JEE Mains · Physics · STD 12 - 13. Nuclei
The binding energy per nucleon of \(^{209}_{83}Bi\) is _______ MeV.
\([\text{Take } m(^{209}_{83}Bi) = 208.980388 \, u, \, m_p = 1.007825 \, u, \, m_n = 1.008665 \, u, \, 1 \, u = 931 \, \text{MeV}/c^2]\)
- A \(7.48\)
- B \(7.84\)
- C \(8.79\)
- D \(6.94\)
Answer & Solution
Correct Answer
(B) \(7.84\)
Step-by-step Solution
Detailed explanation
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…
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