If the binding energy per nucleon in `_(3)^(7) Li and _(2)^(4)He nuclei are 5.60 MeV and 7.06MeV` respectively then in the reaction
`P +_(3)^(7) Li rarr 2 _(2)^(4) He`
energy of proton mnust be

Options

A. `28.24 MeV`
B. `17.28 MeV`
C. `1.46 MeV`
D. `39.2 MeV`

Explanation

To solve the problem, we need to calculate the energy of the proton involved in the reaction \( P + _{3}^{7}Li \rightarrow 2 \, _{2}^{4}He \) using the binding energy per nucleon of the involved nuclei.

### Step 1: Calculate the binding energy of \( _{3}^{7}Li \)
The binding energy of a nucleus can be calculated using the formula:
\[ \text{Binding Energy} = (\text{Number of Nucleons}) \times (\text{Binding Energy per Nucleon}) \]
For \( _{3}^{7}Li \):
- Number of nucleons = 7
- Binding energy per nucleon = 5.60 MeV

So, the total binding energy of \( _{3}^{7}Li \) is:
\[ BE_{Li} = 7 \times 5.60 = 39.20 \, \text{MeV} \]

### Step 2: Calculate the binding energy of \( _{2}^{4}He \)
For \( _{2}^{4}He \):
- Number of nucleons = 4
- Binding energy per nucleon = 7.06 MeV

So, the total binding energy of \( _{2}^{4}He \) is:
\[ BE_{He} = 4 \times 7.06 = 28.24 \, \text{MeV} \]

### Step 3: Calculate the total binding energy of the products
Since there are 2 helium nuclei produced in the reaction, the total binding energy of the products is:
\[ BE_{products} = 2 \times BE_{He} = 2 \times 28.24 = 56.48 \, \text{MeV} \]

### Step 4: Calculate the energy of the proton
In the reaction, the energy of the proton must provide enough energy to account for the difference in binding energy between the reactants and the products. The energy of the proton can be calculated as follows:
\[ E_{proton} = BE_{products} - BE_{Li} \]
Substituting the values we calculated:
\[ E_{proton} = 56.48 \, \text{MeV} - 39.20 \, \text{MeV} = 17.28 \, \text{MeV} \]

### Final Answer
The energy of the proton must be **17.28 MeV**.

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