In the nuclear reaction,$_1H^2 + _1H^2 \to _0n^1 + _2He^3$. If the binding energy of deuteron is $2.23 \ MeV$ and the $Q$-value of the reaction is $3.27 \ MeV$,then the binding energy of $_2He^3$ is ......... $MeV$.

If $M_0$ is the mass of isotope ${ }_{5}^{12} B$,$M_p$ and $M_n$ are the masses of a proton and a neutron respectively,then the nuclear binding energy of the isotope is:

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 $fm$)

The binding energy per nucleon of ${}_3^7Li$ and ${}_2^4He$ nuclei are $5.60\,MeV$ and $7.06\,MeV$ respectively. In the nuclear reaction ${}_3^7Li + {}_1^1H \to 2{}_2^4He + Q$, the value of energy $Q$ released is.........$MeV$.

If the binding energy of the deuterium is $2.23 \, MeV$, the mass defect in $a.m.u.$ is:

Obtain the binding energy of the nuclei $^{56}_{26} Fe$ and $^{209}_{83} Bi$ in units of $MeV$ from the following data:
$m(^{56}_{26} Fe) = 55.934939 \; u$
$m(^{209}_{83} Bi) = 208.980388 \; u$ (in $; MeV$)