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Question

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

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become more than doubled

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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 given by:$eV_0' = h(2
u) - \phi = 2h
u - \phi$We can rewrite this as:$eV_0' = 2(h
u - \frac{\phi}{2})$ --- $(2)$Comparing $(1)$ and $(2)$,since $h
u - \frac{\phi}{2} > h
u - \phi$,it follows that $eV_0' > 2eV_0$. 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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