The binding energy per nucleon of _83^209 Bi is ____ MeV. [Take m (_83^209 Bi )=208.980388 u, m_p =1.007825 u, m_n =1.008665 u, 1 u =931 MeV / c^2 ]

The binding energy per nucleon of $\displaystyle { }_{83}^{209} \mathrm{Bi}$ is $\displaystyle \_\_\_\_$ MeV. [Take $\displaystyle \mathrm{m}\left({ }_{83}^{209} \mathrm{Bi}\right)=208.980388 \mathrm{u}, \mathrm{m}_{\mathrm{p}}=1.007825 \mathrm{u}, \mathrm{m}_{\mathrm{n}}=1.008665 \mathrm{u}, 1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^2$ ]