Question

Difficulty: MediumPhotoelectric Effect and Work Function

In a photoelectric effect experiment using light of a frequency greater than the threshold frequency of a metal surface, doubling the intensity of the incident light doubles the maximum kinetic energy of the emitted photoelectrons.

Answer: Answer

Answer

The statement is false. Doubling the light intensity increases the number of emitted photoelectrons per second but leaves their maximum kinetic energy unchanged.
The statement is false because the maximum kinetic energy of emitted photoelectrons is governed by Einstein's equation Kmax=hfW0K_{\text{max}} = hf - W_0. It depends strictly on the frequency ff of incident light and the metal work function W0W_0. Doubling the light intensity increases the rate of photon bombardment, which increases the number of photoelectrons emitted per unit time (photoelectric current), but leaves the maximum kinetic energy of individual photoelectrons completely unchanged.

Step-by-Step Solution

1
Identify the factors determining individual photon energy and maximum kinetic energy
Individual photon energy is given by E=hfE = hf, and maximum photoelectron kinetic energy is Kmax=hfW0K_{\text{max}} = hf - W_0.
Einstein's photoelectric equation governs the energy exchange between a single incident photon and a single bound electron.
2
Analyze the physical meaning of light intensity in quantum terms
Intensity II is proportional to the number of photons striking the surface per unit time, not the energy of individual photons.
At a constant frequency ff, changing intensity varies photon flux while individual photon energy hfhf stays constant.
3
Evaluate the effect of doubling light intensity on maximum kinetic energy
Doubling intensity doubles the rate of photoemission (photoelectric current) but does not change KmaxK_{\text{max}}.
Because ff and W0W_0 remain constant, KmaxK_{\text{max}} remains strictly unchanged.

Key Concept

Independence of photoelectron kinetic energy from light intensity
Estimated Time:1m 0s
Rate this question