If `K_(1) and K_(2)` are maximum kinetic energies of photoelectrons emitted when light of wavelength `lambda_(1) and lambda_(2)` 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 to the energy of the incident photons and the work function of the metal.

### Step-by-Step Solution:

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

2. **Einstein's Photoelectric Equation**:
The maximum kinetic energy of the emitted photoelectrons can be expressed as:
\[ K = E - \phi_0 \]
where \( E \) is the energy of the incident photon and \( \phi_0 \) is the work function of the metal.

3. **Photon Energy in Terms of Wavelength**:
The energy of a photon can be expressed in terms of its wavelength:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light.

4. **Applying the Equation for Both Wavelengths**:
For \( \lambda_1 \):
\[ K_1 = \frac{hc}{\lambda_1} - \phi_0 \]
For \( \lambda_2 \):
\[ K_2 = \frac{hc}{\lambda_2} - \phi_0 \]

5. **Finding the Relation Between \( K_1 \) and \( K_2 \)**:
We can subtract the two equations:
\[ K_1 - K_2 = \left(\frac{hc}{\lambda_1} - \phi_0\right) - \left(\frac{hc}{\lambda_2} - \phi_0\right) \]
This simplifies to:
\[ K_1 - K_2 = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_2} \]
Factoring out \( hc \):
\[ K_1 - K_2 = hc \left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) \]

6. **Substituting the Relation \( \lambda_1 = 3\lambda_2 \)**:
Substitute \( \lambda_1 \) in the equation:
\[ K_1 - K_2 = hc \left(\frac{1}{3\lambda_2} - \frac{1}{\lambda_2}\right) \]
This can be simplified to:
\[ K_1 - K_2 = hc \left(\frac{1 - 3}{3\lambda_2}\right) = -\frac{2hc}{3\lambda_2} \]

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

8. **Analyzing the Result**:
Since \( \frac{2hc}{3\lambda_2} \) is a positive quantity, we conclude that:
\[ K_1 < K_2 \]
Therefore, we can express this as:
\[ K_1 < \frac{K_2}{3} \]

### Final Result:
Thus, the relationship between the maximum kinetic energies of the photoelectrons is:
\[ K_1 < \frac{K_2}{3} \]