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Difficulty: MediumPhotoelectric Effect and Work Function

In a photoelectric effect experiment, monochromatic light with a frequency greater than the threshold frequency of a cesium metal emitter is directed at the surface. If the intensity of the incident light is doubled while keeping its frequency constant, which of the following statements is correct?

  1. The rate of emitted photoelectrons doubles, but their maximum kinetic energy remains unchanged.Answer
  2. B
    Both the maximum kinetic energy of the photoelectrons and the rate of emission double.
  3. C
    The maximum kinetic energy doubles, while the rate of photoelectron emission remains unchanged.
  4. D
    The maximum kinetic energy increases, but the rate of photoelectron emission decreases.

Answer

The rate of emitted photoelectrons doubles, but their maximum kinetic energy remains unchanged.
According to Einstein's photoelectric equation, Kmax=hf−W0K_{\text{max}} = hf - W_0, the maximum kinetic energy of emitted photoelectrons depends strictly on the frequency of incident light and the work function of the metal emitter. Doubling light intensity at constant frequency increases the number of incident photons per second, which proportionally doubles the number of photoelectrons ejected per second (photoelectric current) without affecting their kinetic energy.

Step-by-Step Solution

1
Relate light intensity to the rate of incident photons.
Light intensity II is directly proportional to the photon flux (number of photons per unit area per unit time).
Intensity is defined by total photon energy per unit area per second (I=Nhf/(At)I = N h f / (A t)).
2
Determine the impact of intensity on the rate of photoelectron emission.
Doubling intensity doubles the number of photons hitting the surface per second, thereby doubling the photoelectron emission rate.
Photoelectric emission follows a 1-to-1 photon-to-electron collision process.
3
Apply Einstein's photoelectric equation to analyze kinetic energy.
The maximum kinetic energy Kmax=hf−W0K_{\text{max}} = hf - W_0 remains constant.
Since frequency ff and work function W0W_0 are kept constant, the energy per photon hfhf and maximum kinetic energy KmaxK_{\text{max}} are unaffected by light intensity.

Key Concept

Independence of photoelectron kinetic energy from light intensity
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Topics

Photoelectric Effect and Work Function

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