Question

If the binding energy per nucleon of deuteron (\( ^1\mathrm{H}^2 \)) is 1.15 MeV and an \(\alpha\)-particle has a binding energy of 7.1 MeV per nucleon, then the energy released per nucleon in the given reaction is \[ ^1\mathrm{H}^2 + ^1\mathrm{H}^2 \rightarrow ^2\mathrm{He}^4 + Q \]

• A. 23.8 MeV
• B. 26.1 MeV
• C. 5.95 MeV
• D. 28.9 MeV

Hint:

To find energy released in nuclear reactions, subtract total binding energy of reactants from that of products. Divide by total number of nucleons to find per nucleon.

Answer 1

The Correct Option is C

Step 1: Total binding energy of the reactants
Each deuteron has 2 nucleons, binding energy per nucleon = 1.15 MeV. So, total for 2 deuterons = \(2 \times 2 \times 1.15 = 4.6 \, \text{MeV}\) Step 2: Binding energy of the product (\(^4\mathrm{He}\))
An alpha particle has 4 nucleons with 7.1 MeV per nucleon: \[ \text{Total} = 4 \times 7.1 = 28.4 \, \text{MeV} \] Step 3: Energy released (Q-value) \[ Q = 28.4 - 4.6 = 23.8 \, \text{MeV} \] Step 4: Energy released per nucleon in the product
Total nucleons in \(^4\mathrm{He}\) = 4 \[ \text{Energy per nucleon} = \frac{23.8}{4} = 5.95 \, \text{MeV} \]