The stopping potential as a function of the frequency of the incident radiation is plotted for two different photoelectric surfaces `A` and `B` . The graphs show that work function of `A` is

Options

A. `15.5 eV`
B. `17.8 eV`
C. `20.6 eV`
D. `24.5 eV`

Answer

C

Explanation

The intercept on the `upsilon` axis of the graph gives the threshold frequency `upsilon_0` It is `5 xx 10^15` Hz.
`therefore` The work function `phi_0 =h upsilon`
`= 6.6 xx 10^(-34) xx 5 xx 10^(15) =33 xx 10^(-19)`
`= (33 xx 10^(-19))/(1.6 xx 10^(-19))` eV
`therefore phi_0 = 20.625 ~~ 20.6eV`