Calculate the binding energy per nucleon of calcium nucleus $(_{20}{Ca}^{40})$. Given, Mass of $_{20}{Ca}^{40} = 39.962589\;u$; Mass of neutron $m_n = 1.008665\;u$; Mass of proton $m_p = 1.007825\;u$; $1\;u = 931\;MeV$.

Given,

Atomic mass of $_{20}{Ca}^{40} =  A  = 39.962589\;u$
Mass of $_1H^1$                        $= m_p = 1.007825\;u$
Mass of $_0 n^1$                       $= m_n = 1.008665\;u$
$1u = 931\;MeV$
B.E. per nucleon, $(\overline{BE}) = ?$

Now,
$B.E. = [Z\;m_p + (A-Z)\;m_n - M] * 931 \;MeV$

$B.E. = [20 * 1.007825 + (40 - 20) * 1.008665 - 39.962589] * 931 = 1793.954141\;MeV$

$\overline{BE} =$ $\frac{B.E.}{A} = \frac{1793.954141}{40}$ $= 44.848853 \; Me V/nucleon$,

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