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
Approach Solution - 1
To determine the wavelength required to double the kinetic energy of the emitted electron, we need to apply the concept of photoelectric effect. According to Einstein's photoelectric equation, the kinetic energy \( K \) of the emitted electron when light of wavelength \( \lambda \) is incident on a metal surface is given by:
\(K = \frac{hc}{\lambda} - \phi\)
where:
- \(h\) is Planck's constant
- \(c\) is the speed of light
- \(\phi\) is the work function of the metal
Given that the initial kinetic energy \( E \) can be expressed as:
\(E = \frac{hc}{\lambda} - \phi\)
We want to double this kinetic energy, so we have:
\(2E = \frac{hc}{\lambda'} - \phi\)
where \( \lambda' \) is the new wavelength. We can set up the equation for \( 2E \) and equate it to the adjusted photoelectric equation:
\(2E = \frac{hc}{\lambda'} - \left( \frac{hc}{\lambda} - E \right)\)
Simplifying this equation:
\(2E = \frac{hc}{\lambda'} + E - \frac{hc}{\lambda}\)
\(2E - E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}\)
\(E = \frac{hc}{\lambda'} - \frac{hc}{\lambda}\)
Now solve for \( \lambda' \):
\(\frac{hc}{\lambda'} = E + \frac{hc}{\lambda}\)
\(\lambda' = \frac{hc}{E + \frac{hc}{\lambda}} = \frac{hc \cdot \lambda}{E \lambda + hc}\)
Thus, the new wavelength required to double the kinetic energy is:
\(\lambda' = \frac{hc \lambda}{E \lambda + hc}\)
This matches the provided option:
\(\frac{hc\lambda}{E\lambda + hc}\)
Therefore, the correct answer is indeed \(\frac{hc\lambda}{E\lambda + hc}\), which is consistent with the principles of the photoelectric effect.
Approach Solution -2
\[ k = \frac{hc}{\lambda} - \Phi = E \] and,
\[ 2k = \frac{hc}{\lambda^2} - \Phi = 2E \]
\[ \Rightarrow \frac{hc}{\lambda} - E = \frac{hc}{\lambda^2} - 2E \]
\[ \Rightarrow \frac{hc}{\lambda^2} = \frac{hc}{\lambda} + E \]
\[ \Rightarrow \lambda^2 = \frac{hc\lambda}{hc + \lambda E} \]
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.