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Photons of energies $1\; eV$ and $2\; eV$ are successively incident on a metallic surface of work function $0.5\; eV$ . The ratio of kinetic energy of most energetic photoelectrons in the two cases will be.
Option: 1 $1:2$
Option: 2 $1:1$
Option: 3 $1:3$
Option: 4 $1:4$

Answers (1)

The ratio of kinetic energies are

$KE=h\nu-work\ function$

$=\frac{1-0.5}{2-0.5}=\frac{0.5}{1.5}=\frac{1}{3}$

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Safeer PP

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