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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

  1. A \(7.92 \times 10^{-19} \mathrm{~J}\)
  2. B \(7.92 \times 10^{-13} \mathrm{~MeV}\)
  3. C \(4.95 \times 10^{-13} \mathrm{eV}\)
  4. D \(7.92 \times 10^{-13} \mathrm{~J}\)
Verified Solution

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 \(\mathrm{C}^{13}\),…