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Standard XII
Chemistry
Emission and Absorption Spectra
Question
If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of
H
e
+
is :
36
A
7
5
A
9
36
A
5
9
A
5
A
36
A
5
B
36
A
7
C
5
A
9
D
9
A
5
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Solution
Verified by Toppr
For Lyman series
h
y
d
r
o
g
e
n
atom, the shortest wavelength corresponds to:
n
1
=
1
and
n
2
=
∞
1
λ
=
R
Z
2
[
1
n
2
1
−
1
n
2
2
]
1
A
=
R
[
1
1
2
−
1
∞
2
]
1
A
=
R
.....(1)
For Paschen series of
H
e
+
ion, the longest wavelength corresponds to :
n
1
=
3
and
n
2
=
4
1
λ
=
R
Z
2
[
1
n
2
1
−
1
n
2
2
]
1
λ
=
1
A
×
2
2
×
[
1
3
2
−
1
4
2
]
λ
=
36
A
7
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