The binding energy per nucleon of ${}_{83}^{209}Bi$ is _______ MeV . [Take $m\left({}_{83}^{209}Bi\right)=208.980388u,{m}_{p}=1.007825u,{m}_{n}=1.008665u$ $1u=931MeV/{c}^{2}$ ] [02 April, 2026 (Shift-2)]

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The binding energy per nucleon of \({}_{83}^{209}Bi\) is _______ MeV . [Take \(m\left({}_{83}^{209}Bi\right)=208.980388u…

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