Question

Energy released when two deuterons $ \text{(H}_2\text{)} $ fuse to form a helium nucleus $ \text{(He}_4\text{)} $ is:

• A. 8.1 MeV
• B. 5.9 MeV
• C. 23.6 MeV
• D. 26.8 MeV

Hint:

The energy released in a nuclear fusion reaction can be calculated using the binding energy per nucleon of the reactants and products.

Answer 1

The Correct Option is C

Inputs: - Binding energy per nucleon of \( \text{H}_2^1 \) = 1.1 MeV
- Binding energy per nucleon of \( \text{He}_4^2 \) = 7.0 MeV
The energy released \( Q \) is calculated as the difference between the total binding energy of the reactants and the total binding energy of the products: \[ E_B = \text{BE}_{\text{reactant}} - \text{BE}_{\text{product}} \] \[ E_B = 1.1 \times 2 + 1.1 \times 2 - 7 \times 4 = 23.6 \, \text{MeV} \] Therefore, the energy released is: \[ Q = 23.6 \, \text{MeV} \]