The threshold wavelength of a metal surface is 500 nm. V is the maximum velocity of photo electrons when the wavelength of incident light is 400 nm. If the maximum velocity of photo electrons from the same metal surface is 2V, the wavelength of incident light is

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350 nm

200 nm

250 nm

300 nm

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} m V^{2} = h c (\frac{1}{λ} - \frac{1}{λ_{0}}) \\ \begin{array}{l} {(\frac{V_{1}}{V_{2}})}^{2} = \frac{(\frac{1}{λ_{1}} - \frac{1}{λ_{0}})}{(\frac{1}{λ_{2}} - \frac{1}{λ_{0}})} \\ {(\frac{1}{2})}^{2} = \frac{(\frac{1}{400} - \frac{1}{500})}{(\frac{1}{λ_{2}} - \frac{1}{500})} \\ \frac{1}{λ_{2}} = \frac{1}{100} - \frac{3}{500} \\ λ_{2} = 250 nm \end{array} \end{array}$

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