K.E. of photo electron is E when incident frequency is `lambda_(1)`. It is 2E when incident frequency is `lambda_(2)`. The relation between the wavelength is

Options

A. `lambda_(2) = lambda_(1)`
B. `lambda_(2) gt 2lambda_(1)`
C. `lambda_(2) lt 2lambda_(1)`
D. `lambda_(2) = 2lambda_(1)`

Explanation

To solve the problem, we need to analyze the kinetic energy of photoelectrons emitted under different incident frequencies (or wavelengths). The relationship between the kinetic energy of the emitted photoelectrons and the incident wavelengths can be derived from the photoelectric effect.

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
According to the photoelectric effect, the energy of the incident photon is given by:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the incident light.

2. **Setting Up the Equations**:
For the first wavelength \( \lambda_1 \), the kinetic energy of the photoelectron is \( E \). The equation can be expressed as:
\[ \frac{hc}{\lambda_1} = \phi + E \quad \text{(1)} \]
where \( \phi \) is the work function of the metal.

For the second wavelength \( \lambda_2 \), the kinetic energy is \( 2E \). The equation becomes:
\[ \frac{hc}{\lambda_2} = \phi + 2E \quad \text{(2)} \]

3. **Dividing the Two Equations**:
To find the relationship between \( \lambda_1 \) and \( \lambda_2 \), we can divide equation (1) by equation (2):
\[ \frac{\frac{hc}{\lambda_1}}{\frac{hc}{\lambda_2}} = \frac{\phi + E}{\phi + 2E} \]
This simplifies to:
\[ \frac{\lambda_2}{\lambda_1} = \frac{\phi + E}{\phi + 2E} \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\phi + 2E}{\phi + E} \]

5. **Analyzing the Right Side**:
The term \( \frac{\phi + 2E}{\phi + E} \) can be analyzed. Since \( E \) is a positive quantity, it follows that:
\[ \phi + 2E > \phi + E \]
Thus, \( \frac{\phi + 2E}{\phi + E} > 1 \).

6. **Conclusion**:
Since \( \frac{\lambda_1}{\lambda_2} > 1 \), we conclude that:
\[ \lambda_1 > \lambda_2 \]
However, we need to express it in terms of a multiple. From the equation:
\[ \frac{\lambda_1}{\lambda_2} = 1 + \frac{E}{\phi + E} \]
Since \( \frac{E}{\phi + E} < 1 \), it follows that:
\[ \lambda_1 < 2 \lambda_2 \]

### Final Relation:
Thus, the relation between the wavelengths is:
\[ \lambda_1 < 2\lambda_2 \]