Question

Two lighter nuclei combine to form a comparatively heavier nucleus by the relation \[ \frac{2}{1}\text{X} + \frac{2}{1}\text{X} = \frac{4}{2}\text{Y} \] The binding energies per nucleon of \( \frac{2}{1}\text{X} \) and \( \frac{4}{2}\text{Y} \) are 1.1 MeV and 7.6 MeV respectively. The energy released in this process is

• A. 26 MeV
• B. 56 MeV
• C. 78 MeV
• D. 108 MeV

Hint:

Always remember to multiply the B.E. per nucleon by the mass number (A) to get the total B.E. for each nucleus. A common mistake is to simply subtract the per-nucleon values without accounting for the total number of particles!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In a nuclear fusion reaction, the energy released ($Q$-value) is the difference between the total binding energy of the product nucleus and the total binding energy of the reacting nuclei. An increase in the total binding energy corresponds to energy being released.
Step 2: Key Formula or Approach:
Total Binding Energy (BE) = (Binding Energy per nucleon) $\times$ (Number of nucleons, A)
Energy released ($Q$) = Total BE of products $-$ Total BE of reactants
Step 3: Detailed Explanation:
For the reactant nucleus ${}_1^2\text{X}$:
Number of nucleons $A = 2$
Binding Energy per nucleon = 1.1 MeV
Total BE of one ${}_1^2\text{X}$ nucleus = $2 \times 1.1 = 2.2$ MeV
Since there are two ${}_1^2\text{X}$ nuclei reacting,
Total BE of reactants = $2 \times 2.2 = 4.4$ MeV
For the product nucleus ${}_2^4\text{Y}$:
Number of nucleons $A = 4$
Binding Energy per nucleon = 7.6 MeV
Total BE of the product ${}_2^4\text{Y}$ = $4 \times 7.6 = 30.4$ MeV
Energy released ($Q$): \[ Q = (\text{Total BE of products}) - (\text{Total BE of reactants}) \] \[ Q = 30.4 \, \text{MeV} - 4.4 \, \text{MeV} = 26 \, \text{MeV} \] Step 4: Final Answer:
The energy released in the process is 26 MeV. The correct option is (A).