A photoelectric surface is illuminated succes | vedclass

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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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $\lambda$:$K_1 = \frac{hc}{\lambda} - \phi$  ..... $(i)$For the second case with wavelength $\frac{\lambda}{2}$:$K_2 = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$  ..... $(ii)$Given that $K_2 = 3K_1$,we substitute equations $(i)$ and $(ii)$:$\frac{2hc}{\lambda} - \phi = 3 \left( \frac{hc}{\lambda} - \phi \right)$Expanding the right side:$\frac{2hc}{\lambda} - \phi = \frac{3hc}{\lambda} - 3\phi$Rearranging the terms to solve for $\phi$:$3\phi - \phi = \frac{3hc}{\lambda} - \frac{2hc}{\lambda}$$2\phi = \frac{hc}{\lambda}$$\phi = \frac{hc}{2\lambda}$

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $3$ times that in the first case,the work function of the surface is:

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