Modern Physics 2 Question 18

17. The figure shows the variation of photocurrent with anode potential for a photosensitive surface for three different radiations. Let Ia,Ib and Ic be the intensities and fa,fb and fc be the frequencies for the curves a,b and c respectively

(2004,2M)

(a) fa=fb and IaIb

(c) fa=fb and Ia=Ib

(b) fa=fc and Ia=Ic

(d) fb=fc and Ib=Ic

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Solution:

  1. Saturation current is proportional to intensity while stopping potential increases with increase in frequency.

Hence, fa=fb while Ia<Ib



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