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Let \mathrm{\mathrm{K}_{1}} and \mathrm{K}_{2} be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength \mathrm{\lambda_{1}} and \mathrm{\lambda_{2}}, respectively are incident on a metallic surface.

\mathrm{\text { If } \lambda_{1}=3 \lambda_{2} \text { then : }}

Option: 1

\mathrm{\mathrm{K}_{1}>\frac{\mathrm{K}_{2}}{3}}


Option: 2

\mathrm{K}_{1}<\frac{\mathrm{K}_{2}}{3}


Option: 3

\mathrm{K}_{1}=\frac{\mathrm{K}_{2}}{3}


Option: 4

\mathrm{K}_{2}=\frac{\mathrm{K}_{1}}{3}


Answers (1)

best_answer

Using Einstein's photoelectric effect eqn
\begin{aligned} &E=\phi+k_{\text {Emx }} \\ \end{aligned}

\frac{h C}{\lambda_{1}} =\Phi +K_{1}-(1)

\frac{h C}{\lambda_{2}}=\phi+K_{2}-(2)

taking the ratio of Eqn (1) with Eqn (2)

And using \lambda_{1}=3 \lambda_{2}

\frac{ \lambda_{2}}{ \lambda_{1}}=\frac{\phi +K_1}{\phi +K_2}=\frac{1}{3}

\Rightarrow 3 K_{1}=K_{2}-2 \phi

\Rightarrow K_{1}<K_{2} / 3 \quad

Hence Option B is correct.

Posted by

Pankaj Sanodiya

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