One requires energy ${E_n}$ to remove a nucleon from a nucleus and an energy ${E_e}$ to remove an electron from the orbit of an atom. Then:

Mass numbers of the elements $A, B, C$ and $D$ are $30, 60, 90$ and $120$ respectively. The specific binding energies of them are $5 \text{ MeV}, 8.5 \text{ MeV}, 8 \text{ MeV}$ and $7 \text{ MeV}$ respectively. In which of the following reaction$(s)$ is energy released?
$1. D \rightarrow 2B$
$2. C \rightarrow B + A$
$3. B \rightarrow 2A$

${ }_{92}^{238} A \rightarrow{ }_{90}^{234} B +{ }_2^4 D + Q$
In the given nuclear reaction,the approximate amount of energy released will be $.....\,MeV$.
[Given: mass of ${ }_{92}^{238} A = 238.05079 \, u$,mass of ${ }_{90}^{234} B = 234.04363 \, u$,mass of ${ }_2^4 D = 4.00260 \, u$,and $1 \, u = 931.5 \, MeV/c^2$]

Find the binding energy per nucleon for $^{120}_{50}Sn$. Given: mass of proton $m_{p} = 1.00783 \, U$,mass of neutron $m_{n} = 1.00867 \, U$,and mass of tin nucleus $m_{Sn} = 119.902199 \, U$. (Take $1 \, U = 931 \, MeV$) (in $, MeV$)

If the binding energy per nucleon in $_3^7Li$ and $_2^4He$ nuclei are $5.60 \, MeV$ and $7.06 \, MeV$ respectively,then in the reaction $p + {}_3^7Li \to 2 {}_2^4He$,the energy of the proton must be ........... $MeV$.

From the given data,the amount of energy required to break the nucleus of aluminium ${ }_{13}^{27} {Al}$ is $x \times 10^{-3} {J}$.
Mass of neutron $= 1.00866 \, {u}$
Mass of proton $= 1.00726 \, {u}$
Mass of aluminium nucleus $= 26.98154 \, {u}$
(Assume $1 \, {u}$ corresponds to $1 \, {J}$ of energy for the purpose of this calculation)
(Round off to the nearest integer)