Calculate the binding energy per nucleon of the nuclei  ${ }_{83}^{209}Bi$  from the following data.

$m_p=1.007825u,m_n=1.008665u,M{(}_{83}^{209}Bi)=208.980388u$
 
 

$\begin{array}{l}\mathrm{\Delta }m=83\times m_p+(209-83)m_n-\left[m\left({ }_{83}B{i}^{209}\right)\right]\\ \mathrm{\Delta }m=83\left[1.007825\right]+126\left[1.008665\right]-208.980388=1.760872\\ \text{ Binding energy }=1.760872\times 931.5\mathrm{MeV}=1640.3 \mathrm{MeV}\\ \text{ Binding energy per nucleon }=\frac{1640.3}{209}=7.85 \mathrm{MeV}\end{array}$