As the mass number increases, the binding energy per nucleon .....

Nuclear binding energy is equivalent to

$A$ nuclide $1$ is said to be the mirror isobar of nuclide $2$ if $Z_1 = N_2$ and $Z_2 = N_1$. $(a)$ What nuclide is a mirror isobar of $_{11}^{23}Na$? $(b)$ Which nuclide out of the two mirror isobars has greater binding energy and why?

The mass of an $\alpha$-particle 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)

The atomic mass of ${ }_{6} C^{12}$ is $12.000000 \ u$ and that of ${ }_{6} C^{13}$ is $13.003354 \ u$. The required energy to remove a neutron from ${ }_{6} C^{13}$,if the mass of a neutron is $1.008665 \ u$,will be: (in $MeV$)