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

Vedclass · Accepted Answer

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

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For the first case with wavelength $\lambda$:$K_1 = \frac{hc}{\lambda} - \phi$ ... $(i)$For the second case with wavelength $\lambda / 3$:$K_2 = \frac{hc}{\lambda / 3} - \phi = \frac{3hc}{\lambda} - \phi$ ... $(ii)$Given that $K_2 = 4K_1$,we substitute the expressions:$\frac{3hc}{\lambda} - \phi = 4 \left( \frac{hc}{\lambda} - \phi \right)$Expanding the equation:$\frac{3hc}{\lambda} - \phi = \frac{4hc}{\lambda} - 4\phi$Rearranging to solve for $\phi$:$4\phi - \phi = \frac{4hc}{\lambda} - \frac{3hc}{\lambda}$$3\phi = \frac{hc}{\lambda}$$\phi = \frac{hc}{3\lambda}$

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