The work function of cesium is 2.27 eV. The cut-off voltage wich stops the emission of electrons from a cesium cathode irradiated with light of 600 nm wavelength is

Options

A. 0.5 V
B. `0.2 V`
C. `-0.5 V`
D. 0.2 V

Answer

C

Explanation

Energy of each photon incident on cesium,
`E = (hc)/(lamda) = (6.6 xx 10^(-34) xx 3 xx 10^(8))/(600 xx 10^(-9)) = 3.3 xx 10^(-19)J`
= 2.06 eV
which is less than the work function of cesium (2.27 eV). Hence there will not be any photocurrent.