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

The binding energy $$B$$ of a nucleus is obtained from the mass-defect formula
$$\Delta m = Z\,m_p + N\,m_n - m_{\text{nucleus}}$$
and $$B = \Delta m \times 931\; \text{MeV}$$ (because $$1\;{\rm u}=931\;{\rm MeV}/c^{2}$$).

For $${}^{209}_{83}\text{Bi}$$:
Atomic number $$Z = 83$$, mass number $$A = 209$$, so the number of neutrons $$N = A-Z = 209-83 = 126$$.

Mass of all free protons:
$$Z\,m_p = 83 \times 1.007825\;{\rm u} = 83.649475\;{\rm u}$$

Mass of all free neutrons:
$$N\,m_n = 126 \times 1.008665\;{\rm u} = 127.091790\;{\rm u}$$

Total mass of separated nucleons:
$$Z\,m_p + N\,m_n = 83.649475 + 127.091790 = 210.741265\;{\rm u}$$

Nuclear mass of $${}^{209}_{83}\text{Bi}$$ (given): $$m_{\text{nucleus}} = 208.980388\;{\rm u}$$

Mass defect:
$$\Delta m = 210.741265 - 208.980388 = 1.760877\;{\rm u}$$

Total binding energy:
$$B = 1.760877 \times 931\;{\rm MeV} = 1639.38\;{\rm MeV}$$

Binding energy per nucleon:
$$\frac{B}{A} = \frac{1639.38\;{\rm MeV}}{209} = 7.84\;{\rm MeV}$$ (rounded to two decimal places).

Hence, the binding energy per nucleon of $${}^{209}_{83}\text{Bi}$$ is $$7.84$$ MeV.

Option B which is: $$7.84$$