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(B) Let $\phi_{0}$ be the work function of the surface of the material. According to Einstein's photoelectric equation,the maximum kinetic energy of t

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

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(B) Let $\phi_{0}$ be the work function of the surface of the material. According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted photoelectrons in the first case is $K_{\max 1} = \frac{hc}{\lambda} - \phi_{0}$.In the second case,the wavelength is $\lambda/2$,so the maximum kinetic energy is $K_{\max 2} = \frac{hc}{\lambda/2} - \phi_{0} = \frac{2hc}{\lambda} - \phi_{0}$.Given that $K_{\max 2} = 3 K_{\max 1}$,we substitute the expressions:$\frac{2hc}{\lambda} - \phi_{0} = 3 \left( \frac{hc}{\lambda} - \phi_{0} \right)$.Expanding the equation: $\frac{2hc}{\lambda} - \phi_{0} = \frac{3hc}{\lambda} - 3\phi_{0}$.Rearranging the terms: $3\phi_{0} - \phi_{0} = \frac{3hc}{\lambda} - \frac{2hc}{\lambda}$.$2\phi_{0} = \frac{hc}{\lambda}$.Therefore,the work function is $\phi_{0} = \frac{hc}{2\lambda}$.

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