Question

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

• A. \(7.48\)
• B. \(7.84\)
• C. \(8.79\)
• D. \(6.94\)

Hint:

Remember:

• Mass defect: \[ \Delta m = (Zm_p + Nm_n) - m_{\text{nucleus}} \]
• Binding energy: \[ \text{BE} = \Delta m \times 931\,\text{MeV} \]
• Binding energy per nucleon: \[ \frac{\text{BE}}{A} \]

Answer 1

The Correct Option is B

Concept: Binding energy of a nucleus is the energy released when nucleons combine to form the nucleus. \[ \text{Binding Energy} = (\Delta m)c^2 \] where: \[ \Delta m = \text{Mass defect} \] Binding energy per nucleon: \[ \text{BE per nucleon} = \frac{\text{Total Binding Energy}}{A} \]

Step 1: Calculate the mass defect. For: \[ ^{209}_{83}\mathrm{Bi} \] Number of protons: \[ Z = 83 \] Number of neutrons: \[ N = 209-83 = 126 \] Mass of separate nucleons: \[ M = 83(1.007825) + 126(1.008665) \] \[ M = 83.649475 + 127.09179 \] \[ M = 210.741265\,u \] Given actual nuclear mass: \[ m = 208.980388\,u \] Mass defect: \[ \Delta m = 210.741265 - 208.980388 \] \[ \Delta m = 1.760877\,u \]

Step 2: Calculate total binding energy. Using: \[ \text{BE} = \Delta m \times 931 \] \[ \text{BE} = 1.760877 \times 931 \] \[ \text{BE} \approx 1639.78\,\text{MeV} \]

Step 3: Calculate binding energy per nucleon. \[ \text{BE per nucleon} = \frac{1639.78}{209} \] \[ \text{BE per nucleon} \approx 7.84\,\text{MeV} \] Therefore, \[ \boxed{7.84\,\text{MeV}} \]