Given that a photon of light of wavelength `10,000A` has an energy equal to 1.23 eV. When light of wavelength `5000A` and intenstiy `I_0` falls on a photoelectric cell, the saturation current is `0.40xx10^(-6)A` and the stopping potential is 1.36V, then the work function is

Options

A. 0.43eV
B. 1.10 eV
C. 1.36 eV
D. 2.47 eV

Answer

B

Explanation

Since `Epropto(1)/(lamda)`, so energy corresponding to `5000A` is
`E=2.46eV`
Now, `hupsilon-W=eV_S`
or `2.46eV-W=1.36eV`
`W=(2.46-1.36)eV=1.1eV`