COMEDK · Physics · 29. Nuclear Physics
The binding energy per nucleon for \(\mathrm{C}^{12}\) is \(7.68 \mathrm{~MeV}\) and that for \(\mathrm{C}^{13}\) is \(7.47 \mathrm{~MeV}\). The energy required to remove a neutron from \(\mathrm{C}^{13}\) is
- A \(7.92 \times 10^{-19} \mathrm{~J}\)
- B \(7.92 \times 10^{-13} \mathrm{~MeV}\)
- C \(4.95 \times 10^{-13} \mathrm{eV}\)
- D \(7.92 \times 10^{-13} \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(7.92 \times 10^{-13} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
The binding energy of a nucleus is given by \(BE = A \times (BE/A)\), where \(A\) is the mass number and \(BE/A\) is the binding energy per nucleon. For \(\mathrm{C}^{12}\), \(BE_{12} = 12 \times 7.68 \text{ MeV} = 92.16 \text{ MeV}\). For \(\mathrm{C}^{13}\),…
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