Question

In the reaction \(^2_1H+^3_1H\) \(\rightarrow\) \(^4_2He+^1_0n,\) if the binding energies of \(^2_1H\), \(^3_1H\) and \(^4_2He\) are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is:

• A. c+a-b
• B. c-a-b
• C. a+b+c
• D. a+b-c

Answer 1

The Correct Option is B

When analyzing the given nuclear reaction:

\[^2_1H + ^3_1H \rightarrow ^4_2He + ^1_0n\]

we need to calculate the energy released using the binding energies of the involved nuclei.

1. Identify the binding energies:

   • Binding energy of ^2_1H: a MeV
   • Binding energy of ^3_1H: b MeV
   • Binding energy of ^4_2He: c MeV

2. Understand the concept of binding energy:

   The energy released in the nuclear reaction can be calculated as the difference between the binding energy of the product and the reactants. The total binding energy of the reactants (^2_1H and ^3_1H) is given as:

   E_{\text{reactants}} = a + b

   The total binding energy of the products (^4_2He and ^1_0n, where the neutron has a negligible binding energy compared to ^4_2He) is:

   E_{\text{products}} = c

3. Calculate the energy released:

   The energy released can be calculated by subtracting the total binding energy of the reactants from that of the products:

   E_{\text{released}} = E_{\text{products}} - E_{\text{reactants}} = c - (a + b)

   Simplifying this gives:

   E_{\text{released}} = c - a - b

4. Conclusion:

   The energy released in this reaction is c - a - b MeV.

Thus, the correct answer is c-a-b.