Question

If the binding energy of $N^{14}$ is $7.5MeV$ per nucleons and that of $N^{15}$ is $7.7MeV$ per nucleon, then the energy is required to remove a neutron from $N^{15}$ is

• A. 5.25 MeV
• B. 0.2 MeV
• C. 10.5 MeV
• D. 0.4 MeV

Answer 1

Answer: (C)

Given, binding energy per nucleon of
$N^{14}=7.5MeV/$nucleon
Binding energy per nucleon of
$N^{15}=7.MeV/$nucleon
$\therefore$ Total $BE$ of $N^{14}=7.5\times 14MeV$
Total $BE$ of $N^{15}=7.7\times 15MeV$
Then, the energy required to remove a neutron from $N^{15}$,
$=B{E}_{\text{total }}{N}^{15}-B{E}_{\text{total }}{N}^{14}$
$=(7.7\times 15-7.5\times 14)MeV=10.5MeV$