The graph between the stopping potential `(V_(0))` and wave number `(1//lamda)` is as shown in the figure.

`phi` is the work function, then:

Options

A. `phi_(1):phi_(2):phi_(3)=1:2:4`
B. `phi_(1):phi_(2):phi_(3)=4:2:1`
C. `tanthetaprophC//e` where `theta` is the slope
D. ultraviolet light can be used to eject photoelectron from metal 2 and metal 3 only

Answer

A, C

Explanation

`phi_(1): phi_(2): phi_(3)=(hc)/(lambda0_(1)):(hc)/(lambdao_(2)):(hc)/(lambdao_(3))`
`:.(hc)/(lambda)=phi+eV_(s)`
`:. V_(S)=(hc)/(elambda)-(phi)/(c)`
`:.` slope `"tan" theta=(hc)/(e)`
`(1)/(lambda0_(1))=0.001 mm^(-1) rArr lambda0_(1)=1000Å`
`(1)/(lambda0_(2))=0.002 nm^(-1) rArr lambda0_(2)=5000Å`
`(1)/(lambda0_(3))=0.004 nm^(-1) rArr lambda0_(3)=2500 Å`
Here ultraviolet light can be used to eject photoelectrons from all 1,2 and 3 metals.