Calculate Binding Energy Per Nucleon of Bismuth Nucleus

To calculate the binding energy per nucleon of the $^{209}_{83}\text{Bi}$ nucleus, we follow these steps:

1. Identify the composition of the nucleus

For $^{209}_{83}\text{Bi}$:

• Atomic number (number of protons), $Z = 83$
• Mass number (total number of nucleons), $A = 209$
• Number of neutrons, $N = A - Z = 209 - 83 = 126$

---

2. Calculate the total mass of individual nucleons

• Total mass of $83$ protons: $M_p = 83 \times 1.007825\text{ u} = 83.649475\text{ u}$

• Total mass of $126$ neutrons: $M_n = 126 \times 1.008665\text{ u} = 127.091790\text{ u}$

• Total mass of individual nucleons ($M_{\text{total}}$): $M_{\text{total}} = M_p + M_n = 83.649475 + 127.091790 = 210.741265\text{ u}$

---

3. Calculate the mass defect ($\Delta m$)

The mass defect is the difference between the total mass of the constituent nucleons and the actual mass of the nucleus: $\Delta m = M_{\text{total}} - m\left(^{209}_{83}\text{Bi}\right)$ $\Delta m = 210.741265\text{ u} - 208.980388\text{ u} = 1.760877\text{ u}$

---

4. Calculate the total Binding Energy ($\text{BE}$)

Given that $1\text{ u} = 931\text{ MeV/c}^2$: $\text{BE} = \Delta m \times 931\text{ MeV}$ $\text{BE} = 1.760877 \times 931 = 1639.376487\text{ MeV}$

---

5. Calculate the Binding Energy per nucleon ($\overline{\text{BE}}$)

$\overline{\text{BE}} = \frac{\text{BE}}{A} = \frac{1639.376487\text{ MeV}}{209} \approx 7.8439\text{ MeV} \approx 7.84\text{ MeV}$

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

Correct Option: B