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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
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Topics: Atoms and Nuclei
(136)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An automobile engine develops 100 kW when rotating at a speed of 1800 rev/min
- X-rays when incident on a metal
- A parallel plate condenser with a dielectric of dielectric constant K between the plates
- A 100Ω resistance and a capacitor of 100Ω reactance are connected in series
- The de-Broglie wavelength of an electron moving with a velocity 1.5×10⁸ ms⁻¹ is equal
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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