Question

Find the Binding energy per nucleon for ${}_{50}{}^{120}\mathrm{Sn}$. Mass of proton ${\mathrm{m}}_{\mathrm{p}}=1.00783\mathrm{U}$, mass of neutron ${\mathrm{m}}_{\mathrm{n}}=1.00867\mathrm{U}$ and mass of tin nucleus ${\mathrm{m}}_{\mathrm{Sn}}=119.902199\mathrm{U}$. (take $1\mathrm{U}=931\mathrm{MeV}$)

• A. $8.0\mathrm{MeV}$
• B. $9.0\mathrm{MeV}$
• C. $7.5\mathrm{MeV}$
• D. $8.5\mathrm{MeV}$

Answer 1

The correct option is D

$8.5\mathrm{MeV}$

The correct option is D.

Explanation for correct option:

$$\begin{gathered} \mathbf{Step}\mathbf{}\mathbf{1}\mathbf{:}\mathbf{}\mathbf{Calculation}\mathbf{}\mathbf{of}\mathbf{}\mathbf{∆}\mathbf{m} \\ \mathrm{B}.\mathrm{E}=∆{\mathrm{mc}}^2 \\ =∆\mathrm{m}\mathrm{x}931 \\ ∆\mathrm{m}=(50\mathrm{x}1.00783)+(70\mathrm{x}1.00867)-(119.902199) \\ =(120.9984–119.902199)\mathrm{U}=\mathbf{1}\mathbf{.}\mathbf{0962}\mathbf{}\mathbf{U}\mathbf{}\mathbf{} \\ \mathbf{Step}\mathbf{}\mathbf{2}\mathbf{:}\mathbf{}\mathbf{Calculation}\mathbf{}\mathbf{of}\mathbf{}\mathbf{binding}\mathbf{}\mathbf{energy}\mathbf{}\mathbf{per}\mathbf{}\mathbf{nucleon} \\ \mathrm{BE}=(1.0962\mathrm{x}931)=1020.5622\mathrm{MeV} \\ \mathbf{}\mathbf{BE}\mathbf{}\mathbf{per}\mathbf{}\mathbf{nucleon}\mathbf{}\approx 1020.5622/120=\mathbf{}\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{MeV} \end{gathered}$$

Therefore, Binding energy per nucleon is found to be $\mathbf{}\mathbf{8}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{MeV}$.

Hence, Option A is correct.

Options A, B and C are incorrect.

Explanation for incorrect options :

The numerical value given in options A, B and C don't match the calculated value.

Hence, A, B and C options are incorrect.