Question

The binding energy per nucleon of ${{C}^{12}}$is ${{E}_{1}}$ and that of ${{C}^{13}}$ is ${{E}_{2}}$. The energy required to remove one neutron from ${{C}^{13}}$ is
$\begin{aligned} & \text{A}\text{. }12{{E}_{1}}+12{{E}_{2}} \\\ & \text{B}\text{. }13{{E}_{2}}-12{{E}_{1}} \\\ & \text{C}\text{. }12{{E}_{2}}-13{{E}_{1}} \\\ & \text{D}\text{. }13{{E}_{2}}+12{{E}_{1}} \\\ \end{aligned}$

Answer 1

The energy required to completely disassemble it into separate protons and neutrons is the binding energy of the nucleus. This value is equivalent to the mass defect of the nucleus. Binding energy per nucleon can be determined by dividing the binding energy with the number of nucleons.

Complete step by step answer:
Binding energy is defined as the energy that holds a nucleus together. This value is equal to the mass defect of the nucleus. Binding energy is the amount of energy required to separate a particle from a system of particles or to disperse all the particles of the system. Binding energy is essentially applicable to sub atomic particles in atomic nuclei, to electrons bound to nuclei in atoms, and to atoms and ions bound together in crystals. The more tightly a system is bound, the stronger the forces that hold it together and the greater the energy required to pull it apart.
Mass defect is expressed as the difference between the predicted mass and the actual mass of an atom’s nucleus. The binding energy of a system can actually appear as extra mass, which accounts for the difference in two masses and results in mass defect.
The binding energy per nucleon of a nucleus is the binding energy divided by the total number of nucleons. It is a measure of stability of the nucleus.
Expression for Binding energy:
$\text{Binding energy = mass defect}\times {{\text{c}}^{2}}$
Where,
$\text{c = speed of light in vacuum}$
We are given that the binding energy per nucleon of ${{C}^{12}}$ is ${{E}_{1}}$ and that of ${{C}^{13}}$ is ${{E}_{2}}$.
Binding energy per nucleon of ${{C}^{12}}$ is ${{E}_{1}}$
Binding energy of nucleus of ${{C}^{12}}$ is, ${{B}_{{{E}_{1}}}}=12\times {{E}_{1}}$
Binding energy per nucleon of ${{C}^{13}}$ is ${{E}_{2}}$
Binding energy of nucleus of ${{C}^{13}}$ is, ${{B}_{{{E}_{2}}}}=12\times {{E}_{2}}$
Energy required to remove one neutron from ${{C}^{13}}$ is,
$\begin{aligned} & \Delta E={{B}_{{{E}_{2}}}}-{{B}_{{{E}_{1}}}} \\\ & \Delta E=13{{E}_{2}}-12{{E}_{1}} \\\ \end{aligned}$
Thus,
The energy required to remove one neutron from ${{C}^{13}}$ is $\Delta E=13{{E}_{2}}-12{{E}_{1}}$
Hence, the correct option is B.

Note:
Binding energy per nucleon can be calculated by dividing the binding energy of the nucleus by the total number of nucleons. In a large nucleus, most of the nucleons lie inside the nucleus and not on the surface. The change in the binding energy, if any, would be negligibly small. Therefore, the binding energy per nucleon is a constant.