For intensity `I` of a light of wavelength `5000 Å` the photoelectron saturation current is `0.40 mu A` and 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

By using `E = W_(0) + K_(max)`
`rArr E = (12375)/(5000) = 2.475 eV` and `K_(max) = eV_(0) = 1.36 eV`
So `2.475 = W_(0) + 1.36 rArr W_(0) = 1.1 eV`