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.

Answer 1

The correct answer is: 2.16 eV.
Wavelength of light produced by the argon laser, \(λ=488nm=488×10^{−9}m\)
Stopping potential of the photoelectrons, \(V_0=0.38V\)
\(1eV=1.6×10^{−19}J\)
\(∴V_0=\frac{0.38}{1.6×10^{-19}}eV\)
Planck’s constant, \(h=6.6×10^{−34}Js\)
Charge on an electron, \(e=1.6×10^{−19}C\)
Speed of light, \(c=3×10^8m/s\)
From Einstein’s photoelectric effect, we have the relation involving the work function \(Φ_0\)
of the material of the emitter as:
\(eV_0=\frac{hc}{λ}-ϕ_0\)
\(ϕ_0=\frac{hc}{λ}-eV_0\)

\(=\frac{6.6×10^{-34}×3×10^8}{1.6×10^{-19}×488×10^{-9}}-\frac{1.6×10^{-19}×0.38}{1.6×10^{-19}}\)

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

Answer 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: 
\(\phi_0 = \frac{hc}{\lambda} - eV_0\)
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: \(\lambda = 488\) nm \(=488 \times 10^{-9}\) m
• The stopping potential for photoelectrons: \(V_0 = 0.38 V\)
   Given constants:
• Planck’s constant: \(h = 6.6 \times 10^{-34}\) Js
• Charge on an electron: \(e = 1.6 \times 10^{-19}\)C
• Speed of light: \(c = 3 \times 10^8\) m/s 
  Now, let's apply the formula: \(\phi_0 = \frac{hc}{\lambda} - eV_0​ \)
  Substitute the given values: \(\phi_0 = \left( \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 488 \times 10^{-9}} \right) - (1.6 \times 10^{-19} \times 0.38)\)
  \(\Rightarrow \phi_0 = 2.54 - 0.38\)
  \(\Rightarrow \phi_0 = 2.16 \text{ eV}\)
• 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.