Calculate the binding energy per nucleon of $_{26}{Fe}^{56}$. Atomic mass of $_{26}{Fe}^{56}$ is $55.9349 \;u$ and that of $_1H^1$ is $1.00783\;u$. Mass of $_0 n^1 = 1.00867\;u$ and $1u = 931\;MeV$.

Given,

Atomic mass of $_{26}{Fe}^{56} = A = 55.9349 \;u$
Mass of $_1H^1$                        $= m_p = 1.00783\;u$
Mass of $_0 n^1$                       $= m_n = 1.00867\;u$
$1u = 931\;MeV$
B.E. per nucleon, $(\overline{BE}) = ?$

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

$B.E. = [26 * 1.00783 + (56 - 26) * 1.00867 - 55.9349] * 931 = 492.29418\;MeV$

$\overline{BE} =$  $\frac{B.E.}{A} = \frac{492.29418}{56}$  $= 8.79096 \; Me V/nucleon        =      1.41 * 10^{-12}\;J$.

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