Question

The work function of a metal 3eV. The colour of the visible light that is required to cause emission of photo electrons is:

• A. Yellow
• B. Blue
• C. Red
• D. Green

Hint:

In the photoelectric effect, light with frequency greater than the threshold frequency causes the emission of photoelectrons.

Answer 1

The Correct Option is B

The photoelectric effect can be explained using the equation: \[ E_{\text{photon}} = h \cdot f = \text{Work Function} + K_{\text{max}} \] Where \( E_{\text{photon}} \) is the energy of the incident light, \( h \) is Planck's constant, \( f \) is the frequency of the light, and \( K_{\text{max}} \) is the maximum kinetic energy of the emitted photoelectrons. For the emission of photoelectrons, the energy of the incoming photon should be equal to or greater than the work function of the metal. We are given that the work function of the metal is 3eV. To cause the emission of photoelectrons, the energy of the photon should be at least 3eV. We know the energy of a photon is related to its frequency by: \[ E = h \cdot f \] Where \( E = 3 \, \text{eV} \). For visible light, the color that corresponds to a frequency that can cause electron emission at 3eV is blue light. Therefore, the color of the light is blue.