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Difficulty: HardPhotoelectric Effect and Work Function

The threshold wavelength for photoelectric emission from a metallic emitter is λ0\lambda_0. When light of wavelength λ03\frac{\lambda_0}{3} illuminates the emitter, photoelectrons are ejected with a maximum kinetic energy of K1K_1. If the incident light is changed to a wavelength of λ04\frac{\lambda_0}{4}, what is the new maximum kinetic energy K2K_2 of the ejected photoelectrons in terms of K1K_1?

  1. 32K1\frac{3}{2} K_1Answer
  2. B
    43K1\frac{4}{3} K_1
  3. C
    2K12 K_1
  4. D
    23K1\frac{2}{3} K_1

Answer

The new maximum kinetic energy K2K_2 is equal to 32K1\frac{3}{2} K_1.
According to Einstein's photoelectric equation, maximum kinetic energy is K=E−W0K = E - W_0. With threshold wavelength λ0\lambda_0, the work function is W0=hcλ0W_0 = \frac{hc}{\lambda_0}. For light of wavelength λ03\frac{\lambda_0}{3}, the photon energy is 3W03W_0, yielding K1=3W0−W0=2W0K_1 = 3W_0 - W_0 = 2W_0. For light of wavelength λ04\frac{\lambda_0}{4}, the photon energy is 4W04W_0, yielding K2=4W0−W0=3W0K_2 = 4W_0 - W_0 = 3W_0. Comparing K2K_2 and K1K_1 gives K2=32K1K_2 = \frac{3}{2} K_1.

Step-by-Step Solution

1
Express the work function W0W_0 in terms of threshold wavelength λ0\lambda_0.
W0=hcλ0W_0 = \frac{hc}{\lambda_0}
The threshold wavelength defines the minimum photon energy required to liberate an electron from the metal surface.
2
Apply Einstein's photoelectric equation for the first incident wavelength λ1=λ03\lambda_1 = \frac{\lambda_0}{3}.
K1=hcλ0/3−W0=3(hcλ0)−W0=3W0−W0=2W0K_1 = \frac{hc}{\lambda_0/3} - W_0 = 3\left(\frac{hc}{\lambda_0}\right) - W_0 = 3W_0 - W_0 = 2W_0
Einstein's photoelectric equation states that maximum kinetic energy equals photon energy minus work function.
3
Apply Einstein's photoelectric equation for the second incident wavelength λ2=λ04\lambda_2 = \frac{\lambda_0}{4}.
K2=hcλ0/4−W0=4(hcλ0)−W0=4W0−W0=3W0K_2 = \frac{hc}{\lambda_0/4} - W_0 = 4\left(\frac{hc}{\lambda_0}\right) - W_0 = 4W_0 - W_0 = 3W_0
Calculate maximum kinetic energy for the shorter incident wavelength.
4
Determine the ratio of K2K_2 to K1K_1.
K2K1=3W02W0=32  ⟹  K2=32K1\frac{K_2}{K_1} = \frac{3W_0}{2W_0} = \frac{3}{2} \implies K_2 = \frac{3}{2} K_1
Express the new kinetic energy in terms of the initial kinetic energy.

Key Concept

Einstein's Photoelectric Equation and Work Function Threshold
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Topics

Photoelectric Effect and Work Function

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