25.1. The Nucleus http://www.ck12.org
Answer: True. Both are nitrogen nuclei. Nitrogen^147 Nis the common form of nitrogen having an equal number of
protons and neutrons, and Nitrogen^157 Nis an isotope of nitrogen having one more neutron than proton.
Binding energy
Experimental evidence shows that the total mass of a stable nucleus is smaller than the sum of the masses of its
constituent particles. The energy equivalent of this mass difference(∆m), or mass deficit, is called thebinding
energyof the nucleus.
Note: Most books give the mass of the atom which includes the mass of the electrons. This is of no concern as long
as the mass of the electrons cancel out during the calculation.
An example will help clarify this idea:
Consider an isotope of hydrogen calleddeuterium
( 2
1 H
)
. It is composed of one proton and one neutron and one
electron.
The mass of the deuterium atom is
mD= 2. 014102 u
The mass of one neutron: 1. 008665 u.
The mass of one hydrogen atom (one proton and one electron): 1. 007825 u.
The total mass of the neutron and hydrogen atom:
mtotal= 1. 008665 u+ 1. 007825 u= 2. 01649 u
Therefore, the mass deficit is:
mtotal−mD=∆m= 201649 u− 2. 014102 u= 0. 002388 u.
It can be experimentally verified that the excess mass (the mass deficit) is given off as energy when nucleons form a
deuterium nucleus. We can compute the energy using the Einstein equation:
E=∆mc^2 →
First, we convert∆minto kilograms and then compute the energy. Lastly, we express the answer in millions of
electron volts (MeV).
∆m= ( 0. 002388 u)1. 6605 × 10 −^27 kg
1 u
= 3. 965274 × 10 −^30 kg→E= 3. 965274 × 10 −^30 kg(
2. 997 × 108
m
s) 2
= 3. 561613 × 10 −^13 J
We have usedc= 2. 997 × 108 msinstead ofc= 3. 00 × 108 msfor the speed of light in the above calculation.
Recall: 1. 6 × 10 −^19 J= 1 eV→( 1. 6 × 10 −^19 J)( 106 ) = ( 1 eV)( 106 )→ 1. 6 × 10 −^13 J= 1 MeV
E=^3.^561613 ×^10
− (^13) J
- 60 × 10 −^13 MeVJ =^2.^226 →^2.^23 MeV
The result is the binding energy for deuterium. This is the amount of energy that is needed to separate the nucleons
from each other.
It is typical to express the total binding energy in terms of the average binding energy per nucleon. Since deuterium
is composed of two nucleons, we have:
Average binding energy per nucleon (see alsoFigure25.2)=^2.^2262 = 1. 113 → 1. 11 MeV
There are 92 naturally occurring elements, and as stated before, the most stable atoms are those that have the same
number of protons as neutrons. These atoms then have the greatest binding energies per nucleon, as shown inFigure