Nuclei `X` decay into nuclei `Y` by emitting `alpha`-aprticles. Energies of `alpha`-particle are found to be only `1 MeV` & `1.4 MeV`. Disregarding the recoil of nuclei `Y`. The energy of `gamma` photon emitted will be

Options

A. `0.8 MeV`
B. `1.4 MeV`
C. `1 MeV`
D. `0.4 MeV`

Explanation

To solve the problem, we need to determine the energy of the gamma photon emitted during the decay of nucleus X into nucleus Y by the emission of alpha particles. We have two energies of the alpha particles emitted: 1 MeV and 1.4 MeV.

### Step-by-Step Solution:

1. **Identify the Energies of Alpha Particles**:
- The energies of the emitted alpha particles are given as:
- \( E_1 = 1 \, \text{MeV} \)
- \( E_2 = 1.4 \, \text{MeV} \)

2. **Calculate the Energy Difference**:
- The energy of the gamma photon emitted during the decay process can be calculated as the difference between the two energies of the alpha particles.
- The formula to find the energy of the gamma photon (\( E_{\gamma} \)) is:
\[ E_{\gamma} = E_2 - E_1 \]
- Substituting the values:
\[ E_{\gamma} = 1.4 \, \text{MeV} - 1 \, \text{MeV} = 0.4 \, \text{MeV} \]

3. **Conclusion**:
- Therefore, the energy of the gamma photon emitted is \( 0.4 \, \text{MeV} \).

### Final Answer:
The energy of the gamma photon emitted is \( 0.4 \, \text{MeV} \).

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