| ⇦ | 
| ⇨ |
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: - With the decrease of current in the primary coil from 2 amperes to zero value
- When a p-n junction is reverse biased, then the current through the junction
- In n-type semiconductor, electrons are majority charge carriers
- A potentiometer wire of length 10 m and resistance 10 Ω per metre is connected in series
- 64 drops of mercury, each charged to a potential of 10 V, are combined
Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- With the decrease of current in the primary coil from 2 amperes to zero value
- When a p-n junction is reverse biased, then the current through the junction
- In n-type semiconductor, electrons are majority charge carriers
- A potentiometer wire of length 10 m and resistance 10 Ω per metre is connected in series
- 64 drops of mercury, each charged to a potential of 10 V, are combined
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