The graph between `1//lambda` and stopping potential (V) of three metals having work- functions `Phi_1`, `Phi_2` and `Phi_3` in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct? (Here, `lambda` is the wavelength of the incident ray). (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 constant and c is the speed of light (d) The violet colour light can eject photoelectrons from metals 2 and 3 .

Options

A. (i),(iii)
B. (i),(iv)
C. (ii),(iii)
D. (i),(ii),(iv)

Answer

A

Explanation

`E =phi +eV_(s)`
`(hc)/(lambda) =phi +eV_(s)`
`V_(s) =(hc)/(e lambda) - (phi)/(e)`
`V_(s) v//s (1)/(lambda)` graph:
`y=(hc)/(e) x - (phi)/(e)` (straight line)
Slope `= tan theta =(hc)/(e)`, (iii) is O.K.
`{:("For plate 1:", "Plate 2:","Plate 3:"),((phi_(1))/(hc) =0.001, (phi_(2))/(hc) =0.002,(phi_(3))/(hc) =0.004):}`
`phi_(1) : phi_(2) : phi_(3) =1 : 2 : 4`, (i) is O.K.
For plate 2, thereshold wavelength, `(1)/(0.002) =500 nm`
For plate 3, threshold wavelength, `(1)/(0.004) =250 nm`
`lambda_("violet") = 400 nm`, so `lambda_("violet") lt lambda_("threshold")` for plate 1, `lambda_("violet") gt lambda_("threshold")` for plate 3
Therefore violet colour light will eject photo-electrons from plate 2 and not plate 3.