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If K_{1} and K_{2} are the maximum kinetic energies of photoelectrons emitted when lights of wavelength \lambda _{1} and \lambda _{2} respectively incident on a metallic surface and \lambda _{1}=3\lambda _{2} , then -

Option: 1

K_{1}> \frac{K_{2}}{3}


Option: 2

K_{1}<\frac{K_{2}}{3}


Option: 3

K_{1}=3K_{2}


Option: 4

K_{2}=3K_{1}


Answers (1)

best_answer

\begin{aligned} & \lambda_1=3 \lambda_2 \\ & \lambda_2=\frac{\lambda_1}{3} \\ & v_2=3 v_1 \\ & \mathrm{~K}_2>3 \mathrm{~K}_1 \\ & \frac{\mathrm{K}_2}{3}>\mathrm{K}_1 \end{aligned}

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