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,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

The maximum K.E. of photoelectrons ejected from a photometer when it i

1.	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,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» Correct Answer - A::C `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)`

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