Question

Let ${{m}_{p}}$ be the mass of the proton, ${{m}_{n}}$ the mass of a neutron, ${{M}_{1}}$ the mass of a ${}_{10}^{20}N$ nucleus and ${{M}_{2}}$ the mass of a ${}_{20}^{40}Ca$ nucleus. Then
$\begin{align}
  & \text{A}\text{. }{{M}_{2}}=2{{M}_{1}} \\
 & \text{B}\text{. }{{M}_{2}}>2{{M}_{1}} \\
 & \text{C}\text{. }{{M}_{2}}<2{{M}_{1}} \\
 & \text{D}\text{. }{{M}_{1}}<10\left( {{m}_{p}}+{{m}_{n}} \right) \\
\end{align}$

Answer 1

- Hint: We know that the number that is given on the head of the element shows atomic mass which is a sum of the mass of proton and neutron. Using this, calculate the mass neutron of element ${}_{10}^{20}N$ and ${}_{20}^{40}Ca$. We also know that mass of the nucleus cannot be equal to the mass of constituent nucleons.

Complete step-by-step solution
Nucleons consist of protons and neutrons. The nucleus is bound together in a nucleus with a very strong attractive force. Energy must be supplied to separate its constituent nucleons. It is observed that the mass of the nucleus is always less than the sum of masses of its constituent nucleons. This binding energy results in mass defect and makes the nucleus stable against the repulsive electrostatic forces between protons.
Neon is a highly stable gas having protons equal to 10 and neutrons equal to 10.
For the nucleus to be stable, ${{M}_{1}} < 10({{m}_{n}}+{{m}_{p}})$.
We know that nuclear binding energy per nucleon increases at lower atomic numbers. The higher the binding energy per nucleon, the greater, is the stability of the nucleus.
Therefore binding energy per nucleon will be higher for calcium hence mass of ${}_{20}^{40}Ca$ i.e. ${{M}_{2}}$ is lesser than the mass of neon i.e. ${}_{10}^{20}N$ i.e. ${{M}_{1}}$.
$\therefore {{M}_{2}} < 2{{M}_{1}}$
Therefore option (C) and (D) is the correct option.

Note: The difference mass is being used as energy that holds nucleons together. Mass defect and binding energy is the main reason for option (C), (D). The binding energy per nucleon is practically constant and is independent of mass number for nuclei $30 < A < 170$ when heavy nucleus $(A=240)$ breaks into lighter nuclei $(A=120)$ , binding energy increase i.e. Nucleons get more tightly bound. When very light nuclei $A < 10$, join to form a heavier nucleus, binding energy increase i.e. Nucleons get more tightly bound.