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

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become more than double.

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of the emitted photoelectrons is given by $K_{max} = h
u - \phi$,where $h$ is Planck's constant,$
u$ is the frequency of incident light,and $\phi$ is the work function of the metal.Since the stopping potential $V_s$ is related to the maximum kinetic energy by $K_{max} = eV_s$,we have $eV_s = h
u - \phi$,or $V_s = \frac{h
u}{e} - \frac{\phi}{e}$.Initially,let the frequency be $
u_1$ and stopping potential be $V_{s1} = \frac{h
u_1}{e} - \frac{\phi}{e}$.When the frequency is doubled,$
u_2 = 2
u_1$,the new stopping potential is $V_{s2} = \frac{h(2
u_1)}{e} - \frac{\phi}{e} = \frac{2h
u_1}{e} - \frac{\phi}{e}$.Comparing $V_{s2}$ with $V_{s1}$,we see that $V_{s2} = 2V_{s1} + \frac{\phi}{e}$.Since $\frac{\phi}{e} > 0$,it follows that $V_{s2} > 2V_{s1}$.Therefore,the stopping potential becomes more than double.

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

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