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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $E_k$ is given by $E_k = \frac{hc}{\lambda} - \phi$,where $\phi$ is the

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$\frac{2 hc}{3 \lambda}$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $E_k$ is given by $E_k = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For the first case with wavelength $\lambda$:$E_1 = \frac{hc}{\lambda} - \phi$ ... $(i)$For the second case with wavelength $\frac{\lambda}{2}$:$E_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$ ... (ii)Given that $E_1 = \frac{1}{4} E_2$,which implies $4E_1 = E_2$ ... (iii)Substituting $(i)$ and (ii) into (iii):$4\left(\frac{hc}{\lambda} - \phi\right) = \frac{2hc}{\lambda} - \phi$$\frac{4hc}{\lambda} - 4\phi = \frac{2hc}{\lambda} - \phi$$\frac{4hc}{\lambda} - \frac{2hc}{\lambda} = 4\phi - \phi$$\frac{2hc}{\lambda} = 3\phi$$\phi = \frac{2hc}{3\lambda}$

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