Question

The mass of \[_{3}L{{i}^{7}}\] nucleus is 0.042 amu less than the sum of masses of its nucleons. Find the binding energy per nucleon.

Answer 1

Hint: We need to understand the nuclear binding energies involved in a nucleus and the relation of this energy with the masses of the nucleus. The source of this energy which is required throughout the life of the nucleus will help us solve this question.

Complete answer:
The nucleus is the central part of an atom which consists of protons and neutrons. We can see from the types of charges possessed by these nucleons that the nucleus should have a strong force to withstand the Coulombic repulsions from these.
Also, the force which is due to the proton-proton repulsion is the life-long force which needs to be tackled always. The strong nuclear forces come into play in this case.
The nuclear forces are those forces which are powerful enough to keep the proton-proton-neutron system under such a small dimension to the scale of fermi. The nuclear force is a direct result of the energy mass equivalence. The mass of the nucleons when taken separately in an atom will be greater than the mass of the nucleus. This mass defect is converted to energy which is used for nuclear stability.
So, for the given \[_{3}L{{i}^{7}}\], the binding energy can be given as the difference in the mass of the nucleus and the total mass of individual nucleons as –
\[\begin{align}
  & E=m{{c}^{2}} \\
 & \Rightarrow BE=[p{{m}_{p}}+n{{m}_{n}}-M]{{c}^{2}} \\
 & \Rightarrow BE=\Delta M{{c}^{2}} \\
 & \Rightarrow BE=[0.042\times 1.66\times {{10}^{-27}}kg]{{c}^{2}} \\
 & \therefore BE=6.27\times {{10}^{-12}}J \\
 & \text{but,} \\
 & 1eV=1.6\times {{10}^{-19}}J \\
 & \Rightarrow BE=\dfrac{6.27\times {{10}^{-12}}J}{1.6\times {{10}^{-19}}} \\
 & \therefore BE=39.22MeV \\
\end{align}\]
Now, we can find the binding energy per nucleon by dividing by the number of nucleons in the Lithium nucleus as –
\[\begin{align}
  & B{{E}_{pernucleon}}=\dfrac{BE}{A} \\
 & \Rightarrow B{{E}_{pernucleon}}=\dfrac{39.22MeV}{7} \\
 & \therefore B{{E}_{pernucleon}}=5.6MeV \\
\end{align}\]
The binding energy per nucleon for a \[_{3}L{{i}^{7}}\] nucleus is 5.6MeV.
This is the required solution.

Note:
The binding energy for the nucleus for all the elements are almost the same, varying between five to eight mega electron volts. The nuclear force is the strongest known natural force and it makes the elements stable and thus, nature is stable.