Question

A metal surface is illuminated by a light of a given intensity and frequency to cause photoemission. If the intensity illumination is reduced to one-fourth of its original value, then the maximum KE of the emitted photoelectrons will become
A. ${(1/16)^{th}}$ of the original value
B. Unchanged
C. Twice the original value
D. Four times the original value

Answer 1

Hint: To find the answer of this question we take the formula: ${\text{KE = h}}\upsilon - \phi$, where( KE) is the Kinetic energy of photoelectrons, ($\upsilon$ )is the frequency, ($\phi$) is the work function and (h) is the Planck’s constant. Do not depend upon the intensity of illumination while it depends upon the frequency of radiation.

Complete step-by-step answer:
To calculate the required maximum kinetic energy we are using the formula:
${\text{KE = h}}\upsilon - \phi$……………………..(i)
Where, KE is the kinetic energy of the photoelectrons.
$\nu$ is the frequency of incident radiation.
$\phi$ is the work function and h is the Planck's constant.
From equation (i) we know that the work function of a sample has some fixed value, so from here we have an idea of frequency dependence of the kinetic energy of photoelectrons.
Intensity is nothing but the amount of energy per unit area per unit time of a given frequency of radiation so if intensity gets to change it does not mean that the frequency will also change.
Therefore, according to the question if the intensity of illumination is reduced to one-fourth of its original value, then the maximum kinetic energy of emitted photoelectrons will become unchanged.
Hence, Option (B) is the correct answer.

Note: In order to answer such kinds of questions one should have a basic understanding of the different terminology involved in this topic like -Work function, frequency, Intensity of radiation, etc. Students must also have to practice the different kinds of curves drawn between these physical quantities for a better in-depth understanding of this topic.