The graph of binding energy per nucleon $B_N$ versus mass number $A$ is given. In which process will energy be released?

$1 \text{ a.m.u.}$ is equivalent to

If $M(A, Z)$,$M_p$,and $M_n$ represent the masses of the nucleus ${}_{Z}^{A}X$,proton,and neutron in $u$ units respectively $(1u = 931.5 \, MeV/c^2)$,and $BE$ represents the binding energy in $MeV$,then which of the following relations is correct?

$A$ nucleus $^{A}_{Z} X$ has mass represented by $M(A, Z)$. If $M_p$ and $M_n$ denote the mass of a proton and a neutron respectively, and $B.E.$ is the binding energy in $MeV$, then:

Consider the following nuclear reaction : $^{238}\text{U} \rightarrow ^{234}\text{Th} + ^4\text{He}$. Take masses of $^{238}\text{U}$, $^{234}\text{Th}$ and $^4\text{He}$ as $238.050 \text{ u}$, $234.043 \text{ u}$ and $4.003 \text{ u}$, respectively. The $Q$ value for the reaction, in $\text{keV}$, is : [Given : $1 \text{ u} = 931.5 \text{ MeV/c}^2$]

The mass of a ${}_{3}^{7}Li$ nucleus is $0.042 \, u$ less than the sum of the masses of all its nucleons. The binding energy per nucleon of ${}_{3}^{7}Li$ nucleus is nearly...........$MeV$.