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 :

Question

Aicharya · Accepted Answer

Expression for maximum kinetic energy can be written as, ${(K E)}_{m a x} = E - ϕ$

${(K E)}_{m a x, 1} = E_{1} - ϕ = 3.8 - 0.6 = 3.2 eV$ , ${(K E)}_{m a x, 2} = E_{2} - ϕ = 1.4 - 0.6 = 0.8 eV$

$\frac{{(K E)}_{m a x, 1}}{{(K E)}_{m a x, 2}} = \frac{3.2}{0.8} = 4 \Rightarrow 4 = \frac{\frac{1}{2} {m v}_{1}^{2}}{\frac{1}{2} m v_{2}^{2}}$

$\Rightarrow \frac{v_{1}^{2}}{v_{2}^{2}} = 4$ $\Rightarrow \frac{v_{1}}{v_{2}} = 2 : 1$

The light of two different frequencies whose photons have energies 3 . 8 eV and 1 . 4 eV respectively,…

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