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(B) The energy equation in the photoelectric effect is given by Einstein's photoelectric equation:$e V_0 = h 
u - \phi_0 \Rightarrow V_0 = \frac{h

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become more than double

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(B) The energy equation in the photoelectric effect is given by Einstein's photoelectric equation:$e V_0 = h 
u - \phi_0 \Rightarrow V_0 = \frac{h 
u}{e} - \frac{\phi_0}{e}$ $\ldots$ $(i)$Where $V_0$ is the stopping potential,$h$ is Planck's constant,$
u$ is the frequency,and $\phi_0$ is the work function.When the frequency is doubled,the new frequency is $
u' = 2
u$.The new stopping potential $V_0'$ is given by:$V_0' = \frac{h(2
u)}{e} - \frac{\phi_0}{e} = \frac{2h
u}{e} - \frac{\phi_0}{e}$We can rewrite this as:$V_0' = 2\left(\frac{h
u}{e} - \frac{\phi_0}{e}ight) + \frac{\phi_0}{e}$Substituting equation $(i)$ into this expression:$V_0' = 2V_0 + \frac{\phi_0}{e}$Since $\frac{\phi_0}{e} > 0$,it follows that $V_0' > 2V_0$.Therefore,the stopping potential will become more than double.

In a photoelectric effect experiment,if the frequency of light is doubled,the stopping potential will

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