Correct Answer - C
The energy of incident photosn is given by
`hv=eV_(s) +phi_(0) =2+5=7eV` (`V_(S)` is stopping potential and `phi_(0)` is work function)
Saturation current `=10^(-6) A=(etaP)/(hv) e =(10^(-5)P)/(7xxe) e` (`eta` is photo emission efficiency)
`:. P=7W`.