MolSynapse
question bank
All chapters
NEET
JEE
Practice
MolSynapse
/
Question bank
/
Atomic Structure
/
Hydrogen spectrum and spectral series
JEE Advanced level
One or more correct options
Difficulty 3 of 3
as110
Use
R
H
=
1.097
×
1
0
7
m
−
1
. Hydrogen atoms are excited to
n
=
3
and then emit all possible lines. Select the correct statements.
A
Two of the emitted lines belong to the Balmer series
B
The shortest wavelength emitted is about
102.6
nm
C
The longest wavelength emitted is about
656.3
nm
D
Three different lines are emitted
Answer: B, C, D
Lines:
3
→
2
,
3
→
1
,
2
→
1
. Shortest is
3
→
1
:
λ
=
9/
(
8
R
H
)
=
102.6
nm
. Longest is
3
→
2
:
λ
=
36/
(
5
R
H
)
=
656.3
nm
. Only
3
→
2
is Balmer, so 'two' is wrong.
Practice Atomic Structure
More on Hydrogen spectrum and spectral series
Previous question
Next question
More questions on Hydrogen spectrum and spectral series
The lines of the Lyman series in the hydrogen spectrum lie in the
NEET level
A large collection of hydrogen atoms is excited to the n = 4 level. The maximum number of different spectral lines that can be emitted as th...
NEET level
All lines of the Balmer series of hydrogen result from electronic transitions that end at the level with n equal to
NEET level
Which Balmer transition gives the line of longest wavelength?
NEET level
Which of the following electronic transitions in a hydrogen atom emits a photon of the highest energy?
NEET level
The lines of the Paschen series of hydrogen lie in the
NEET level