The binding energy per nucleon of a nucleus ${}_Z X^A$ at rest is $6 \ MeV$. It undergoes $\beta^-$ decay as shown below:
${}_Z X^A \to {}_{Z+1} Y^A + {}_{-1}^0 e + \bar{\nu}$
The total kinetic energy $(K.E.)$ of the products is $3 \ MeV$. The binding energy per nucleon of $Y$ (in $MeV$) is:

$M_n$ and $M_p$ represent the mass of a neutron and a proton, respectively. If an element having atomic mass $M$ has $N$ neutrons and $Z$ protons, then the correct relation is:

The mass of a proton is $1.0073 \, u$ and that of a neutron is $1.0087 \, u$. The binding energy of $_2He^4$ is .... ($u = \text{atomic mass unit}$,mass of helium nucleus = $4.0015 \, u$)

The energy equivalent of $1 \,g$ of substance is:

$A$ plot of the number of neutrons $(N)$ against the number of protons $(Z)$ for stable nuclei exhibits upward deviation from linearity for atomic number $Z > 20$. For an unstable nucleus having an $N/Z$ ratio less than $1$,the possible mode$(s)$ of decay is(are):
$(A)$ $\beta^{-}$-decay ($\beta$ emission)
$(B)$ Orbital or $K$-electron capture
$(C)$ Neutron emission
$(D)$ $\beta^{+}$-decay (positron emission)

For a nucleus ${ }_Z^A X$ having mass number $A$ and atomic number $Z$:
$A.$ The surface energy per nucleon $(b_s) = a_1 A^{2/3}$
$B.$ The Coulomb contribution to the binding energy $b_c = -a_2 \frac{Z(Z-1)}{A^{4/3}}$
$C.$ The volume energy $b_v = a_3 A$
$D.$ Decrease in the binding energy is proportional to surface area.
$E.$ While estimating the surface energy,it is assumed that each nucleon interacts with $12$ nucleons,($a_1, a_2$ and $a_3$ are constants)
Choose the most appropriate answer from the options given below: