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Question 19

The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation:

In the photoelectric effect, when light of frequency $$\nu$$ falls on a metal surface with work function $$\phi$$, Einstein's photoelectric equation gives the maximum kinetic energy of the emitted photoelectrons as $$K_{\text{max}} = h\nu - \phi$$, where $$h$$ is Planck's constant.

The stopping potential $$V_0$$ is defined by $$eV_0 = K_{\text{max}} = h\nu - \phi$$, which gives $$V_0 = \frac{h\nu - \phi}{e}$$.

From this relation, the stopping potential depends on the frequency $$\nu$$ of the incident radiation and the work function of the metal. It does not depend on the intensity, amplitude, or phase of the incident light. Increasing the intensity increases the number of photoelectrons (i.e., the photocurrent) but not the maximum kinetic energy or the stopping potential.

Therefore, the stopping potential depends on the frequency of the incident electromagnetic radiation.

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