JEE MAIN - Physics (2022 - 24th June Evening Shift - No. 17)
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 :
1 : 1
2 : 1
4 : 1
1 : 4
Explanation
$$3.8 = 0.6 + {1 \over 2}mv_1^2$$
$$1.4 = 0.6 + {1 \over 2}mv_2^2$$
$$ \Rightarrow {{v_1^2} \over {v_2^2}} = {{3.2} \over {0.8}} = {4 \over 1}$$
$$ \Rightarrow {{{v_1}} \over {{v_2}}} = {2 \over 1}$$
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