If the binding energy per nucleon in $Li^7$ and $He^4$ nuclei are respectively $5.60 \ MeV$ and $7.06 \ MeV$, then the energy of the reaction $Li^7 + p \to 2He^4$ is ......... $MeV$.

The electrostatic energy of $Z$ protons uniformly distributed throughout a spherical nucleus of radius $R$ is given by $E = \frac{3}{5} \frac{Z(Z-1) e^2}{4 \pi \varepsilon_0 R}$. The measured masses of the neutron,${ }_1^1 H$,${ }_7^{15} N$,and ${ }_8^{15} O$ are $1.008665 \ u$,$1.007825 \ u$,$15.000109 \ u$,and $15.003065 \ u$,respectively. Given that the radii of both the ${ }_7^{15} N$ and ${ }_8^{15} O$ nuclei are the same,$1 \ u = 931.5 \ MeV/c^2$ ($c$ is the speed of light),and $e^2 / (4 \pi \varepsilon_0) = 1.44 \ MeV \ fm$. Assuming that the difference between the binding energies of ${ }_7^{15} N$ and ${ }_8^{15} O$ is purely due to the electrostatic energy,the radius of either of the nuclei is $(1 \ fm = 10^{-15} \ m)$: (in $fm$)

$A$ nucleus of mass $M$ emits a $\gamma$-ray photon of frequency $\nu$. The loss of internal energy by the nucleus is:

The mass defect of $ { }_{2}^{4} He $ is $ 0.03 \ u $. The binding energy per nucleon of helium (in $ MeV $ ) is

What is the fundamental reason for the chemical energy produced in a chemical process?

For the decay of a nucleus,which of the following is $NOT$ a possible reason?