| ⇦ | 
| ⇨ |
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: - In Young’s double slit experiment, the locus of the point P lying in a plane
- Light of wavelength λ from a point source falls on a small circular obstacle
- Two masses of 9 g and 6 g are connected with a string which is passing over a frictionless
- The relationship between decay constant λ and half-life T of a radioactive substance is
- A very small circular loop of radius a is initially (at t=0) coplanar and concentric
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In Young’s double slit experiment, the locus of the point P lying in a plane
- Light of wavelength λ from a point source falls on a small circular obstacle
- Two masses of 9 g and 6 g are connected with a string which is passing over a frictionless
- The relationship between decay constant λ and half-life T of a radioactive substance is
- A very small circular loop of radius a is initially (at t=0) coplanar and concentric
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