Question

If the binding energy per nucleon in \( \text{Li}^7 \) and \( \text{He}^4 \) nuclei are respectively \( 5.60 \, \text{MeV} \) and \( 7.06 \, \text{MeV} \), then the energy of the reactor \[ \text{Li}^7 + p \to \text{He}^4 + 2e \] is

• A. \( 19.6 \, \text{MeV} \)
• B. \( 2.4 \, \text{MeV} \)
• C. \( 8.4 \, \text{MeV} \)
• D. \( 17.3 \, \text{MeV} \)

Hint:

To find the energy released in a nuclear reaction, subtract the binding energy of the reactants from the binding energy of the products.

Answer 1

The Correct Option is D

The energy released in the reaction can be calculated by considering the binding energies of the nuclei before and after the reaction. The total energy released is \( 17.3 \, \text{MeV} \).

Step 2: Conclusion.
The energy of the reactor is \( 17.3 \, \text{MeV} \), corresponding to option (d).