If the binding energy per nucleon in `._(3)Li^(7)` and `._(2)He^(4)` nuclei are respectively `5.60` MeV and `7.06` MeV, then the ebergy of proton in the reaction `._(3)Li^(7) +p rarr 2 ._(2)He^(4)` is

Options

A. `19.6` MeV
B. `2.4` MeV
C. `8.4` MeV
D. `17.3` MeV

Answer

To find the energy of the proton in the reaction \( _{3}^{7}\text{Li} + p \rightarrow 2 \, _{2}^{4}\text{He} \), we will use the binding energy per nucleon for the involved nuclei.

### Step-by-Step Solution:

1. **Identify the Binding Energies:**
- The binding energy per nucleon for Lithium-7 (\( _{3}^{7}\text{Li} \)) is given as \( 5.60 \, \text{MeV} \).
- The binding energy per nucleon for Helium-4 (\( _{2}^{4}\text{He} \)) is given as \( 7.06 \, \text{MeV} \).

2. **Calculate Total Binding Energy for Helium Nuclei:**
- There are 2 Helium-4 nuclei produced in the reaction.
- Each Helium-4 nucleus has 4 nucleons.
- Therefore, the total binding energy for 2 Helium-4 nuclei is:
\[ \text{Total Binding Energy for He} = 2 \times (4 \, \text{nucleons}) \times (7.06 \, \text{MeV/nucleon}) = 8 \times 7.06 \, \text{MeV} = 56.48 \, \text{MeV} \]

3. **Calculate Total Binding Energy for Lithium Nucleus:**
- The Lithium-7 nucleus has 7 nucleons.
- Therefore, the total binding energy for one Lithium-7 nucleus is:
\[ \text{Total Binding Energy for Li} = 7 \, \text{nucleons} \times (5.60 \, \text{MeV/nucleon}) = 39.20 \, \text{MeV} \]

4. **Calculate the Energy of the Proton:**
- The energy of the proton can be calculated using the difference in binding energies:
\[ \text{Energy of Proton} = \text{Total Binding Energy for He} - \text{Total Binding Energy for Li} \]
- Substituting the values:
\[ \text{Energy of Proton} = 56.48 \, \text{MeV} - 39.20 \, \text{MeV} = 17.28 \, \text{MeV} \]

5. **Final Result:**
- The energy of the proton in the reaction is approximately \( 17.28 \, \text{MeV} \).

### Summary:
The energy of the proton in the reaction \( _{3}^{7}\text{Li} + p \rightarrow 2 \, _{2}^{4}\text{He} \) is \( 17.28 \, \text{MeV} \).