MolSynapse
question bank
All chapters
NEET
JEE
Practice
MolSynapse
/
Question bank
/
Atomic Structure
/
Hydrogen spectrum and spectral series
JEE Main level
Single correct option
Difficulty 2 of 3
as075
For the same transition
n
=
3
→
n
=
2
, the ratio of the wavelength emitted by
He
X
+
to that emitted by hydrogen is
A
4
B
4
1
C
2
1
D
2
Answer: B
1/
λ
∝
Z
2
for the same
n
i
,
n
f
, so
λ
H
e
+
/
λ
H
=
1/4
.
Practice Atomic Structure
More on Hydrogen spectrum and spectral series
Previous question
Next question
More questions on Hydrogen spectrum and spectral series
Use R_H = 1.097x10^7 m^-1. The wavelength of the first line (longest wavelength) of the Lyman series of hydrogen is
JEE Main level
The ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series of hydrogen is
JEE Main level
A large number of hydrogen atoms are excited to the n=5 level. How many of the emitted lines belong to the Balmer series?
JEE Main level
Which transition in He+ emits radiation of the same wavelength as the n=2-> n=1 line of hydrogen?
JEE Main level
Use R_H = 1.097x10^7 m^-1. The shortest wavelength in the Paschen series of hydrogen is nearest to
JEE Main level
Use R_H = 1.097x10^7 m^-1. Calculate the wavelength, in nm (nearest integer), of the hydrogen line produced by the transition n=4-> n=2.
JEE Main level