The figure shows the variation of stopping potential $V_{o}$ with the frequency $\nu$ of the incident radiations for two photosensitive metals P and Q. Which metal has smaller threshold wavelength? Justify your answer.

From the graph, we can see that, Q metal will be of smaller threshold wavelength as frequency $(\nu)$ is inversely proportional to $\lambda$,

hence, $\lambda =\frac{c}{\nu_{o}}$

Where C is the speed of light

$\nu_{o}$ is frequency

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