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

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$h(f_1 - f_0)$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of emitted photoelectrons is given by:$K_{max} = hf - \phi$where $h$ is Planck's constant,$f$ is the frequency of incident light,and $\phi$ is the work function of the metal.In the photoelectric effect,the stopping potential $V$ is related to the maximum kinetic energy by the relation $K_{max} = eV$,where $e$ is the charge of an electron.Thus,$eV = hf - \phi$,or $V = \frac{h}{e}f - \frac{\phi}{e}$.From the graph,when $f = f_0$ (threshold frequency),the stopping potential $V = 0$. Therefore,$0 = \frac{h}{e}f_0 - \frac{\phi}{e}$,which implies $\phi = hf_0$.Substituting this into the expression for $K_{max}$ at frequency $f_1$:$K_{max} = hf_1 - hf_0 = h(f_1 - f_0)$.Therefore,the correct option is $C$.

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