Light of wavelength `4000 Å` falls on a photosensitive metal and a negative `2 V` potential stops the emitted electrons. The work function of the material ( in eV) is approximately `( h = 6.6 xx 10^(-34) Js , e = 1.6 xx 10^(-19) C , c = 3 xx 10^(8) ms^(-1))`

Options

A. `1.1`
B. `2.0`
C. `2.2`
D. `3.1`

Answer

A

Explanation

Energy of incident light `E( eV) = (12375)/(4000) = 3.09 eV`
Stopping potential is `- 2V` so `K_(max) = 2 eV`
Hence by using `E = W_(0) + K_(max) rArr W_(0) = 1.09 ~~ 1.1` eV