A photon of wavelength `1000 Å` has energy `12.3 eV`. If light of wavelength `5000 Å`, having intensity I, falls on a metal surface, the saturaion current is `0.40 mu A` and the stopping potential is `1.36 V`. The work function of metal is

Options

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

Answer

C

Explanation

`E =(12.3)/(5) =2.46 eV` , `K_(max) =1.36 eV`
`phi = E-W=2.46 -1.36 = 1.10 eV`