Question

The maximum wavelength of light which causes photoelectric emission from a photosensitive metal surface is \( \lambda_0 \). Two light beams of wavelengths \( \frac{\lambda_0}{3} \) and \( \frac{\lambda_0}{9} \) incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is:

• A. \( 3:4 \)
• B. \( 1:3 \)
• C. \( 1:2 \)
• D. \( 2:3 \)

Hint:

For photoelectric emission, the kinetic energy is given by \( K.E. = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \), and velocity is proportional to the square root of kinetic energy.

Answer 1

The Correct Option is C

Step 1: Apply Energy and Photoelectric Equation The kinetic energy of the emitted photoelectron is given by: \[ K.E. = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \] Since \( K.E. \propto v^2 \), the velocity is given by: \[ v = \sqrt{\frac{2K.E.}{m}} \] Step 2: Compute Kinetic Energy For \( \lambda = \frac{\lambda_0}{3} \): \[ K.E._1 = \frac{hc}{\frac{\lambda_0}{3}} - \frac{hc}{\lambda_0} \] \[ K.E._1 = \frac{3hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{2hc}{\lambda_0} \] For \( \lambda = \frac{\lambda_0}{9} \): \[ K.E._2 = \frac{hc}{\frac{\lambda_0}{9}} - \frac{hc}{\lambda_0} \] \[ K.E._2 = \frac{9hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{8hc}{\lambda_0} \] Step 3: Compute Velocity Ratio \[ \frac{v_1}{v_2} = \sqrt{\frac{K.E._1}{K.E._2}} = \sqrt{\frac{2hc/\lambda_0}{8hc/\lambda_0}} \] \[ = \sqrt{\frac{1}{4}} = \frac{1}{2} \] Thus, the correct answer is \( 1:2 \).