# Find the binding energy per neuclon for $_{50}^{120}Sn$. Mass of proton $m_{p}=1.00783U$, mass of neutron $m_{n}=1.00867U$ and mass of tin nucleus $m_{Sn}=119.902199U$. (take 1U = 931 MeV) Option: 1 7.5 MeV Option: 2 9.0 MeV Option: 3 8.0 MeV Option: 4 8.5 MeV

$\\ \text{ The number of protons in} \ _{50} \mathrm{Sn}^{120}=50 and the number of neutrons =120-50=70.$

$\begin{array}{l} =[50 \times 1.00783 \mathrm{u}+70 \times 1.00867 \mathrm{u}-119.902199 \mathrm{u}] \mathrm{c}^{2}=(1.096201 \mathrm{u}) \mathrm{c}^{2} \\ =(1.096201 \mathrm{u})(931 \mathrm{MeV} / \mathrm{u})=1020.563131 \mathrm{MeV} \end{array}$

$\text { Binding energy per nucleon }=\frac{1020.563131}{120}=8.504 \mathrm{MeV} \approx 8.5 \mathrm{MeV}$

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