Question

Two photons of energies 2.5 eV and 5.5 eV incident on a metal surface of work function 1.5 eV. The ratio of the maximum speeds of the photoelectrons emitted from the metal surface is:

• A. \( 1:4 \)
• B. \( 1:2 \)
• C. \( 1:1 \)
• D. \( 5:11 \)

Hint:

The speed of photoelectrons depends on the kinetic energy, which in turn depends on the photon energy minus the work function. The ratio of speeds is the square root of the ratio of kinetic energies.

Answer 1

The Correct Option is B

The maximum kinetic energy of the photoelectrons is given by the equation: \[ K_{\text{max}} = E_{\text{photon}} - \phi \] where \( E_{\text{photon}} \) is the energy of the photon, and \( \phi \) is the work function. For the two photons, the maximum kinetic energies are: \[ K_{\text{max1}} = 2.5 - 1.5 = 1 \, \text{eV}, \quad K_{\text{max2}} = 5.5 - 1.5 = 4 \, \text{eV} \] The speed of the photoelectron is given by: \[ v = \sqrt{\frac{2 K_{\text{max}}}{m_e}} \] Thus, the ratio of speeds is: \[ \frac{v_1}{v_2} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]