separated protons and neutrons is greater by an amount
than the mass of the stable
nucleus. The difference in mass
is known as the
mass defect of the
nucleus.
As Example 2 shows, the binding energy of a nucleus can be determined from
the mass defect according to Equation 31.3:

Example 2 
 
The Binding Energy of the Helium
Nucleus  
The most abundant isotope of helium has a nucleus whose mass is . For this
nucleus, find (a) the mass defect and (b) the binding energy.
Reasoning The symbol indicates
that the helium nucleus contains protons and neutrons. To obtain the mass
defect ,
we first determine the sum of the individual masses of the separated
protons and neutrons. Then we subtract from this sum the mass of the nucleus.
Finally, we use Equation 31.3 to calculate the binding energy from the
value for .
Solution

a. 
Using data from Table 311, we find that the sum of the
individual masses of the nucleons is
This value is
greater than the mass of the intact nucleus, and the
mass defect is


b. 
According to Equation 31.3, the binding energy is
Usually, binding
energies are expressed in energy units of electron volts
instead of joules :

 
In this result, one million electron volts is denoted by the unit
MeV. The value of 28.3 MeV is more than two million times greater than the
energy required to remove an orbital electron from an atom. 

In calculations such as that in Example 2, it is customary to use the
atomic mass unit (u)
instead of the kilogram. As introduced in Section 14.1, the atomic mass unit is
onetwelfth of the mass of a
atom of carbon. In terms of this unit, the mass of
a
atom is
exactly 12 u. Table 311 also gives the masses of the electron, the proton, and
the neutron in atomic mass units. For future calculations, the energy equivalent
of one atomic mass unit can be determined by observing that the mass of a proton
is
or 1.0073
u, so that
and
In electron volts, therefore, one atomic mass
unit is equivalent to
Data tables for isotopes, such as that in Appendix F, give masses in atomic
mass units. Typically, however, the given masses are not nuclear masses. They
are
atomic masses—that is, the masses of
neutral atoms, including the mass of the orbital electrons. Example 3 deals
again with th
nucleus and shows how to take into account the effect of the orbital electrons
when using such data to determine binding energies.

Example 3 
 
The Binding Energy of the Helium Nucleus,
Revisited  
The atomic mass of is 4.0026 u, and the atomic mass of is 1.0078 u.
Using atomic mass units instead of kilograms, obtain the binding energy of
the
nucleus.
Solution The sum of the
individual masses is
The mass defect is . Since 1 u is equivalent to
931.5 MeV, the binding energy is , which matches that obtained in Example
2. 

Check Your Understanding
2 
The following table gives values for the mass defect for four hypothetical
nuclei: A, B, C, and D. Which statement is true regarding the stability of
these nuclei? (a) Nucleus D is the most stable, and A is the least stable.
(b) Nucleus C is stable, whereas A, B, and D are not stable. (c) Nucleus A
is the most stable, and D is not stable. (d) Nuclei A and B are stable,
but B is more stable than A. (The answer
is given at the end of the book.)

A 
B 
C 
D 
Mass defect, 


0 kg 
 
Background: The mass
defect of a nucleus is the sum of the individual masses of the separated
nucleons minus the mass of the intact nucleus. The binding energy of a
nucleus is the energy required to separate it into its constituent protons
and neutrons. The greater the binding energy, the more stable is the
nucleus. The binding energy and the mass defect are related.
For similar
questions (including calculational counterparts), consult SelfAssessment
Test 31.1,
which is described at the end of Section
31.5. 

Copyright © 2007 John Wiley & Sons,
Inc. All rights reserved. 