Question

Find the binding energy per nucleon of ${ }_{79}^{197} Au$ if its atomic mass is $196.96 \,u$. Where $m _{ p }=1.007277 \,u \,\,\& \,\,m _{ n }=1.00866\, u$ (in $MeV /$ nucleus )

Answer 1

$\Delta m =79 m _{ p }+118 M _{ n }-$ (atomic mass $-79 m _{ e }$ )
$=79 \times 1.007277+118 \times 1.00866+79\times 0.00055-196.96000$
$=1.680213\, amu$
Hence $\frac{\text { B.E. }}{ A }=\frac{1.6802 \times 931.5}{197} MeV$
Ans. $=7.94 \,MeV /$ nucleon