The graph between the stopping potential `(V_(0)) and ((1)/(lambda)) ` is shown in the figure `phi_(1) , phi_(2) and phi_(3)` are work function , which of the following is /are correct

Options

A. Ratio of work functions `phi_(1) : phi_(2) : phi_(3) = 1 : 2 : 4`
B. Ratio of work functions `phi_(1) : phi_(2) : phi_(3) = 4 : 2 : 1`
C. `tan theta` is directly proportional to `hc//e`, where h is Planck's constant and c is the speed of light can eject photoelectrons from metals 2 and 3.
D. The vilolet colour light can eject photoelectron from metals 2 and 3.

Answer

A, C

Explanation

From the relation,
`eV=(hc)/lambda - phi`
`implies V=((hc)/e)(1/lambda)-phi/e`
This is equation of straight line.
Slope is `tan theta= (hc)/e`
`phi_(1):phi_(2):phi_(3)=(hc)/lambda_(01): (hc)/lambda_(02): (hc)/lambda_(03)=1:2:4`
`1/lambda_(01)=0.001 nm^(-1) implies lambda_(01)=10000Å`
`1/lambda_(02)=0.002nm^(-1)implies lambda_(02)=5000 Å`
`1/lambda_(03)=0.004 nm^(-1) implies lambda_(03)=2500Å`
Violet colour has wavelength `4000 Å`.
So violet colour can eject photoelectrons from metal-1 and metal -2.