A metal surface of work function 3 eV is illuminated by photons of energy 2 eV. The kinetic energy of emitted photo-electrons will be
- 1 eV
- 2 eV
- 3 eV
- 0
The Correct Option is D
Solution and Explanation
As, Einstein?s photoelectric equation $E_0 = W_0 + KE_0$
So, $KE_0 = E_0 - W_0$
$\Rightarrow KE_0 = 2 - 3 = -1 \,eV$
Since, kinetic energy is a scalar quantity i.e., $KE > 0$, positive.
So, the photoelectron does not ejected and $KE = 0$
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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.