Question:

When light of wavelength 300 nm falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however light of 600 nm wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters:

Updated On: Jun 7, 2022
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The Correct Option is C

Solution and Explanation

From the formula of work function $ {{W}_{0}}=\frac{hc}{\lambda } $ here $ {{\lambda }_{0}} $ is threshold wavelength $ \frac{{{W}_{01}}}{{{W}_{02}}}=\frac{{{\lambda }_{02}}}{{{\lambda }_{01}}}=\frac{600}{300}=\frac{2}{1} $
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Concepts Used:

Photoelectric Effect

When light shines on a metal, electrons can be ejected from the surface of the metal in a phenomenon known as the photoelectric effect. This process is also often referred to as photoemission, and the electrons that are ejected from the metal are called photoelectrons.

Photoelectric Effect Formula:

According to Einstein’s explanation of the photoelectric effect :

The energy of photon = energy needed to remove an electron + kinetic energy of the emitted electron

i.e. hν = W + E

Where,

  • h is Planck’s constant.
  • ν is the frequency of the incident photon.
  • W is a work function.
  • E is the maximum kinetic energy of ejected electrons: 1/2 mv².

Laws of Photoelectric Effect:

  1. The photoelectric current is in direct proportion to the intensity of light, for a light of any given frequency; (γ > γ Th).
  2. There exists a certain minimum (energy) frequency for a given material, called threshold frequency, below which the discharge of photoelectrons stops completely, irrespective of how high the intensity of incident light is.
  3. The maximum kinetic energy of the photoelectrons increases with the increase in the frequency (provided frequency γ > γ Th exceeds the threshold limit) of the incident light. The maximum kinetic energy is free from the intensity of light. 
  4. The process of photo-emission is an instantaneous process.