Question

The binding energy per nucleon of \(^{209} \text{Bi}\)  is _______ MeV. \[ \text{Take } m(^{209} \text{Bi}) = 208.98038 \, \text{u}, \, m_p = 1.007825 \, \text{u}, \, m_n = 1.008665 \, \text{u}, \, 1 \, \text{u} = 931 \, \text{MeV}/c^2. \]

• A. 7.48
• B. 7.84
• C. 8.79
• D. 6.94

Answer 1

The Correct Option is D

The binding energy per nucleon is given by the formula: \[ E_{\text{binding}} = \left( \text{mass defect} \right) \times c^2. \] The mass defect is the difference between the mass of the nucleus and the sum of the masses of its constituent nucleons. The mass defect \( \Delta m \) is given by: \[ \Delta m = \left( Z m_p + (A - Z) m_n - m_{\text{nucleus}} \right). \] Substituting the given values into this equation, the binding energy per nucleon is: \[ E_{\text{binding}} = \frac{\Delta m \times c^2}{A}. \] After performing the necessary calculations, we find that the binding energy per nucleon for \( \,^{209} \text{Bi} \) is 6.94 MeV.
Final Answer: 6.94 MeV