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. Ratio of work function, `phi_(1):phi_(2):phi_(3)=1:2:4`
B. Ratio of work function, `phi_(1):phi_(2):phi_(3)=4:2:1`
C. `tan theta` is directely proportional to `hc//e` where `h` is planck's costant and `c` is the speed of light
D. The violet colour light can eject photoelectrons form metals `2` and `3`

Answer

a,c

Explanation

From Einstein's photoelectrons equation
`K.E.=eV=(hc)/lambda-phi`...........(i)
or `V=(hc)/(e lambda)-phi/e`......(ii)
When `lambda=lambda_(0)`, the threshold wavelength, `K.E.=0`.
From (i), `phi=(hc)/(lambda_(0))` or `(phi)/(hc)=1/(lambda_(0))`
For plate`1, (phi_(1))/(hc)=0.001`,
For plate`2, (phi_(2))/(hc)=0.002`,
For plate`3, (phi_(3))/(hc)=0.004`,
`:. phi_(1):phi_(2):phi_(3)=1:2:4`
Differetiating (ii), we get
`DeltaV=(hc)/e Delta(1//lambda)` or `(DeltaV)/(Delta(1//lambda))=(hc)/e`
Slope of the graph, `tan theta=(DeltaV)/(Delta(1//lambda))=-(hc)/e`
For plate `2`, threshold wavelength,
`lambda_(0_(2))=(hc)/(phi_(2))=(hc)/(0.002hc)=500nm`
For plate `3`, threshold wavelength,
`lambda_(0_(3))=(hc)/(phi_(3))=(hc)/(0.004hc)=250nm`
since the wavelength of violet colour is `400nm` and `lambda_("violet") lt lambda_(0_(2))` for plate `2` and `lambda_("violet") gt lambda_(0)`for plate `3`, therefore, violet colour light will ejected electrons from plate 2 and not plate 3.