Question

Light of wavelength \( \lambda_A \) and \( \lambda_B \) falls on two identical metal plates A and B respectively. The maximum kinetic energy of photoelectrons is \( K_A \) and \( K_B \) respectively. Given that \( \lambda_A = 2\lambda_B \), which one of the following relations is true?

• A. \( K_A<\frac{K_B}{2} \)
• B. \( 2K_A = K_B \)
• C. \( K_A = 2K_B \)
• D. \( K_A>2K_B \)

Hint:

For the photoelectric effect, remember:
- Energy of the photon is \( h\nu = \frac{hc}{\lambda} \).
- Maximum kinetic energy: \( K_{\max} = h\nu - \phi \).
- Shorter wavelength means higher frequency and greater kinetic energy.

Answer 1

The Correct Option is A

Step 1: Using Einstein's photoelectric equation:
\[ K_{\max} = h\nu - \phi \] where \( \nu = \frac{c}{\lambda} \).
Step 2: Given \( \lambda_A = 2\lambda_B \), we get:
\[ \nu_A = \frac{c}{2\lambda_B} = \frac{\nu_B}{2} \]
Step 3: Substituting in the equation:
\[ K_A = h\nu_A - \phi = \frac{h\nu_B}{2} - \phi \]
Step 4: Since \( K_B = h\nu_B - \phi \), we conclude:
\[ K_A<\frac{K_B}{2} \]