| ⇦ | 
| ⇨ |
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: - The electirc field in a certain region is acting radially outward and is given by E=Ar.
- In R-L-C series circuit, the potential differences across each element is 20 V.
- Masses of 3 wires of same metal are in the ratio 1:2:3 and their lengths
- If the density of earth is doubled keeping its radius constant then acceleration
- In the reaction ₉₂U²³⁴→₈₇Y²²², how many α-particle and β-particles are emitted
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The electirc field in a certain region is acting radially outward and is given by E=Ar.
- In R-L-C series circuit, the potential differences across each element is 20 V.
- Masses of 3 wires of same metal are in the ratio 1:2:3 and their lengths
- If the density of earth is doubled keeping its radius constant then acceleration
- In the reaction ₉₂U²³⁴→₈₇Y²²², how many α-particle and β-particles are emitted
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