Question

For greater stability, a nucleus should have greater value of binding energy per nucleon. Why?

Answer 1

Hint: We know in real life that anything that is tightly packed or bounded the more tough it is to separate, similarly anything that is loosely packed or bounded the more easy it is to separate. This is true in the case of nucleus as well. The more tightly a nucleus is bound the more energy it requires to disperse and the more loosely the nucleus is bound the less energy it requires to disperse. From this we could know about the nuclear forces and the stability of the nucleus.

Formula Used:

Complete step by step answer:
Nucleus is in the core of the atom and it is composed of neutrons and protons. There is an energy that binds the nucleus and protons together, this energy is known as binding energy. In other words we define the binding energy of the nucleus as the energy that is required to completely disassemble or separate the nucleus into neutrons and protons. Generally, the determination of the binding energy of a nucleus can be done from its rest mass. The two are connected through the famous Einstein equation$E=(\mathrm{\Delta }m)c^{{}^2}$. Here: E = Energy; $\mathrm{\Delta }m$= change in mass; c = speed of light. The mass of a bound system is smaller than its separate constituents; the more tightly the nucleus is bounded the smaller would be its mass. Suppose if we pull apart a nuclide (a type of nucleus having a specific number of protons and neutrons), the work done to overcome the nuclear forces holding the nuclide together puts a certain amount of energy into the system. The energy that is put into the system is equal to the binding energy BE. The pieces (Neutrons and protons) are at rest when separated. Therefore, the energy put into them increases their total rest mass as compared with what it was when they were glued together as a complete nucleus. The increase in the rest mass is given as $\mathrm{\Delta }m={\displaystyle \frac{BE}{c^2}}$. Here: $\mathrm{\Delta }m$= change in mass; BE = Binding energy; c = speed of light. This difference in mass is known as mass defect. Suppose a nucleus (${}^{A}Y$ ) has P protons and Neutrons, so the difference in mass would be$\mathrm{\Delta }m=(P{m}_p+N{m}_n)-m_{tot}$.
 Thus
$BE=\mathrm{\Delta }m{c}^2=((P{m}_p+N{m}_n)-m_{tot})\times c^2$
Where;
$\text{BE = Binding Energy;}$ $\mathrm{\Delta }m$= Change in Mass;
$\text{c = Speed of light (3}\times \text{1}{\text{0}}^8\text{m}{\text{s}}^{-1}\text{);}$
$m_p\text{ = Mass of proton;}$
$m_n\text{ = Mass of Neutron;}$
$m_{tot}\text{ = Total Mass;}$

Final Answer: We can see form the equation$(BE=((P{m}_p+N{m}_n)-m_{tot})\times c^2)$ that,$(BE\alpha (P{\text{m}}_{\text{p}}\text{ + N}{\text{m}}_{\text{n}}\text{) - }{\text{m}}_{\text{tot}};\alpha =proportionality-sign)$ the binding energy of the nucleus is directly proportional to the stability of the nucleus and the stability of the nucleus or nuclei is dependent upon the number of neutrons and protons.

Note: Here we have to describe the elements present inside the nucleus and how their quantities are related with the binding energy of the nucleus and also provide a proper reasoning for higher binding energy leads to greater stability of the nucleus.