A photon of energy 10.2 eV corresponds to light of wavelength `lamda_(0)`. Due to an electron transition from n=2 to n=1 in a hydrogen atom, light of wavelength `lamda` is emitted. If we take into account the recoil of the atom when the photon is emitted.

Options

A. `lambda lt lambda_(0)`
B. `lambda gt lambda_(0)`
C. `lambda=lambda_(0)`
D. None of these

Answer

To solve the problem, we need to analyze the situation involving the emission of a photon during an electron transition in a hydrogen atom, taking into account the recoil of the atom.

### Step-by-Step Solution:

1. **Photon Energy Calculation**:
The energy of the photon emitted during the transition from n=2 to n=1 in a hydrogen atom is given as 10.2 eV. We can express this energy in terms of wavelength using the equation:
\[ E = \frac{hc}{\lambda_0} \]
where \(E\) is the energy of the photon, \(h\) is Planck's constant, and \(c\) is the speed of light.

2. **Determine the Wavelength**:
Rearranging the equation gives us:
\[ \lambda_0 = \frac{hc}{E} \]
Substituting \(E = 10.2 \, \text{eV}\) (and converting eV to Joules if necessary) allows us to find \(\lambda_0\).

3. **Photon Emission and Recoil**:
When the photon is emitted, the hydrogen atom recoils. The momentum conservation principle states:
\[ mv = \frac{h}{\lambda} \]
where \(m\) is the mass of the hydrogen atom, \(v\) is its recoil velocity, and \(\lambda\) is the wavelength of the emitted photon.

4. **Energy Conservation**:
The total energy before and after the emission must be conserved. Initially, the energy is just the energy of the photon:
\[ 10.2 \, \text{eV} = \frac{1}{2} mv^2 + \frac{hc}{\lambda} \]
Here, \(\frac{1}{2} mv^2\) is the kinetic energy of the recoiling atom.

5. **Relating Wavelengths**:
Since the recoil of the atom causes a change in the energy of the emitted photon, we can conclude that:
\[ \frac{hc}{\lambda} < 10.2 \, \text{eV} \]
This implies that:
\[ \lambda > \lambda_0 \]
Thus, the wavelength of the emitted photon \(\lambda\) is greater than the original wavelength \(\lambda_0\).

### Conclusion:
The final conclusion is that the wavelength of the emitted photon (\(\lambda\)) is greater than the initial wavelength (\(\lambda_0\)) due to the recoil of the hydrogen atom.