Thank you for reporting, we will resolve it shortly
Q.
Calculate the binding energy per nucleon of ${ }_{20} Ca ^{40}$ nucleus $M \left({ }_{20} Ca ^{40}\right)=39.962589 u , m _{ n }=1.008665 u$ $m _{ p }=1.007825 u :$
Nuclei
Solution:
Number of protons $=20$
Number of neutrons $=40-20=20$
Total mass of 20 protons and 20 Neutrons
$=20 m _{ p }+20 m _{ n }=20\left( m _{ p }+ m _{ m }\right)$
$=20(1.007825+1.008665) $
$=40.32984 $
Mass defect, $ \Delta m =40.3298-39.962589 $
$=0.367211 u $
Total $ B . E =0.367211 u $
$=341.873441 mev$
$B.E / N =\frac{341.873441}{41}$
$=8.54 MeV / N$