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
You need to login to perform this action.
You will be redirected in
3 sec