For zero photoelectric current,the stopping p | vedclass

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = h
u - \Phi_0

Vedclass · Accepted Answer

Increases with an increase in the frequency of incident light.

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by $K_{max} = h
u - \Phi_0$,where $h$ is Planck's constant,$
u$ is the frequency of incident light,and $\Phi_0$ is the work function of the metal.Stopping potential $V_s$ is related to maximum kinetic energy by $K_{max} = eV_s$.Therefore,$eV_s = h
u - \Phi_0$,which implies $V_s = \frac{h}{e}
u - \frac{\Phi_0}{e}$.This equation shows that the stopping potential $V_s$ is a linear function of the frequency $
u$ of the incident light.As the frequency $
u$ of the incident light increases,the stopping potential $V_s$ also increases.

For zero photoelectric current,the stopping potential is:

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