For photoelectric effect in sodium, fig. shows the plot of cut-off voltage versus frequency of the incident radiation. Caclculate:

(i) the threshold frequency
(ii) the work function for sodium. (Use `h=6.6xx10^(-34)Js`)

Answer

`(i) 4.5xx10^(14)Hz, (ii) 1.86 eV`

Explanation

(i) From the graph, threshold frequency is that frequency for which cut-off potential is zero. So, threshold frequency `v_(0)=4.5xx10^(14)Hz`.
(ii) Work function `phi_(0)=hv_(0)`
`=6.6xx10^(-34)xx4.5xx10^(14)=2.97xx10^(-19)J`
`=(2.97xx10^(19))/(1.6xx10^(-19)) eV=1.86 eV`