Examkin

Question

Difficulty: HardPhotoelectric Effect and Work Function

For a given clean photosensitive metal surface, doubling the intensity of incident monochromatic radiation of frequency ff (where f>f0f > f_0) doubles the stopping potential required to reduce the photoelectric current to zero.

Answer: Answer

Answer

The statement is False. Stopping potential depends exclusively on the frequency of the incident radiation and the work function of the metal emitter, remaining independent of light intensity.
The statement is False because stopping potential VsV_s is defined by eVs=hf−W0e V_s = hf - W_0. Doubling the intensity of the incident radiation increases the number of emitted photoelectrons per second (photocurrent) but does not alter the energy per photon hfhf or the maximum kinetic energy of the photoelectrons. Therefore, the stopping potential needed to halt the current remains unchanged.

Step-by-Step Solution

1
Analyze the physical meaning of light intensity in quantum theory.
Increasing the light intensity increases the photon flux (number of photons per unit area per second), thereby increasing the rate of electron emission (photocurrent).
Intensity corresponds to the rate of photon arrival, not the energy of individual photons.
2
Relate photon energy and work function to stopping potential using Einstein's photoelectric equation.
eVs=Kmax=hf−W0e V_s = K_{\text{max}} = hf - W_0, where VsV_s is the stopping potential, ff is frequency, and W0W_0 is work function.
Each emitted electron absorbs energy from a single photon.
3
Evaluate the effect of doubling radiation intensity on stopping potential.
Because ff and W0W_0 are constant, KmaxK_{\text{max}} and VsV_s remain unaltered.
Stopping potential is governed by photon frequency rather than total light intensity.

Key Concept

Independence of photoelectron maximum kinetic energy and stopping potential from incident light intensity
Rate this question

Topics

Photoelectric Effect and Work Function

More questions from this topic