Question

A certain metallic surface is illuminated with monochromatic light of wavelength λ. The stopping potential for photo-electric current for this light is 3V₀. If the same surface is illuminated with light of wavelength 2λ, the stopping potential is V₀. The threshold wavelength for this surface for photo-electric effect is
1. \(\dfrac{\lambda}{6}\)
2. 6λ 
3. 4λ 
4. \(\dfrac{\lambda}{4}\)

Answer 1

Correct Answer - Option 3 : 4λ 

CONCEPT:

• Photoelectric effect: The photoelectric effect is a phenomenon where electrons are ejected from a metal surface when the light of sufficient frequency is incident on it. These ejected electrons are called photoelectrons.
  • When a photon falls on the metal surface, the photon’s energy is transferred to the electron.
  • Some part of the energy gets utilized in removing the electron from the metal surface, and the remaining goes into giving kinetic energy for the ejected electron.
• Work function (ϕ): It is the minimum amount of energy required to cause photo-emission of electrons from a metal surface when light is incident on it.
• Einstein’s Photoelectric Equation gives the relationship between work function (ϕ), the kinetic energy of the emitted electron, and the energy of the photon (E).

⇒ E = ϕ + hν

Substituting the value \(ν = \frac{C}{λ}\) the above equation can be rewritten as

\(⇒ KE = \frac{hc}{λ} -\phi\)

• if a charged particle q is accelerated through a potential V then K E  = Vq,  then the above equation becomes

\(⇒ qV = \frac{hc}{λ} -\phi\)

Where ν = frequency of incident light, h = Plancks constant, ϕ  = Work function, V =  Stopping Potential

EXPLANATION:

• The Kinetic energy of charged particle accelerated under an applied potential is given by

\(⇒ qV = \frac{hc}{λ} -\phi\)

The above equation can be rewritten as

\(⇒ \frac{hC}{λ} =qV +\phi \)       ----(1)

For the first case, for λ and  3V₀ the above equation can be written as

\(⇒ \dfrac{hc}{λ}=\phi + e \ 3V_0\) -----(2)

For the second case, for 2λ and V0  equation 1 can be written as

\(⇒ \dfrac{hc}{2λ}=\phi + eV_0\)    -----(3)

Equation 3 - Equation 2  gives 

\(⇒ \frac{hc}{λ} -\frac{hc}{2λ} = \phi -\phi +3eV_{0} -eV_{0}\)

\(⇒ \frac{hc}{2λ} = 2eV_{0}\)

\(⇒ eV_{0} = \frac{hc}{4λ}\)

Substituting the value \(\phi =\frac{hc}{λ_{0}}\)  and\( eV_{0} = \frac{hc}{4λ}\) in equation 3 it can be rewritten as   (λ₀ =Theshold wavelength)

\(⇒ \frac{hc}{2λ}=\frac{hc}{λ_{0}} + \frac{hc}{4λ}\)

\(⇒ \frac{hc}{2λ} -\frac{hc}{4λ}=\frac{hc}{4λ} =\frac{hc}{λ_{0}}\)

From the above equation λ₀ = 4λ 

• Hence, option 3 is the answer