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Question

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].

76364_0484535e665c4a08aeedb0437c78ca84.png


A
Ratio of work functions ϕ1:ϕ2:ϕ3=1:2:4.
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B
Ratio of work functions ϕ1:ϕ2:ϕ3=4:2:1.
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C
tan θ is directly proportional to hc/e, where h is Plancks constant and c is the speed of light.
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D
The violet colour light can eject photoelectrons from metals 2 and 3.
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Solution

The correct options are
A Ratio of work functions $$\phi _{1} : \phi _{2} : \phi _{3}=1 : 2 : 4.$$
C tan $$\theta$$ is directly proportional to $$hc/e$$, where $$h$$ is Plancks constant and $$c$$ is the speed of light.
According to Einstein's equation, $$\dfrac{hc}{\lambda }-\phi= $$ eV
or, $$V=\dfrac{hc}{e\lambda }-\dfrac{\phi }{e}$$
Now the ratio, plate 1 : Plate 2 : Plate 3
$$\displaystyle \frac{\phi _{1}}{e}=0.001$$, $$ \dfrac{\phi _{2}}{e}=0.002 $$, $$\dfrac{\phi _{3}}{e}=0.004$$
$$\phi _{1}:\phi _{2}:\phi _{3}=1:2:3$$
For plate 2, threshold wavelength, $$\lambda = \dfrac{hc}{\phi_2} = \dfrac{hc}{0.002hc}= \dfrac{1000}{2} = 500 \ nm$$
For plate 3, threshold wavelength, $$\lambda = \dfrac{hc}{\phi _{3}}=\dfrac{hc}{0.004 hc}=\dfrac{1000}{4}=$$250nm
Since violet colour light $$\lambda $$ is 400 nm, so $$\lambda_{violet} $$ < $$\lambda _{threshold}$$ for plate 2.
So, violet colour light will eject photoelectrons from plate 2 and not from plate 3.

Physics
NCERT
Standard XII

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