Answer
Binding energy per nucleon, $\Delta E_{be/n} = 7.52 MeV$
Work Step by Step
The equation of binding energy is
$\Delta E_{be} = \Sigma mc^2 - Mc^2 $
Where
$ \Sigma mc^2 = [Zm_H + (A-Z) m_n $
Substitute into $ \Delta E_{be}$ equation
$\Delta E_{be} = [Zm_H + (A-Z) m_n - M_{Am}] c^2 $
$[Zm_H + (A-Z) m_n - M_{Am}] c^2 $
For Americium $^{244}_{95}Am$,
$ M_{Am} = 244.064 279 u $
$A = 244$
$Z = 95$
and using
$m_H = 1.007 825 u$
$m_n = 1.008 665 u $
Now we can substitute all values into the equation
$\Delta E_{be} = [(95)(1.007 825 u) + (244-95) (1.008 665 u) -244.064 279 u] c^2 $
$\Delta E_{be} = (1.970181 u)c^2$
But $1 \space u = 931.494013 \space MeV/c^2$
So
$\Delta E_{be} = (1.970181 u) (931.494013 \space MeV/c^2) c^2$
$\Delta E_{be} = 1835.212 MeV$
Now the question wants binding energy per nucleon, and there is 244 nucleons in Americium, so
$\Delta E_{be/n} = \frac{\Delta E_{be}}{A}$
$\Delta E_{be/n} = \frac{1835.212 MeV}{244}$
$\Delta E_{be/n} = 7.52 MeV$
