Question

Given the nuclear reaction: \[ \text{H}_2^2 + \text{H}_1^2 \rightarrow \text{He}_4 \] Binding energy per nucleon of H\(_1^2\) and H\(_2^4\) are 1.1 MeV and 7 MeV respectively. Find energy released in the nuclear reaction given above.

• A. 2.2 MeV
• B. 24.1 MeV
• C. 5.9 MeV
• D. 3 MeV

Answer 1

The Correct Option is A

Step 1: Understanding binding energy.
Binding energy per nucleon represents the energy required to break a nucleus into its constituent protons and neutrons. For a reaction, the energy released can be calculated by finding the difference in the binding energies of the reactants and products. Step 2: Calculation of binding energy.
- Binding energy of H\(_1^2\) (2 protons) = \( 1.1 \times 2 = 2.2 \) MeV.
- Binding energy of H\(_2^4\) (4 nucleons) = \( 7 \times 4 = 28 \) MeV.
- Total binding energy of reactants = \( 2.2 + 28 = 30.2 \) MeV.
The product is He\(_4\), with 4 nucleons, so the binding energy of He\(_4\) = \( 7 \times 4 = 28 \) MeV. Step 3: Energy released.
Energy released = Binding energy of products - Binding energy of reactants \[ = 28 \, \text{MeV} - 30.2 \, \text{MeV} = -2.2 \, \text{MeV} \] Therefore, the correct energy released is 2.2 MeV (taking the magnitude of the difference).
Final Answer: (A) 2.2 MeV