Question:
Find the Binding energy per neucleon for ${ }_{50}^{120} \mathrm{Sn}$. Mass of proton $m_{p}=1.00783 \mathrm{U}$, mass of neutron $m_{n}=1.00867 \mathrm{U}$ and mass of tin nucleus $m_{\text {Sn }}=119.902199 \mathrm{U}$.
$($ take $1 \mathrm{U}=931 \mathrm{MeV})$
Correct Option: , 4
Solution:
(4) Mass defect,
$\Delta m=\left(50 m_{p}+70 m_{n}\right)-\left(m_{s n}\right)$
$=(50 \times 1.00783+70 \times 1.008)-(119.902199)$
$=1.096$
Binding energy $=(\Delta m) C^{2}=(\Delta m) \times 931=1020.56$
$\frac{\text { Binding energy }}{\text { Nucleon }}=\frac{1020.5631}{120}=8.5 \mathrm{MeV}$
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