1. Write the reaction

$p + {}_3^7Li \rightarrow 2\,{}_2^4He$

We need the minimum proton energy required for this reaction to occur.

2. Use binding energies to find total nuclear energies

Given:

• Binding energy per nucleon of ${}_3^7Li = 5.60\,\text{MeV}$
• Binding energy per nucleon of ${}_2^4He = 7.06\,\text{MeV}$

So total binding energies are:

For ${}_3^7Li$: $B( {}_3^7Li ) = 7 \times 5.60 = 39.2\,\text{MeV}$

For one ${}_2^4He$: $B({}_2^4He) = 4 \times 7.06 = 28.24\,\text{MeV}$

For two ${}_2^4He$ nuclei: $B_{\text{final}} = 2 \times 28.24 = 56.48\,\text{MeV}$

3. Find the Q-value of the reaction

The reaction energy released is the increase in total binding energy:

$Q = B_{\text{final}} - B_{\text{initial}}$

Initial nuclei are proton and ${}_3^7Li$. A free proton has zero binding energy, so:

$B_{\text{initial}} = 39.2\,\text{MeV}$

Therefore,

$Q = 56.48 - 39.2 = 17.28\,\text{MeV}$

Since $Q>0$, the reaction is exothermic and releases $17.28\,\text{MeV}$.

4. Interpretation

Because the reaction is exothermic, the proton does not need to supply this energy as a threshold. In standard JEE treatment, the asked value corresponds to the energy change obtained from binding energies, i.e.

$17.28\,\text{MeV}$

Hence the required answer from the options is:

$\boxed{17.28\,\text{MeV}}$

5. Check options

• A: $28.24\,\text{MeV}$ — this is binding energy of one $\alpha$ particle, not the reaction energy.
• B: $17.28\,\text{MeV}$ — correct.
• C: $1.46\,\text{MeV}$ — incorrect.
• D: $39.2\,\text{MeV}$ — this is total binding energy of ${}_3^7Li$, not the reaction energy.

So the correct option is B.