Question

The work function of a photosensitive metal surface is \( 1.1 \) eV. Two light beams of energies \( 1.5 \) eV and \( 2 \) eV incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is?

• A. \( 3:4 \)
• B. \( 1:1 \)
• C. \( 2:3 \)
• D. \( 4:9 \)

Hint:

In the photoelectric effect, the maximum velocity of emitted electrons is proportional to the square root of their kinetic energy: \( v_{\max} \propto \sqrt{K_{\max}} \).

Answer 1

The Correct Option is C

Step 1: Use Einstein's photoelectric equation 
The maximum kinetic energy (\( K_{\max} \)) of the emitted photoelectrons is given by: \[ K_{\max} = h\nu - \phi \] where: - \( h\nu \) is the energy of the incident photon, - \( \phi \) is the work function of the metal. 

Step 2: Compute the kinetic energy for each incident photon 
For \( h\nu_1 = 1.5 \) eV: \[ K_{\max,1} = 1.5 - 1.1 = 0.4 \text{ eV} \] For \( h\nu_2 = 2 \) eV: \[ K_{\max,2} = 2 - 1.1 = 0.9 \text{ eV} \] 

Step 3: Compute the ratio of maximum velocities 
The maximum velocity (\( v_{\max} \)) of the emitted photoelectrons is related to the kinetic energy by: \[ K_{\max} = \frac{1}{2} m v_{\max}^2 \] \[ v_{\max} \propto \sqrt{K_{\max}} \] Thus, the ratio of maximum velocities is: \[ \frac{v_{\max,1}}{v_{\max,2}} = \sqrt{\frac{K_{\max,1}}{K_{\max,2}}} \] \[ = \sqrt{\frac{0.4}{0.9}} \] \[ = \frac{\sqrt{4}}{\sqrt{9}} = \frac{2}{3} \] Thus, the required ratio is \( 2:3 \). \