| ⇦ | 
| ⇨ |
The ratio of longest wavelength corresponding to Lyman and Blamer series in hydrogen spectrum is
Options
(a) 3 / 23
(b) 7 / 29
(c) 9 / 13
(d) 5 / 27
Correct Answer:
5 / 27
Explanation:
For Lyman series (2 → 1)
1/λL = R [1 – 1/2] = 3R/4
For Balmer series (3 → 2)
1/λB = R [1/4 – 1/9] = 5R/36
λL / λB = 4/3R / 36/5R = 4/36 (5/3) = 5/27
Related Questions: - A spring of spring constant 5 x 10³ Nm⁻¹ is streched intially by 5 cm from
- which of the following can not be emitted by radioactive substances during their decay?
- In an interference experiment, third bright fringe is obtained at a point
- A boat of anchor is rocked by waves whose crests are 100 m apart and velocity
- The maximum particle velocity in a wave motion is half the wave velocity.
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A spring of spring constant 5 x 10³ Nm⁻¹ is streched intially by 5 cm from
- which of the following can not be emitted by radioactive substances during their decay?
- In an interference experiment, third bright fringe is obtained at a point
- A boat of anchor is rocked by waves whose crests are 100 m apart and velocity
- The maximum particle velocity in a wave motion is half the wave velocity.
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply