The binding energy of a nucleon in a nucleus is of the order of a few

$1 \text{ amu}$ is equal to

The energy equivalent to a mass of $1 \,kg$ is

The binding energy per nucleon $v/s$ mass number curve for nuclei is shown in the figure. $w, x, y$ and $z$ are four nuclei indicated on the curve. The process that would release energy is:

Let ${m_p}$ be the mass of a proton,${m_n}$ the mass of a neutron,${M_1}$ the mass of a $_{10}^{20}Ne$ nucleus,and ${M_2}$ the mass of a $_{20}^{40}Ca$ nucleus. Then:

If the binding energy per nucleon for $\,_3^7\,Li\,\,$ and $\,_2^4\,\,He$ are $5.60 \,MeV$ and $7.06 \,MeV$ respectively, then the energy of the proton in the reaction $p\,\, + \,\,_3^7\,\,Li\,\, \to \,\,2\,_2^4\,\,He$ must be .......... $MeV$.