If `K_(1) and K_(2)` are maximum kinetic energies of photoelectrons emitted when light of wavelength `lambda_(2) and lambda_(1)` respectively are incident on a metallic surface. `If lambda_(1)=3lambda_(2)` then

Options

A. `K_(1)gt((K_(2))/3)`
B. `K_(1)lt((K_(2))/3)`
C. `K_(1)=3K_(2)`
D. `K_(2)=3K_(1)`

Explanation

To solve the problem, we will use Einstein's photoelectric equation, which relates the maximum kinetic energy of photoelectrons emitted from a metallic surface to the wavelength of the incident light and the work function of the metal.

### Step-by-Step Solution:

1. **Understand the Problem Statement:**
We are given two wavelengths of light, \( \lambda_1 \) and \( \lambda_2 \), and the relationship \( \lambda_1 = 3\lambda_2 \). We need to find the relationship between the maximum kinetic energies of the emitted photoelectrons, \( K_1 \) and \( K_2 \), corresponding to these wavelengths.

2. **Write Einstein's Photoelectric Equation:**
The maximum kinetic energy \( K \) of the emitted photoelectrons can be expressed as:
\[ K = \frac{hc}{\lambda} - \phi_0 \]
where \( h \) is Planck's constant, \( c \) is the speed of light, \( \lambda \) is the wavelength of the incident light, and \( \phi_0 \) is the work function of the metal.

3. **Apply the Equation for \( \lambda_1 \):**
For the wavelength \( \lambda_1 \):
\[ K_1 = \frac{hc}{\lambda_1} - \phi_0 \]

4. **Apply the Equation for \( \lambda_2 \):**
For the wavelength \( \lambda_2 \):
\[ K_2 = \frac{hc}{\lambda_2} - \phi_0 \]

5. **Substitute the Relationship Between Wavelengths:**
Given \( \lambda_1 = 3\lambda_2 \), we can substitute this into the equation for \( K_1 \):
\[ K_1 = \frac{hc}{3\lambda_2} - \phi_0 \]

6. **Express \( K_1 \) and \( K_2 \) in Terms of \( \lambda_2 \):**
Now we have:
\[ K_1 = \frac{hc}{3\lambda_2} - \phi_0 \]
\[ K_2 = \frac{hc}{\lambda_2} - \phi_0 \]

7. **Find the Difference \( K_1 - K_2 \):**
To find the relationship between \( K_1 \) and \( K_2 \), we subtract the two equations:
\[ K_1 - K_2 = \left(\frac{hc}{3\lambda_2} - \phi_0\right) - \left(\frac{hc}{\lambda_2} - \phi_0\right) \]
Simplifying this gives:
\[ K_1 - K_2 = \frac{hc}{3\lambda_2} - \frac{hc}{\lambda_2} \]
\[ K_1 - K_2 = \frac{hc}{3\lambda_2} - \frac{3hc}{3\lambda_2} = -\frac{2hc}{3\lambda_2} \]

8. **Rearranging the Equation:**
This implies:
\[ K_1 = K_2 - \frac{2hc}{3\lambda_2} \]

9. **Conclusion:**
Since \( K_1 \) is less than \( K_2 \) by a positive quantity, we can conclude that:
\[ K_2 > K_1 \]

### Final Result:
Thus, the relationship between the maximum kinetic energies of the photoelectrons is:
\[ K_2 > K_1 \]