Question

Calculate mass defect and binding energy per nucleon of ${}_{10}^{20}Ne,$ given
Mass of ${}_{10}^{20}Ne=19.992397$u
Mass of ${}_1^{1}H=1.007825$u
Mass of ${}_0^{1}n=1.008665$u.

Answer 1

The required nuclear reaction is : $10{}_0^{1}n+10{}_1^{1}H\to {}_{10}^{20}Ne$

Mass defect $\Delta M=10(m{)}_{{}_0^{1}n}+10(m{)}_{{}_1^{1}H}-(m{)}_{{}_{10}^{20}Ne}$

$\therefore$ $\Delta M=10\left(1.008665\right)+10\left(1.007825\right)-19.992397=0.172503u$

Binding energy $E=\Delta M\times 931.5MeV$

$\therefore$ $E=0.172503\times 931.5=160.69MeV$

Binding energy per nucleon $u=\frac{E}{20}=\frac{160.69}{20}=8.03MeV$