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

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

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(B) According to Einstein's photoelectric equation,the maximum kinetic energy $E$ is given by: $E = \frac{hc}{\lambda} - \phi$ ... $(i)$,where $\phi$ is the work function.When the wavelength is reduced by $50 \%$,the new wavelength $\lambda' = \lambda - 0.5\lambda = 0.5\lambda = \frac{\lambda}{2}$.The new maximum kinetic energy $E' = 3E$.Substituting these into the photoelectric equation: $3E = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$ ... $(ii)$.From equation $(i)$,we have $E = \frac{hc}{\lambda} - \phi$. Substituting this into equation $(ii)$:$3(\frac{hc}{\lambda} - \phi) = \frac{2hc}{\lambda} - \phi$$\frac{3hc}{\lambda} - 3\phi = \frac{2hc}{\lambda} - \phi$$\frac{3hc}{\lambda} - \frac{2hc}{\lambda} = 3\phi - \phi$$\frac{hc}{\lambda} = 2\phi$$\phi = \frac{hc}{2\lambda}$.

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