The kinetic energy of emitted electron is E when the light incident on the metal has wavelength λ. To double the kinetic energy, the incident light must have wavelength :
\(\frac{hc}{E\lambda - hc}\)
\(\frac{hc\lambda}{E\lambda + hc}\)
\(\frac{h\lambda}{E\lambda + hc}\)
\(\frac{hc\lambda}{E\lambda - hc}\)
The Correct Option is B
Solution and Explanation
\(k = \frac{hc}{\lambda} - \Phi = E\)
and,
\(2k = \frac{hc}{\lambda^2} - \Phi = 2E\)
\(⇒\)\(\frac{hc}{\lambda} - E = \frac{hc}{\lambda^2} - 2E\)
\(⇒\) \(\frac{hc}{\lambda^2} = \frac{hc}{\lambda} + E\)
\(⇒\) \(\lambda^2 = \frac{hc\lambda}{hc + \lambda E}\)
So, the correct option is (B).
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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.