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 surface is 1.9 eV,

Options

A. The maximum `K.E`. Of photo electrons when it is irradiated with `500nm` photons will be `0.42 eV`
B. The maximum `K.E`. In case `(a)` will be `1.725eV`
C. The longest wavelength which will eject the photo electrons from the surface is nearly `610nm`
D. The maximum `K.E`. Will increase if the intesity of radiation is increased

Answer

A, C

Explanation

`KE_(max)=(hc)/(elambda)-WrArr(hc)/(elambda)=1+1.9=2.9eV`
`rArr (hc)/e=2.9xx400 eV nm` for `lambda=500 nm,E=(hc)/(e500)=(2.9xx4cancel(0)cancel(0))/(5cancel(0)cancel(0))=2.32eV` i.e `KE_(max)=2.32-1.9=0.42eV`
`therefore lambda_(0)=(hc)/(eW)=(2.9xx400)/1.9=610 A^(0)`