The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :
Solution
<p>$3.8 = 0.6 + {1 \over 2}mv_1^2$</p>
<p>$1.4 = 0.6 + {1 \over 2}mv_2^2$</p>
<p>$\Rightarrow {{v_1^2} \over {v_2^2}} = {{3.2} \over {0.8}} = {4 \over 1}$</p>
<p>$\Rightarrow {{{v_1}} \over {{v_2}}} = {2 \over 1}$</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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