One million electron volt $(1\,MeV)$ is equal to

Find the binding energy per nucleon for $^{120}_{50}Sn$. Given: mass of proton $m_{p} = 1.00783 \, U$,mass of neutron $m_{n} = 1.00867 \, U$,and mass of tin nucleus $m_{Sn} = 119.902199 \, U$. (Take $1 \, U = 931 \, MeV$) (in $, MeV$)

The mass of the oxygen isotope $_8O^{17}$ is $M_0$. If $M_p$ and $M_n$ are the masses of a proton and a neutron respectively,the nuclear binding energy of the isotope is ......

The mass of a nucleus $_Z^AX$ is denoted by $M(A, Z)$. If $M_p$ and $M_n$ are the masses of a proton and a neutron respectively, the binding energy of this nucleus is given by:

From the given data,the amount of energy required to break the nucleus of aluminium ${ }_{13}^{27} {Al}$ is $x \times 10^{-3} {J}$.
Mass of neutron $= 1.00866 \, {u}$
Mass of proton $= 1.00726 \, {u}$
Mass of aluminium nucleus $= 26.98154 \, {u}$
(Assume $1 \, {u}$ corresponds to $1 \, {J}$ of energy for the purpose of this calculation)
(Round off to the nearest integer)

$A$ nucleus of mass $M + \Delta m$ is at rest and decays into two daughter nuclei of equal mass $\frac{M}{2}$ each. The speed of light is $c$. The speed of the daughter nuclei is: