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.

 

 
 
 
 
 

Answers (1)
S safeer

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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