Question

For the reaction find energy released ²₁H + ²₁H → ⁴₂He BE/A for \(^2_1H = 1.1 \, \text{MeV}\) and for \(^4_2He = 7.2 \, \text{MeV}\):

• A. 24.4 MeV
• B. 6.2 MeV
• C. 6.1 MeV
• D. 8 MeV

Hint:

The energy released in nuclear reactions can be calculated by finding the difference between the binding energies of the reactants and the products. The energy per nucleon can be found by dividing the energy released by the number of nucleons.

Answer 1

The Correct Option is A

The equation for the Q-value is given by: \[ Q = B_{E_{\text{He}}} - B_{E_{\text{H}}} \] Substitute the values: \[ Q = 4 \times 7.2 - 2 \times (2 \times 1.1) \] Simplify: \[ Q = 28.8 - 4.4 = 24.4 \, \text{MeV} \] Thus, the final value of \( Q \) is: \[ Q = 24.4 \, \text{MeV} \]