Answer
(a) $\Delta E_{\mathrm{be}}= Z \Delta_{h}+N \Delta_{n}-\Delta$
(b) $\Delta E_{\text {be }} = 7.92 \mathrm{MeV}$
Work Step by Step
(a) If the masses are given in atomic mass units, then mass excesses are defined by $\Delta_{H}=\left(m_{H}-1\right) c^{2}, \Delta_{n}=\left(m_{n}-1\right) c^{2},$ and $\Delta=(M-A) c^{2} .$
This means
$$m_{H} c^{2}=\Delta_{H}+c^{2}, m_{n} c^{2}=\Delta_{n}+c^{2},$$ and
$$M c^{2}=\Delta+A c^{2} .$$
Thus,
$$
\Delta E_{\mathrm{be}}=\left(Z \Delta_{H}+N \Delta_{n}-\Delta\right)+(Z+N-A) c^{2}=\boxed{Z \Delta_{h}+N \Delta_{n}-\Delta}
$$
Where $A = Z+N$
(b) For ${ }_{79}^{197} \mathrm{Au}, Z=79$ and $N=197-79=118 .$ Hence,
$$
\Delta E_{\mathrm{be}}=(79)(7.29 \mathrm{MeV})+(118)(8.07 \mathrm{MeV})-(-31.2 \mathrm{MeV})=1560 \mathrm{MeV}
$$
This means the binding energy per nucleon is
$$\Delta E_{\text {be }}=(1560 \mathrm{MeV}) / 197=\boxed{7.92 \mathrm{MeV}}$$
