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, energy is released when the total binding energy of the product nucleus is greater than the total binding energy of the reactant nuclei. The energy released (\(Q\)-value) is the difference between these two.

Step 2: Key Formula or Approach
1. Total Binding Energy (B.E.) = (B.E. per nucleon) \(\times\) (Number of nucleons, \(A\)) 2. Energy released = (Total B.E. of products) \(-\) (Total B.E. of reactants)

Step 3: Detailed Explanation
1. Reactants (\(2\) nuclei of \(_1^2\text{X}\)): - Each nucleus has \(A = 2\). - B.E. per nucleon = 1.1 MeV. - Total B.E. of reactants = \(2 \times (2 \times 1.1) = 2 \times 2.2 = 4.4\) MeV. 2. Product (\(1\) nucleus of \(_2^4\text{Y}\)): - Nucleus has \(A = 4\). - B.E. per nucleon = 7.6 MeV. - Total B.E. of product = \(4 \times 7.6 = 30.4\) MeV. 3. Energy Released: - \(\Delta E = 30.4 - 4.4 = 26\) MeV.

Step 4: Final Answer
The energy released in the process is 26 MeV.