If the binding energy per nucleon in `L i^7` and `He^4` nuclei are respectively `5.60 MeV` and `7.06 MeV`. Then energy of reaction `L i^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

Explanation

To solve the problem, we need to calculate the energy of the reaction \( \text{Li}^7 + p \rightarrow 2 \text{He}^4 \) using the binding energies per nucleon for lithium-7 and helium-4. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the Binding Energy of \( \text{Li}^7 \)

The binding energy per nucleon for \( \text{Li}^7 \) is given as \( 5.60 \, \text{MeV} \). Since \( \text{Li}^7 \) has 7 nucleons (3 protons and 4 neutrons), we can calculate the total binding energy:

\[ \text{Binding Energy of } \text{Li}^7 = \text{Number of nucleons} \times \text{Binding Energy per nucleon} \]
\[ = 7 \times 5.60 \, \text{MeV} = 39.20 \, \text{MeV} \]

### Step 2: Calculate the Binding Energy of \( \text{He}^4 \)

The binding energy per nucleon for \( \text{He}^4 \) is given as \( 7.06 \, \text{MeV} \). Since \( \text{He}^4 \) has 4 nucleons (2 protons and 2 neutrons), we can calculate the total binding energy:

\[ \text{Binding Energy of } \text{He}^4 = \text{Number of nucleons} \times \text{Binding Energy per nucleon} \]
\[ = 4 \times 7.06 \, \text{MeV} = 28.24 \, \text{MeV} \]

### Step 3: Calculate the Total Binding Energy for Two \( \text{He}^4 \) Nuclei

Since the reaction produces two \( \text{He}^4 \) nuclei, we need to multiply the binding energy of one \( \text{He}^4 \) nucleus by 2:

\[ \text{Total Binding Energy for } 2 \text{He}^4 = 2 \times 28.24 \, \text{MeV} = 56.48 \, \text{MeV} \]

### Step 4: Calculate the Energy of the Reaction

The energy of the reaction can be found by taking the difference between the total binding energy of the products and the reactants:

\[ \text{Energy of Reaction} = \text{Total Binding Energy of Products} - \text{Total Binding Energy of Reactants} \]
\[ = 56.48 \, \text{MeV} - 39.20 \, \text{MeV} = 17.28 \, \text{MeV} \]

### Final Answer

The energy of the reaction \( \text{Li}^7 + p \rightarrow 2 \text{He}^4 \) is \( 17.28 \, \text{MeV} \).

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