Answer
Binding Energy per nucleon, $\Delta E_{be/n} = 7.38 MeV$
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_{Eu}] c^2 $
For Rutherfordium $^{259}_{104}Rf$,
$ M_{Rf} = 259.10563 u $
$A = 259$
$Z = 104$
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} = [(104)(1.007 825 u) + (259-104) (1.008 665 u) - 259.10563 u] c^2 $
$\Delta E_{be} = (2.051245 u)c^2$
But $1 \space u = 931.494013 \space MeV/c^2$
So
$\Delta E_{be} = (2.051245) (931.494013 \space MeV/c^2) c^2$
$\Delta E_{be} = 1910.722 MeV$
Now the question wants binding energy per nucleon, and there is 259 nucleons in Rutherfordium, so
$\Delta E_{be/n} = \frac{\Delta E_{be}}{A}$
$\Delta E_{be/n} = \frac{1910.722 MeV}{259}$
$\Delta E_{be/n} = 7.38 MeV$
