The stopping potential for photoelectrons:

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Question

(C) According to Einstein's photoelectric equation, $K_{max} = h
u - \Phi_0$, where $K_{max} = eV_s$.Here, $V_s$ is the stopping potential, $h$ is Pl

Vedclass · Accepted Answer

Depends on both the frequency of the incident light and the nature of the cathode material.

Vedclass · Answer

(C) According to Einstein's photoelectric equation, $K_{max} = h
u - \Phi_0$, where $K_{max} = eV_s$.Here, $V_s$ is the stopping potential, $h$ is Planck's constant, $
u$ is the frequency of incident light, and $\Phi_0$ is the work function of the material.Rearranging the equation: $eV_s = h
u - \Phi_0$, which gives $V_s = \frac{h}{e}
u - \frac{\Phi_0}{e}$.This shows that the stopping potential $V_s$ depends on the frequency $
u$ of the incident light and the work function $\Phi_0$, which is a characteristic property of the cathode material.Therefore, the stopping potential depends on both the frequency of the incident light and the nature of the cathode material.

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