| ⇦ |
| ⇨ |
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 spherical drop of mercury having a potential of 2.5 v is obtained as a result
- Two pendulums A and B are oscillating simultaneously with time period
- If the lattice parameter for a crystalline structure is 3.6 Å, then the atomic radius
- The energy required to charge a parallel plate condenser of plate separation d
- If alpha, beta and gamma rays carry same momentum, which has the longest
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A spherical drop of mercury having a potential of 2.5 v is obtained as a result
- Two pendulums A and B are oscillating simultaneously with time period
- If the lattice parameter for a crystalline structure is 3.6 Å, then the atomic radius
- The energy required to charge a parallel plate condenser of plate separation d
- If alpha, beta and gamma rays carry same momentum, which has the longest
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