Question

If the binding energy per nucleon in \( ^7_3Li \) and \( ^4_2He \) nuclei are respectively \(5.60\,MeV\) and \(7.06\,MeV\), then energy of \(p\) in the reaction \( p + ^7_3Li \rightarrow 2\,^4_2He \) is

• A. \(12.28\,MeV\)
• B. \(13.28\,MeV\)
• C. \(28.28\,MeV\)
• D. \(17.28\,MeV\)

Hint:

In nuclear reactions, energy released = final total binding energy − initial total binding energy.

Answer 1

The Correct Option is D

Step 1: Concept of binding energy.
Binding energy of a nucleus is the energy required to break it into its constituent nucleons. Total binding energy is given by:
\[ BE = (\text{binding energy per nucleon}) \times (\text{number of nucleons}) \]

Step 2: Calculate total binding energy of \( ^7_3Li \).
\[ BE_{Li} = 5.60 \times 7 = 39.2\,MeV \]

Step 3: Calculate total binding energy of \( ^4_2He \).
\[ BE_{He} = 7.06 \times 4 = 28.24\,MeV \]
Since two helium nuclei are formed:
\[ BE_{\text{final}} = 2 \times 28.24 = 56.48\,MeV \]

Step 4: Initial total binding energy.
Proton has no binding energy, so:
\[ BE_{\text{initial}} = 39.2\,MeV \]

Step 5: Energy released in reaction.
\[ Q = BE_{\text{final}} - BE_{\text{initial}} \]
\[ Q = 56.48 - 39.2 = 17.28\,MeV \]

Step 6: Interpretation.
This energy appears as kinetic energy of products, hence corresponds to energy involved in reaction.

Step 7: Final conclusion.
\[ \boxed{17.28\,MeV} \] Hence, correct answer is option (D).