Question:

Light of wavelength 488nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38V. Find the work function of the material from which the emitter is made.

Updated On: Jul 8, 2024
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Approach Solution - 1

The correct answer is: 2.16 eV.
Wavelength of light produced by the argon laser, λ=488nm=488×109mλ=488nm=488×10^{−9}m
Stopping potential of the photoelectrons, V0=0.38VV_0=0.38V
1eV=1.6×1019J1eV=1.6×10^{−19}J
V0=0.381.6×1019eV∴V_0=\frac{0.38}{1.6×10^{-19}}eV
Planck’s constant, h=6.6×1034Jsh=6.6×10^{−34}Js
Charge on an electron, e=1.6×1019Ce=1.6×10^{−19}C
Speed of light, c=3×108m/sc=3×10^8m/s
From Einstein’s photoelectric effect, we have the relation involving the work function Φ0Φ_0
of the material of the emitter as:
eV0=hcλϕ0eV_0=\frac{hc}{λ}-ϕ_0
ϕ0=hcλeV0ϕ_0=\frac{hc}{λ}-eV_0

=6.6×1034×3×1081.6×1019×488×1091.6×1019×0.381.6×1019=\frac{6.6×10^{-34}×3×10^8}{1.6×10^{-19}×488×10^{-9}}-\frac{1.6×10^{-19}×0.38}{1.6×10^{-19}}

=2.540.38=2.16eV=2.54-0.38=2.16eV
Therefore, the material with which the emitter is made has the work function of 2.16 eV.

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Approach Solution -2

To find the work function (in eV) of the material from which the emitter is made, we'll use the formula derived from Einstein's photoelectric effect. The formula is: 
ϕ0=hcλeV0\phi_0 = \frac{hc}{\lambda} - eV_0
Now, let's break down the steps to solve the problem: 
The work function represents the minimum energy required to release electrons from a metal's surface. To solve this question, let's list the given values:

  • The argon laser produces light with a wavelength: λ=488\lambda = 488 nm =488×109=488 \times 10^{-9} m
  • The stopping potential for photoelectrons: V0=0.38VV_0 = 0.38 V
     Given constants:
  • Planck’s constant: h=6.6×1034h = 6.6 \times 10^{-34} Js
  • Charge on an electron: e=1.6×1019e = 1.6 \times 10^{-19}C
  • Speed of light: c=3×108c = 3 \times 10^8 m/s 
    Now, let's apply the formula: ϕ0=hcλeV0\phi_0 = \frac{hc}{\lambda} - eV_0​
    Substitute the given values: ϕ0=(6.6×1034×3×1081.6×1019×488×109)(1.6×1019×0.38)\phi_0 = \left( \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 488 \times 10^{-9}} \right) - (1.6 \times 10^{-19} \times 0.38)
    ϕ0=2.540.38\Rightarrow \phi_0 = 2.54 - 0.38
    ϕ0=2.16 eV\Rightarrow \phi_0 = 2.16 \text{ eV}
  • Thus, the material used for the emitter has a work function of 2.16 eV. It's important to consider this when connecting different metals in electronic circuits, as electrons tend to move from lower to higher work function metals. This happens because some electrons are more weakly bound than others, and the work function defines which electrons are involved.
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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:

  1. The photoelectric current is in direct proportion to the intensity of light, for a light of any given frequency; (γ > γ Th).
  2. 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.
  3. 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. 
  4. The process of photo-emission is an instantaneous process.