The maximum kinetic energies of photoelectron | vedclass

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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 wo

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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 $\lambda_1$, $K_1 = \frac{hc}{\lambda_1} - \Phi$.For wavelength $\lambda_2$, $K_2 = \frac{hc}{\lambda_2} - \Phi$.Given $\lambda_1 = 3\lambda_2$, we substitute this into the equation for $K_1$:$K_1 = \frac{hc}{3\lambda_2} - \Phi$.Since $\frac{hc}{\lambda_2} = K_2 + \Phi$, we have $K_1 = \frac{K_2 + \Phi}{3} - \Phi = \frac{K_2}{3} + \frac{\Phi}{3} - \Phi = \frac{K_2}{3} - \frac{2\Phi}{3}$.Since $\Phi > 0$, it follows that $K_1 < \frac{K_2}{3}$.

The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when light of wavelength $\lambda_1$ and $\lambda_2$ respectively are incident on a metallic surface. If $\lambda_1 = 3\lambda_2$ then

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