• # question_answer Calculate the binding energy per nucleon of nucleus ${}_{20}C{{a}^{40}}.$ Given ${{m}_{n}}$ and ${{m}_{p}}$ are 1.008665 u and 1.007825 u respectively and m  $({}_{20}C{{a}^{40}})=39.962589$ u.

In ${}_{20}C{{a}^{40}}$ nucleus, number of protons =20 and number of neutrons $=40-20=20$ Mass of 20 neutrons and 20 protons $=20({{m}_{n}}+{{m}_{p}})$ $=20\times 1.008665+20\times 1.007825$ = 40.3298u Mass defect, $\Delta m=40.3298-39.962589$ = 0,367211u Total binding energy $=0.367211\times 931$ = 341.873441 MeV Binding Eenergy per nucleon,             ${{E}_{bn}}=\frac{341.873441}{40}$ = 8.547 MeV/nucleon