Q151 / 180 · COMEDK 2024 (Evening Shift) · 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$4.95 \times 10^{-13} \mathrm{eV}$
- Correct answerD$7.92 \times 10^{-13} \mathrm{~J}$
Solution
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 $\mathrm{C}^{13}$, $BE_{13} = 13 \times 7.47 \text{ MeV} = 97.11 \text{ MeV}$.
The energy required to remove a neutron from $\mathrm{C}^{13}$ is the difference in binding energies: $\Delta E = BE_{13} - BE_{12} = 97.11 \text{ MeV} - 92.16 \text{ MeV} = 4.95 \text{ MeV}$.
Converting this energy to Joules using $1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}$:
$\Delta E = 4.95 \times 1.6 \times 10^{-13} \text{ J} = 7.92 \times 10^{-13} \text{ J}$.
Answer: $7.92 \times 10^{-13} \mathrm{~J}$
For $\mathrm{C}^{12}$, $BE_{12} = 12 \times 7.68 \text{ MeV} = 92.16 \text{ MeV}$.
For $\mathrm{C}^{13}$, $BE_{13} = 13 \times 7.47 \text{ MeV} = 97.11 \text{ MeV}$.
The energy required to remove a neutron from $\mathrm{C}^{13}$ is the difference in binding energies: $\Delta E = BE_{13} - BE_{12} = 97.11 \text{ MeV} - 92.16 \text{ MeV} = 4.95 \text{ MeV}$.
Converting this energy to Joules using $1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J}$:
$\Delta E = 4.95 \times 1.6 \times 10^{-13} \text{ J} = 7.92 \times 10^{-13} \text{ J}$.
Answer: $7.92 \times 10^{-13} \mathrm{~J}$
Hence, the correct answer is option: 'D'.