The energy of the reaction `Li^(7)+prarr2He^(4)` is ( the binding energy per nucleon in `Li^(7)` and `He^(4)` nuclei are `5.60` and `7.60 MeV` respectively.)

Options

A. (a)19.6 MeV
B. (b)2.4 MeV
C. (c)8.4 MeV
D. (d)17.28 MeV

Answer

To solve the problem, we need to calculate the energy of the reaction \( \text{Li}^7 + \text{p} \rightarrow 2 \text{He}^4 \) using the binding energy per nucleon of the involved nuclei.

### Step-by-Step Solution:

1. **Calculate the Binding Energy of Lithium-7 (\( \text{Li}^7 \))**:
- The binding energy per nucleon for \( \text{Li}^7 \) is given as \( 5.60 \, \text{MeV} \).
- The total number of nucleons in \( \text{Li}^7 \) is 7.
- Therefore, the total binding energy for \( \text{Li}^7 \) is:
\[ BE_{\text{Li}^7} = 7 \times 5.60 \, \text{MeV} = 39.20 \, \text{MeV} \]

2. **Calculate the Binding Energy of Helium-4 (\( \text{He}^4 \))**:
- The binding energy per nucleon for \( \text{He}^4 \) is given as \( 7.60 \, \text{MeV} \).
- The total number of nucleons in \( \text{He}^4 \) is 4.
- Therefore, the total binding energy for one \( \text{He}^4 \) nucleus is:
\[ BE_{\text{He}^4} = 4 \times 7.60 \, \text{MeV} = 30.40 \, \text{MeV} \]

3. **Calculate the Total Binding Energy for Two Helium-4 Nuclei**:
- Since the reaction produces two \( \text{He}^4 \) nuclei, the total binding energy for two \( \text{He}^4 \) nuclei is:
\[ BE_{\text{total He}} = 2 \times 30.40 \, \text{MeV} = 60.80 \, \text{MeV} \]

4. **Calculate the Energy of the Reaction**:
- The energy released in the reaction can be calculated by subtracting the binding energy of the reactants from the binding energy of the products:
\[ E_{\text{reaction}} = BE_{\text{total He}} - BE_{\text{Li}^7} \]
\[ E_{\text{reaction}} = 60.80 \, \text{MeV} - 39.20 \, \text{MeV} = 21.60 \, \text{MeV} \]

### Final Answer:
The energy of the reaction \( \text{Li}^7 + \text{p} \rightarrow 2 \text{He}^4 \) is \( 21.60 \, \text{MeV} \).

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