The nucleus having the highest binding energy per nucleon is

The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation energies of the nuclei $_{20}^{41} Ca$ and $_{13}^{27} Al$ from the following data:
$m(_{20}^{40} Ca) = 39.962591 \; u$
$m(_{20}^{41} Ca) = 40.962278 \; u$
$m(_{13}^{26} Al) = 25.986895 \; u$
$m(_{13}^{27} Al) = 26.981541 \; u$
(Given mass of neutron $m_n = 1.008665 \; u$)

The mass of a nucleus,a neutron,and a proton are $M$,$M_n$,and $M_p$ respectively. Then which of the following is true?

Nucleus $A$ having $Z=17$ and an equal number of protons and neutrons has $1.2 \, MeV$ binding energy per nucleon. Another nucleus $B$ of $Z=12$ has a total of $26$ nucleons and $1.8 \, MeV$ binding energy per nucleon. The difference in binding energy of $B$ and $A$ will be $........... \, MeV$.

If the binding energy per nucleon of a nuclide is high,then:

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?