Two identical metal plates show photoelectric effect. Light of wavelength `lamda_A` falls on plate A and `lamda_B` fall on plate B `lama_A=2lamda_B`, The maximum KE of the photoelectrons are `K_A` and `K_B`, respectively, Which one of the following is true?

Options

A. `2K_A=K_B`
B. `K_A=2K_B`
C. `(K_AltK_B/(2))`
D. `K_Agt2K_B`

Explanation

To solve the problem, we need to analyze the photoelectric effect for two plates A and B with different wavelengths of light falling on them. The relationship between the kinetic energy of the emitted photoelectrons and the wavelength of the incident light is given by the photoelectric equation:

\[ K = \frac{hc}{\lambda} - \Phi \]

where:
- \( K \) is the maximum kinetic energy of the emitted photoelectrons,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light,
- \( \Phi \) is the work function of the metal.

### Step-by-Step Solution:

1. **Identify the Wavelengths:**
We have two wavelengths:
- For plate A: \( \lambda_A \)
- For plate B: \( \lambda_B \)
Given that \( \lambda_A = 2\lambda_B \).

2. **Write the Kinetic Energy Equations:**
For plate A, the maximum kinetic energy \( K_A \) can be expressed as:
\[ K_A = \frac{hc}{\lambda_A} - \Phi \]
For plate B, the maximum kinetic energy \( K_B \) can be expressed as:
\[ K_B = \frac{hc}{\lambda_B} - \Phi \]

3. **Substitute \( \lambda_A \) in Terms of \( \lambda_B \):**
Since \( \lambda_A = 2\lambda_B \), we can substitute this into the equation for \( K_A \):
\[ K_A = \frac{hc}{2\lambda_B} - \Phi \]

4. **Express \( K_A \) and \( K_B \) in Terms of \( \lambda_B \):**
Now we can rewrite both kinetic energy equations:
- For plate A:
\[ K_A = \frac{hc}{2\lambda_B} - \Phi \]
- For plate B:
\[ K_B = \frac{hc}{\lambda_B} - \Phi \]

5. **Compare \( K_A \) and \( K_B \):**
To compare \( K_A \) and \( K_B \), we can subtract \( K_A \) from \( K_B \):
\[ K_B - K_A = \left(\frac{hc}{\lambda_B} - \Phi\right) - \left(\frac{hc}{2\lambda_B} - \Phi\right) \]
Simplifying this gives:
\[ K_B - K_A = \frac{hc}{\lambda_B} - \frac{hc}{2\lambda_B} = \frac{hc}{2\lambda_B} \]

6. **Conclusion:**
Since \( K_B - K_A = \frac{hc}{2\lambda_B} > 0 \), it follows that:
\[ K_B > K_A \]
Therefore, the maximum kinetic energy of the photoelectrons from plate B is greater than that from plate A.

### Final Answer:
**\( K_B > K_A \)**