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The graph between 1//lambda and stopping...

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

`phi_1:phi_2:phi_3=1:2:4`

B

`phi_1:phi_2:phi_3=4:2:1`

C

`tanthetaprop((hc))/(e)`

D

Ultraviolet light can be used to emit photoelectrons from metal 2 and metal 3 only

Text Solution

Verified by Experts

The correct Answer is:
A, C
`phi_a:phi_2:phi_3=eV_(01):eV_(02):eV_(03)`
`V_(01):V_(02):V_(03)=0.001:0.002:0.003=1:2:4`
therefore, option (a) is correct.
By Einstein's photoelectric equation, `(hc)/(lamda)-phi=eV`
`impliesV=(hc)/(elamda)-(phi)/(e)` ..(i)
Comparing Eq. (i) by `y=mx+c`, we get the slope of the line `m=(hc)/(e)=tantheta`
`implies` Option (c ) is correct.
From the graph it is clear that,
`(1)/(lamda_(01))=0.001nm^(-1)`
`implieslamda_(01)=(1)/(0.001)=1000nm`
Also, `(1)/(lamda_(02))=0.002nm^(-1)implieslamda_(02)=500nm`
and `lamda_(03)=250nm`
Violet color light will have wavelength less that 400 nm. Therefore, this light will be unable to show photoelectric effect on plate 3 `implies` Option (d) is wrong.
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