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)`
`E = (12375)/(5000) = 2.475 eV " and " K_(max) = eV_0 = 1.36 eV`
So, ` 2.75 = W_0 + 1.36 rArr W_0 = 1.1 eV`
so 2.475 ` = W_0 + 1.36 rArr W_0 = 1.1 eV`