The maximum kinetic energies of photoelectron | vedclass

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Question

(B) 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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$K_1 < \frac{K_2}{3}$

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(B) 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 of the metal.For wavelength $\lambda_1$,$K_1 = \frac{hc}{\lambda_1} - \phi$ $(i)$For wavelength $\lambda_2$,$K_2 = \frac{hc}{\lambda_2} - \phi$ (ii)Given $\lambda_1 = 3 \lambda_2$,substitute this into equation $(i)$:$K_1 = \frac{hc}{3 \lambda_2} - \phi$ (iii)From equation (ii),$\frac{hc}{\lambda_2} = K_2 + \phi$. Substituting this into equation (iii):$K_1 = \frac{1}{3} (K_2 + \phi) - \phi$$K_1 = \frac{K_2}{3} + \frac{\phi}{3} - \phi$$K_1 = \frac{K_2}{3} - \frac{2\phi}{3}$Since the work function $\phi > 0$,it follows that $K_1 < \frac{K_2}{3}$.

The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when lights of wavelengths $\lambda_1$ and $\lambda_2$ are incident on a metallic surface. If $\lambda_1 = 3 \lambda_2$,then:

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