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JEE Advanced 2020 Full Test 2 Paper 1
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© examsnet.com
Question : 18
Total: 54
The peak emission from a black body at a certain temperature occurs at a wavelength of
6000
Å
. On increasing its temperature, the total radiation emitted is increased 16 times. These radiations are allowed to fall on a metal surface. Photoelectrons emitted by the peak radiation at higher temperature can be bought to rest by applying a potential equivalent to the excitation potential corresponding to the transition for the level
n
=
4
to
n
=
2
in the Bohr's hydrogen atom. The work function of the metal is given by
α
100
e
V
where
α
is the numerical constant. The value of
α
is found
53
×
x
,
find
x
?
[Take: he
=
12420
e
V
−
Å
]
Your Answer:
Validate
Solution:
From Wien's displacement law and Stefen Boltzman law the wavelength of the peak radiation is halved. So,
λ
=
3000
Å
h
c
λ
−
α
100
=
13.6
[
1
4
−
1
16
]
h
c
λ
−
α
100
=
2.55
12420
3000
−
α
100
=
2.55
α
100
=
4.14
−
2.55
=
1.59
α
=
159
=
53
x
x
=
3
© examsnet.com
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