The binding energy per nucleon of \(^{209}_{83}Bi\) is _______ MeV.
\([\text{Take } m(^{209}_{83}Bi) = 208.980388 \, u, \, m_p = 1.007825 \, u, \, m_n = 1.008665 \, u, \, 1 \, u = 931 \, \text{MeV}/c^2]\)

For the nucleus \(^{209}_{83}Bi\), the number of protons is \(Z = 83\) and the number of neutrons is \(N = 209 - 83 = 126\).

Mass of \(83\) protons \(= 83 \times 1.007825 = 83.649475 \, u\)

Mass of \(126\) neutrons \(= 126 \times 1.008665 = 127.091790 \, u\)

Total mass of the nucleons \(= 83.649475 + 127.091790 = 210.741265 \, u\)

The mass defect \(\Delta m\) is given by:

\(\Delta m = \text{Total mass of nucleons} - \text{Mass of nucleus}\)

\(\Delta m = 210.741265 - 208.980388 = 1.760877 \, u\)

The total binding energy (\(BE\)) is:

\(BE = \Delta m \times 931 \, \text{MeV}\)

\(BE = 1.760877 \times 931 = 1639.376 \, \text{MeV}\)

The binding energy per nucleon is:

\(\dfrac{BE}{A} = \dfrac{1639.376}{209} \approx 7.84 \, \text{MeV}\)

Answer: \(7.84\)