In the photoelectric effect, the stopping pot | vedclass

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Question

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

Vedclass · Accepted Answer

$(A)$ and $(C)$ both

Vedclass · Answer

(D) According to Einstein's photoelectric equation, the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = h
u - \Phi$, where $h$ is Planck's constant, $
u$ is the frequency of incident light, and $\Phi$ is the work function of the emitter material.Since the stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = eV_s$, we have $eV_s = h
u - \Phi$, which implies $V_s = \frac{h
u}{e} - \frac{\Phi}{e}$.From this equation, it is clear that the stopping potential $V_s$ depends on the frequency of the incident light $(
u)$ and the work function $(\Phi)$, which is a property of the emitter material.It does not depend on the intensity of the incident light.Therefore, the correct answer is $(A)$ and $(C)$ both.

In the photoelectric effect, the stopping potential depends on:

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