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The surface of a metal is illuminated alternately with photons of energies $E_{1} = 5 e V$ and $E_{2} = 2 e V$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal in (eV) is _____

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answer is 1.

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Detailed Solution

$\frac{1}{2} m v_{1}^{2} = 5 - W_{0} a n d \frac{1}{2} m v_{2}^{2} = 2 - W_{0}$

Divide above equations

$4 = \frac{5 - w_{0}}{2 - w_{0}} \Rightarrow 8 - 4 w_{0} = 5 - w_{0} \Rightarrow w_{0} = 1$

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