Question

How much energy is required to release one proton?

Answer 1

Hint: To answer this question we should know about the binding energy of a nucleus, the mass of proton and neutron. We should also know about mass defects. The energy required to break the nucleus into individual components, protons and neutrons is known as binding energy. We will determine the mass defect by considering a nucleus having one proton and one neutron. Then we will determine the binding energy of that nucleus.

Formula used: ${\text{B}}{\text{.E}}{\text{.}}\,{\text{ = }}\,{{\delta m}} \times {\text{931}}{\text{.5}}\,{\text{MeV}}$

Complete step by step answer:
The amount of energy released when nuclei are formed from protons and neutrons is known as binding energy. So, to remove a proton from the nucleus we have to give the energy equal to the binding energy.
The formula of binding energy or we can say the energy required to remove a proton from the nucleus is as follows:
$ \Rightarrow {\text{B}}{\text{.E}}{\text{.}}\,{\text{ = }}\,{{\delta m}} \times {\text{931}}{\text{.5}}\,{\text{MeV}}$
Where,
B.E. is the binding energy of the nucleus
${{\delta m}}$ is the mass defect

By the neutrons and protons, the nucleus forms. The separate neutron and proton have higher energy than the nucleus because by the formation of the nucleus, the neutron and proton get stabilized. So, the difference in energy of protons and neutrons of an atom and its nucleus is known as mass defect.
$\Rightarrow$ Mass defect = (neutron + proton) – nucleus
When we remove a proton from the nucleus having one proton and one neutron, the mass of the nucleus will decrease by the mass of one proton so, the mass defect will be,
$\Rightarrow$ Mass defect = mass of nucleus – mass of proton
$\Rightarrow$ Mass defect = mass of neutron + mass of proton – mass of proton
$\Rightarrow$ Mass defect = mass of neutron
The mass of the neutron is $1.008665$ amu, so the mass defect ${{\delta m}}$ will be $1.008665$ amu.

So, the binding energy of the nucleus is as follows:
$ \Rightarrow {\text{B}}{\text{.E}}{\text{.}}\,{\text{ = }}\,{{\delta m}} \times {\text{931}}{\text{.5}}\,{\text{MeV}}$
On substituting $1.008665$ amu for mass defect ${{\delta m}}$,
$ \Rightarrow {\text{B}}{\text{.E}}{\text{.}}\,{\text{ = }}\,1.008665 \times {\text{931}}{\text{.5}}\,{\text{MeV}}$
$ \Rightarrow {\text{B}}{\text{.E}}{\text{.}}\,{\text{ = }}\,939.06\,{\text{MeV}}$
So, the binding energy of the nucleus is $939.06$MeV so we have to give $939.06$MeV energy to remove a proton from the nucleus.

Therefore, the energy is required to release one proton from a nucleus having one proton and one neutron is $939.06$ MeV.

Note: The exact mass of the nucleus is never equal to the sum of the mass of protons and neutrons of that nucleus. The mass of the nucleus is slightly less than the sum of the exact mass of protons and neutrons of that nucleus. The mass defect has a fixed value for each nucleus. In the nucleus of Fe, the nucleons are most tightly bound so Fe is the most stable nucleus.