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. `2K_(1)=K_(2)`
B. `K_(1)=2K_(2)`
C. `K_(1)lt(K_(2))/(3)`
D. `K_(1)lt2K_(2)`

Answer

To solve the problem, we need to establish the relationship between the maximum kinetic energies \( K_1 \) and \( K_2 \) of photoelectrons emitted when light of wavelengths \( \lambda_1 \) and \( \lambda_2 \) are incident on a metallic surface, given that \( \lambda_1 = 3\lambda_2 \).

### Step-by-Step Solution:

1. **Understanding the Photoelectric Effect**:
The kinetic energy \( K \) of photoelectrons emitted can be expressed using the equation:
\[ K = \frac{hc}{\lambda} - \phi \]
where \( h \) is Planck's constant, \( c \) is the speed of light, \( \lambda \) is the wavelength of the incident light, and \( \phi \) is the work function of the metal.

2. **Setting Up the Equations**:
For the two wavelengths, we can write:
- For \( \lambda_1 \):
\[ K_1 = \frac{hc}{\lambda_1} - \phi \]
- For \( \lambda_2 \):
\[ K_2 = \frac{hc}{\lambda_2} - \phi \]

3. **Substituting the Given Relationship**:
We know that \( \lambda_1 = 3\lambda_2 \). Substituting this into the equation for \( K_1 \):
\[ K_1 = \frac{hc}{3\lambda_2} - \phi \]

4. **Expressing \( K_1 \) in Terms of \( K_2 \)**:
Now we can express \( K_1 \) in terms of \( K_2 \):
- From the equation for \( K_2 \):
\[ K_2 = \frac{hc}{\lambda_2} - \phi \]
- Rearranging gives:
\[ \phi = \frac{hc}{\lambda_2} - K_2 \]
- Substituting this expression for \( \phi \) into the equation for \( K_1 \):
\[ K_1 = \frac{hc}{3\lambda_2} - \left(\frac{hc}{\lambda_2} - K_2\right) \]
- Simplifying this gives:
\[ K_1 = \frac{hc}{3\lambda_2} - \frac{hc}{\lambda_2} + K_2 \]
- Combining the terms:
\[ K_1 = K_2 - \frac{hc}{\lambda_2} + \frac{hc}{3\lambda_2} \]
- This simplifies to:
\[ K_1 = K_2 - \frac{2hc}{3\lambda_2} \]

5. **Finding the Relationship**:
Now we can express \( K_2 \) in terms of \( K_1 \):
\[ K_2 = K_1 + \frac{2hc}{3\lambda_2} \]
Rearranging gives:
\[ K_2 - 3K_1 = \frac{2hc}{3\lambda_2} \]

6. **Conclusion**:
Since \( \phi \) (the work function) is always greater than zero, we can conclude:
\[ K_2 - 3K_1 > 0 \implies K_2 > 3K_1 \]
Thus, we find that:
\[ K_1 < \frac{K_2}{3} \]

### Final Answer:
The relationship between the maximum kinetic energies is:
\[ K_1 < \frac{K_2}{3} \]