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The light of two different frequencies whose photons have energies 2.4 eV and 1.2 eV respectively, illuminate a metallic surface whose work function is 0.9 eV successively. The ratio of square of maximum speeds of emitted electrons for the two frequencies will be

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5 : 1

3 : 1

2 : 1

1 : 3

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{\frac{1}{2} m v_{1}^{2}}{\frac{1}{2} m v_{2}^{2}} = \frac{E_{1} - W_{0}}{E_{2} - W_{0}} \\ \Rightarrow \frac{v_{1}^{2}}{v_{2}^{2}} = \frac{2.4 - 0.9}{1.2 - 0.9} = \frac{1.5}{0.3} = \frac{5}{1} \end{array}$

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