Question

Find the binding energy per nucleon of ${}_{79}{}^{197}$Au if its atomic mass is 196.96 u.

Answer 1

Given:
Atomic mass of Au, A = 196.96
Atomic number of Au, Z = 79
Number of neutrons, N = 118
Binding energy, $B=(\mathrm{Z}{m}_p+N{m}_n-M)c^2$
Here, m_p = Mass of proton
M = Mass of nucleus
m_n = Mass of neutron
c = Speed of light
On substituting the respective values, we get
$$\begin{gathered} B=\left[\right(79\times 1.007276+118\times 1.008665)\mathrm{u}-196.96\mathrm{u}]c^2 \\ =\left(198.597274-196.96\right)\times 931\mathrm{MeV} \\ =1524.302094\mathrm{MeV} \end{gathered}$$
Binding energy per nucleon $=\frac{1524.3}{197}=7.737\mathrm{MeV}$