Figure shows the graph of photocurrent versus potential of the anode with respect to the cathode.The intensity of the incident radiation is halved and freqency doubled. The stopping potential and the saturation current will now be respectively.

Options

A. `-6.0 V` and `2.0 mu A`-
B. `-6.0 V` and `4.0 mu A`
C. `-6.0 V` and `1.0 mu A`
D. None of these

Answer

D

Explanation

`(dN)/(dt)=(IA)/(hf)=(1)/(4)(dN_(0))/(dt)`
`v_(s)=(hf)/(e)-(phi)/(e)`
`3=(hf)/(e)-(phi)/(e)`
`v'=(2hf)/(e)-(phi)/(e)=2xx3+(phi)/(e)=6+(phi)/(e)gt6`