# Find the Binding energy per neucleon

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 $1 \mathrm{U}=931 \mathrm{MeV})$

1. (1) $7.5 \mathrm{MeV}$

2. (2) $9.0 \mathrm{MeV}$

3. (3) $8.0 \mathrm{MeV}$

4. (4) $8.5 \mathrm{MeV}$

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 $=(\Delta m) C^{2}=(\Delta m) \times 931=1020.56$

$\frac{\text { Binding energy }}{\text { Nucleon }}=\frac{1020.5631}{120}=8.5 \mathrm{MeV}$