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  • Find the binding energy per neuclon for . Mass of proton 

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

Answers (1)

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\\ \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}

Posted by

Deependra Verma

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