Question

When monochromatic light falls on a photosensitive material, the number of photoelectrons emitted per second is n and their maximum kinetic energy is ${{K}_{\max }}$. If the intensity of the incident light is doubled, then:
A. n is doubled the ${{K}_{\max }}$ remains same
B. ${{K}_{\max }}$is doubled but n remains same
C. Both n and ${{K}_{\max }}$ are doubled
D. Both n and ${{K}_{\max }}$ are halved

Answer 1

Hint: In this question we are dealing with monochromatic light. By monochromatic light, we mean where the optical spectrum contains only a single optical frequency. Even we can call the light sources can also be called monochromatic, in case there is emission of monochromatic light. To move with the question further we have to deal with the relation of finding the maximum kinetic energy of the emitted photoelectrons.

Complete step by step answer:
We should know that the maximum kinetic energy of emitted photoelectrons is given by ${{K}_{\max }}=hv-\phi $.
In the above formula,
${{K}_{\max }}$represents the maximum kinetic energy
h represents the Planck’s constant
and v is represents the frequency
Hence, it is dependent mainly on the energy (that is frequency) of the incident photon. On doubling the intensity, ${{K}_{\max }}$remains the same.
Every photon having energy more than the work function of the metal has the capacity to emit one of the photoelectrons. Hence, on doubling the intensity, the number of photoelectrons emitted in unit time (that is n) doubles.
Hence, the correct answer is Option A.

Note: We should know that work function is the minimum thermodynamic work which is needed to remove an electron from a solid to a point in case of the vacuum immediately when taken outside the solid surface.
We should know that work function is one of the fundamental electronic properties of bare and coated metallic surfaces. As already seen in the answer, the work function is represented as $\phi $.