The maximum K.E. of photoelectrons ejected from a photometer when it is irradiated with radiation of a wavelength 400nm is 1eV. If the threshold energy of the suface is 1.9 eV,

Options

A. the maximum K.E. of photoelectrons when it is irradiated with `500nm` photons will be `0.42 eV`
B. the maximum K.E. of photoelectron when it is viradiated with `500nm` photons will be `0.43 eV`
C. maximum K.E. will increases if the intensity of radiation is increased
D. the longest wavelength which will eject the photoelectrons from the surface is nearly `610nm`

Answer

a,d

Explanation

Here, `lambda=400nm, K_(max)=1eV, phi_(0)=1.9 eV`
`K_(max)=(hc)/lambda-phi_(0)` or `hc=(K_(max)+phi_(0))lambda`
`=(1+1.9)eVxx400nm=2.9xx400eV nm`
`lambda'=500nm`, then
`K_(max)=(hc)/(lambda')-phi_(0)=(2.9xx400eVnm)/(500nm)-1.9eV`
`=2.32eV-1.9eV=0.42 eV`
Longest wavelength `=(hc)/(phi_(0))=(2.9xx400eVnm)/(1.9eV)`
`=610nm`