Q 12-13-119JEE MainJEE Main 2020 (4 Sep, Shift 2)Medium
Find the binding energy per nucleon for ${}^{120}_{50}\text{Sn}$. Mass of proton $m_p = 1.00783\ \text{u}$, mass of neutron $m_n = 1.00867\ \text{u}$ and mass of tin nucleus $m_{Sn} = 119.902199\ \text{u}$. (take $1\ \text{u} = 931\ \text{MeV}$)
Answer: (D) $8.5\ \text{MeV}$
$Z = 50$, $N = 70$. Mass defect:
$$\Delta m = 50(1.00783) + 70(1.00867) - 119.902199 = 1.096201\ \text{u}$$
$$\frac{B}{A} = \frac{1.096201\times931}{120} \approx 8.5\ \text{MeV}$$
Solution by Sreeraj P, M.Sc Physics