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Question

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

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

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy $K$ is given by $K = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For the first case with wavelength $\lambda$: $K_1 = \frac{hc}{\lambda} - \phi$.For the second case with wavelength $\lambda/2$: $K_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$.Given that $K_2 = 3K_1$,we substitute the expressions:$\frac{2hc}{\lambda} - \phi = 3 \left( \frac{hc}{\lambda} - \phi \right)$.Expanding the equation: $\frac{2hc}{\lambda} - \phi = \frac{3hc}{\lambda} - 3\phi$.Rearranging the terms to solve for $\phi$: $3\phi - \phi = \frac{3hc}{\lambda} - \frac{2hc}{\lambda}$.$2\phi = \frac{hc}{\lambda}$.Therefore,$\phi = \frac{hc}{2\lambda}$.

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ and $\lambda/2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is three times that in the first case,the work function of the surface is

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