A gas absorbs a photon of 300 nm and then reemitts two photons. One photon has a wavelength 600 nm. The wavelength of second photon is -

Options

A. 300 nm
B. 400 nm
C. 500 nm
D. 600 nm

Answer

To solve the problem, we need to find the wavelength of the second photon emitted by the gas after it absorbs a photon of 300 nm and emits one photon of 600 nm.

### Step-by-Step Solution:

1. **Understanding Energy Absorption and Emission**:
- When the gas absorbs a photon of wavelength \( \lambda_0 = 300 \, \text{nm} \), it gains energy. This energy is then re-emitted as two photons, one with a wavelength \( \lambda_1 = 600 \, \text{nm} \) and another with an unknown wavelength \( \lambda_2 \).

2. **Using the Energy-Wavelength Relationship**:
- The energy of a photon can be expressed as:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

3. **Setting Up the Energy Equation**:
- The energy absorbed by the gas when it absorbs the photon of 300 nm is equal to the total energy of the two emitted photons:
\[ E_{\text{absorbed}} = E_1 + E_2 \]
- In terms of wavelengths, this can be written as:
\[ \frac{hc}{300} = \frac{hc}{600} + \frac{hc}{\lambda_2} \]

4. **Canceling Constants**:
- Since \( hc \) is common in all terms, we can cancel it out:
\[ \frac{1}{300} = \frac{1}{600} + \frac{1}{\lambda_2} \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{1}{\lambda_2} = \frac{1}{300} - \frac{1}{600} \]

6. **Finding a Common Denominator**:
- The common denominator for 300 and 600 is 600:
\[ \frac{1}{300} = \frac{2}{600} \]
- Therefore:
\[ \frac{1}{\lambda_2} = \frac{2}{600} - \frac{1}{600} = \frac{1}{600} \]

7. **Solving for \( \lambda_2 \)**:
- Taking the reciprocal gives:
\[ \lambda_2 = 600 \, \text{nm} \]

### Conclusion:
The wavelength of the second photon emitted by the gas is \( \lambda_2 = 600 \, \text{nm} \).