Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda$1 and $\lambda$2, respectively are incident on a metallic surface. If $\lambda$1 = 3$\lambda$2 then :
Solution
<p>$${K_1} = {{hc} \over {{\lambda _1}}} - \phi = {{hc} \over {3{\lambda _2}}} - \phi $$ ..... (i)</p>
<p>and ${K_2} = {{hc} \over {{\lambda _2}}} - \phi$ ..... (ii)</p>
<p>from (i) and (ii) we can say</p>
<p>$3{K_1} = {K_2} - 2\phi$</p>
<p>${K_1} < {{{K_2}} \over 3}$</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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