Given that a photon of light of wavelength 10,000Å has an energy equl to 1.23eV. When light of wavelength 5000Å and intensity `I_(0)` falls on a photoelectric cell and the saturation current is `0.40xx10^(-6)` ampere and the stopping potential is 1.36 volt, then (i)what is the work function? (ii) If intensity of light is made `4I_(0)`, what should be the saturation current and stopping potential?

Answer

(i) 1.1 eV (ii) `1.6xx10^(-6) A`, unchanged

Explanation

(i) `(E_(2))/(E_(1))=(hc//lambda_(2))/(hc//lambda_(1))=(lambda_(1))/(lambda_(2))=10000/5000=2`
or `E_(2)=2E_(1)=2xx1.23=2.46 eV`
`:.` Energy of incident photon, E=2.46 eV, Given, stopping potential =1.36 V, therefore, Kinetic energy of photoelectrons, `K=1.36 eV`.
So, work function is
`phi_(0)=E-K=2.46-1.36=1.1 eV`
(ii) When the intensity of the light is made four times, the stopping potential remains unchanged (i.e., 1.36 eV) but saaturation current will become fout times `=4xx0.40xx10^(-6)=1.6xx10^(-6)A`