Obtain the binding energy (in $MeV$) of a nitrogen nucleus $\left(^{14}_{7} N \right)$ given $m\left(^{14}_{7} N \right)=14.00307 \; u$. (Given: $m_{H} = 1.007825 \; u$,$m_{n} = 1.008665 \; u$) (in $; MeV$)

We are given the following atomic masses:
$^{238}_{92}U = 238.05079 \; u$
$^{4}_{2}He = 4.00260 \; u$
$^{234}_{90}Th = 234.04363 \; u$
$^{1}_{1}H = 1.00783 \; u$
$^{237}_{91}Pa = 237.05121 \; u$
Here,the symbol $Pa$ represents the element protactinium $(Z=91)$.
$(a)$ Calculate the energy released during the alpha decay of $^{238}_{92}U$.
$(b)$ Show that $^{238}_{92}U$ cannot spontaneously emit a proton.

The masses of neutron, proton and deuteron in amu are $1.00893$, $1.00813$ and $2.01473$, respectively. The packing fraction of the deuteron in amu is

${ }_{92}^{238} A \rightarrow{ }_{90}^{234} B +{ }_2^4 D + Q$
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 $1 \, u = 931.5 \, MeV/c^2$]

An alpha particle $\left({ }^{4} He\right)$ has a mass of $4.00300 \ amu$. $A$ proton has a mass of $1.00783 \ amu$ and a neutron has a mass of $1.00867 \ amu$. The binding energy of an alpha particle estimated from these data is closest to: (in $MeV$)

Assertion : Binding energy (or mass defect) of hydrogen nucleus is zero.
Reason : Hydrogen nucleus contains only one nucleon.