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(C) According to Einstein's photoelectric equation,the energy of an incident photon is given by $E = \frac{hc}{\lambda} = W + K_{\max}$.Here,$W$ is th

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is $\left(\frac{1}{e}\right)$ times the maximum kinetic energy of the emitted photoelectrons

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(C) According to Einstein's photoelectric equation,the energy of an incident photon is given by $E = \frac{hc}{\lambda} = W + K_{\max}$.Here,$W$ is the work function and $K_{\max}$ is the maximum kinetic energy of the emitted photoelectrons.The stopping potential $V_s$ is defined as the potential required to stop the most energetic photoelectrons,such that $K_{\max} = eV_s$.Therefore,$eV_s = K_{\max}$,which implies $V_s = \frac{K_{\max}}{e}$.Thus,the stopping potential is $\left(\frac{1}{e}\right)$ times the maximum kinetic energy of the emitted photoelectrons.

In an experiment with the photoelectric effect,the stopping potential:

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