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.
D. The violet colour light can eject photoelectrons from metlas `2` and `3`

Answer

A, C

Explanation

`(hc)/(lambda)-phi=eV`
`{:("For plate1","plate"2,"Plate"2,,),((phi_(1))/(e)=0.001,(phi_(2))/(2)=0.002,(phi_(3))/(e)=0.004,,):}`
`phi_(1):phi_(2):phi_(3)=1:2:4`
For plate `2`, threshold wavelength
`lambda=(hc)/(phi_(2))=(hc)/(0.002hc)=(1000)/(2)=500 nm`
For plate `3` threshold wavelength
`lambda=(hc)/(phi_(3))=(hc)/(0.004hc)=(1000)/(4)=250 nm`
Since violet colour light `lambda` is `400nm`, so `lambda_("violet") lt lambda_("threshold")` for plate `2`.
So, violet colour light will eject photo-electrons from plate `2` and not from plate `3`.