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.

Answer:

            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