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IIT JEE Advanced 2009 Question Paper 2
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Question : 2
Total: 57
Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions
ϕ
p
= 2.0 eV,
ϕ
q
= 2.5 eV and
ϕ
r
= 3.0 eV, q r respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I–V graph for the experiment is (Take hc = 1240 eV nm)
Validate
Solution:
We have
E =
(
h
c
λ
)
J ⇒ E =
(
1240
λ
n
m
)
eV
Therefore,
λ
1
= 550 nm,
E
1
= 2.25 eV
λ
2
= 450 nm ,
E
2
= 2.75 eV
λ
3
= 350 nm ,
E
3
= 3.5 eV
Also,
ϕ
p
= 2 eV, all λ’s cause emissions.
ϕ
q
= 2.5 eV, last two λ’s cause emissions.
ϕ
r
= 3 eV, only the last λ causes emissions.
That is,
I
p
>
I
q
>
I
r
.
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