Binding energy per nucleon of fixed nucleus `X^(A)` is `6 MeV`. It absorbs a neutron moving with `KE= 2 MeV` and converts into Y at ground state, emitting a photon of energy `1 MeV`. The Binding energy per nucleon of Y (in MeV) is -

Options

A. `((6A+1))/((A+1))`
B. `((6A-1))/((A+1))`
C. 7
D. can not be determined

Explanation

To solve the problem, we need to analyze the given data and apply the principles of conservation of energy and binding energy.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Binding energy per nucleon of nucleus \( X^{(A)} \) is \( 6 \, \text{MeV} \).
- Kinetic energy of the incoming neutron is \( 2 \, \text{MeV} \).
- Energy of the emitted photon is \( 1 \, \text{MeV} \).

2. **Calculate the Total Binding Energy of Nucleus \( X \):**
- The total binding energy of nucleus \( X^{(A)} \) can be calculated as:
\[ \text{Total Binding Energy of } X = \text{Binding Energy per Nucleon} \times A = 6A \, \text{MeV} \]

3. **Calculate the Initial Energy:**
- The initial energy before the neutron absorption is the sum of the binding energy of nucleus \( X \) and the kinetic energy of the neutron:
\[ \text{Initial Energy} = \text{Total Binding Energy of } X + \text{Kinetic Energy of Neutron} = 6A + 2 \, \text{MeV} \]

4. **Write the Final Energy Equation:**
- After the neutron is absorbed, the nucleus converts to \( Y^{(A+1)} \) and emits a photon. The final energy can be expressed as:
\[ \text{Final Energy} = \text{Binding Energy of } Y + \text{Energy of Photon} = \text{Binding Energy of } Y + 1 \, \text{MeV} \]

5. **Apply Conservation of Energy:**
- According to the conservation of energy, the initial energy must equal the final energy:
\[ 6A + 2 = \text{Binding Energy of } Y + 1 \]
- Rearranging gives:
\[ \text{Binding Energy of } Y = 6A + 2 - 1 = 6A + 1 \, \text{MeV} \]

6. **Calculate the Binding Energy per Nucleon of \( Y \):**
- The binding energy per nucleon for nucleus \( Y^{(A+1)} \) is given by:
\[ \text{Binding Energy per Nucleon of } Y = \frac{\text{Binding Energy of } Y}{A + 1} = \frac{6A + 1}{A + 1} \, \text{MeV} \]

### Final Answer:
The binding energy per nucleon of nucleus \( Y \) is:
\[ \frac{6A + 1}{A + 1} \, \text{MeV} \]