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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
- A photoelectric surface is illuminated successively by monochromatic light of wavelength
- Two particles of masses m₁ m₂ move with initial velocities u₁ and u₂
- A body is thrown with a velocity of 9.8 m/s making angle of 30° with the horizontal
- The heat dissipated in a resistance can be obtained by measuring of resistance,
- What should be the velocity(v) of a sound source moving towards a stationary observer
Topics: Atoms and Nuclei (136)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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