A beam of light consisting of two wavelengths $λ_{1} = 496 n m$ and $λ_{2} = 620 n m$ falls on a metal surface. The maximum speeds of photoelectrons corresponding to $λ_{1}$ and $λ_{2}$ are $u_{1}$ and $u_{2}$ respectively. If $u_{1} : u_{2} = \sqrt{2} : 1$ and $hc = 1240 eV . nm,$ the work function of the metal is $\frac{x}{2} e V .$ Then the value of $x$ is ____

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Detailed Solution

$\frac{1}{2} m u_{1}^{2} = \frac{h c}{λ_{1}} - ω_{0}______(1)$

$\frac{1}{2} m u_{2}^{2} = \frac{h c}{λ_{2}} - ω_{0}______(2)$

$\frac{(1)}{(2)} \Rightarrow 2 = \frac{\frac{1240}{496} - ω_{0}}{\frac{1240}{620} - ω_{0}} \Rightarrow \frac{1240}{310} - 2 ω_{0} = \frac{1240}{496} - ω_{0} \Rightarrow ω_{0} = 1240 [\frac{1}{310} - \frac{1}{496}]$

$ω_{0} = 1.5 e v = \frac{x}{2} \Rightarrow x = 3 e v$

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