In the case of the photoelectric effect, on i | vedclass

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Question

(C) According to Einstein's photoelectric equation, $K_{max} = h
u - \phi_0$, where $K_{max}$ is the maximum kinetic energy of emitted photoelectrons

Vedclass · Accepted Answer

Stopping potential increases.

Vedclass · Answer

(C) According to Einstein's photoelectric equation, $K_{max} = h
u - \phi_0$, where $K_{max}$ is the maximum kinetic energy of emitted photoelectrons, $h$ is Planck's constant, $
u$ is the frequency of incident light, and $\phi_0$ is the work function of the metal.Since $K_{max} = eV_s$ (where $e$ is the charge of an electron and $V_s$ is the stopping potential), we have $eV_s = h
u - \phi_0$.Rearranging for the stopping potential, $V_s = \frac{h}{e}
u - \frac{\phi_0}{e}$.As the frequency $
u$ of the incident light increases, the term $\frac{h}{e}
u$ increases, which leads to an increase in the stopping potential $V_s$.The photoelectric current depends on the intensity of the incident light, not its frequency, provided the frequency is above the threshold frequency.

In the case of the photoelectric effect, on increasing the frequency of incident light:

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