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(C) The maximum kinetic energy for the photoelectrons is given by Einstein's photoelectric equation:$K_{\max} = h
u - \phi$where $
u$ is the frequen

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become more than double

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(C) The maximum kinetic energy for the photoelectrons is given by Einstein's photoelectric equation:$K_{\max} = h
u - \phi$where $
u$ is the frequency of incident light and $\phi$ is the work function of the metal.Since $K_{\max} = eV_0$,we have:$eV_0 = h
u - \phi$ ......... $(1)$When the frequency is doubled $(
u' = 2
u)$,the new stopping potential $V_0'$ is:$eV_0' = h(2
u) - \phi = 2h
u - \phi$We can rewrite this as:$eV_0' = 2(h
u - \phi) + \phi$Since $eV_0 = h
u - \phi$,we have $h
u = eV_0 + \phi$. Substituting this into the equation for $eV_0'$:$eV_0' = 2(eV_0 + \phi) - \phi = 2eV_0 + 2\phi - \phi = 2eV_0 + \phi$Dividing by $e$,we get:$V_0' = 2V_0 + \frac{\phi}{e}$Since $\phi > 0$,it is clear that $V_0' > 2V_0$. Therefore,the stopping potential becomes more than double.

If the frequency of light in a photoelectric experiment is doubled,the stopping potential will

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