In the photoelectric effect experiment, if th | vedclass

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(C) According to Einstein's photoelectric equation: $K_{max} = h v - \Phi_0$, where $K_{max}$ is the maximum kinetic energy of the emitted photoelectr

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increases

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(C) According to Einstein's photoelectric equation: $K_{max} = h v - \Phi_0$, where $K_{max}$ is the maximum kinetic energy of the emitted photoelectrons, $h$ is Planck's constant, $v$ is the frequency of incident radiation, and $\Phi_0$ is the work function of the metal.Since $K_{max} = e V_s$, where $e$ is the charge of an electron and $V_s$ is the stopping potential, we can write: $e V_s = h v - \Phi_0$.Rearranging for the stopping potential: $V_s = \frac{h}{e} v - \frac{\Phi_0}{e}$.From this linear equation, it is evident that the stopping potential $(V_s)$ is directly proportional to the frequency $(v)$ of the incident radiation.Therefore, if the frequency $(v)$ is increased, the stopping potential $(V_s)$ also increases.

In the photoelectric effect experiment, if the frequency of incident radiation $(v)$ is increased, keeping all other factors constant, what happens to the stopping potential $(V_s)$, given $(v > v_0)$ where $(v_0)$ is the threshold frequency?

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