Given that a photon of light of wavelength 10,000Å has an energy equal 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?

Options

A. `0.43 eV`
B. `1.10 eV`
C. `1.36 eV`
D. `2.72 eV`

Answer

b

Explanation

Work function `phi_(0)` is independent of maximum intensity of incident light.
`E=(hc)/lambda` or `hc=Elambda`
`=(1.23xx1.6xx10^(-19))xx(10000xx10^(-10))`
`=1.23xx1.6xx10^(-25)`
`=(1.23xx1.6xx10^(-25))/(1.6xx10^(-19))=1.23xx10^(-6) eV m`
`eV_(0)=(hc)/(lambda_(1))-phi_(0)`
or `phi_(0)=(hc)/(lambda_(1))-eV_(0)=(1.23xx10^(-6)eVm)/((5000xx10^(-10)))-1.36 eV`
`=2.46 -1.36=1.0 eV`