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In the spectrum of hydrogen, the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is
detailed solution
Correct option is B
The wavelength of a spectral line in the Lyman series is1λL=R112-1n2,n=2,3,4,……and that in the Balmer series is 1λB=R122-1n2,n=3,4,5 For the longest wavelength in the Lyman series, n=2∴ 1λL=R112-122=R11-14=R4-14=3R4 or λL=43RFor the longest wavelength in the Balmer series, n = 3∴ 1λB=R122-132=R14-19=R9-436=5R36 or λB=365R Thus, λLλB=43R365R=43R×5R36=527Talk to our academic expert!
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