There is a square plate of side 1m, the threshold wavelength of plate is 6000Å. Light of uniform intensity 1 watt/m2 is incident normally on plate. The wavelength of light increases linearly from 4000Å to 8000Å as we move from bottom to top. Assume that each capable photon ejects an electron, if the number of electrons ejected per second is 1019K, find K.

Wavelength of light at height y λ=[4000+(8000−4000)y]Å=4×10−7(1+y)   meter.Consider a strip of thickness dy at height y Energy incident on element per sec = Intensity x strip area = (I) ldyNo. of photons incident per second on element  dn=Iℓdyhcλ=Iℓdyhc4×10−7(1+y) Total no. of electrons ejected due to photons per sec  n=Iℓhc4×10−7∫0l/2 (1+y)dy=1(1)4×10−76.6×10−343×10858=2019.8×10198≈10198 So k = 8[Note: No electrons are ejected from upper half, because wavelength of incident wave becomes  greater than threshold wavelength]