The binding energy per nucleon of ^ 209 _ 83 Bi is _______ MeV....

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: $7.84$