Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ1 and λ2, respectively are incident on a metallic surface. If λ1 = 3λ2 then:
\(K_1>\frac{K_2}{3}\)
\(K_1< \frac{K_2}{3}\)
\(K_1=\frac{K_2}{3}\)
\(K_2=\frac{K_1}{3}\)
The Correct Option is B
Solution and Explanation
The correct answer is (B) : \(K_1< \frac{K_2}{3}\)
\(K_1=\frac{hc}{λ_1}−\phi=\frac{hc}{3λ_2}−\phi….(i)\)
and
\(K_2=\frac{hc}{λ_2}−\phi….(ii)\)
from (i) and (ii) we can say
3K1 = K2 – 2φ
\(K_1<\frac{K_2}{3}\)
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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:
- The photoelectric current is in direct proportion to the intensity of light, for a light of any given frequency; (γ > γ Th).
- 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.
- 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.
- The process of photo-emission is an instantaneous process.