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If the series limit wavelength of the Lyman series for hydrogen atom is 912 , then the series limit wavelength for the Balmer series for the hydrogen atom is

a
912Å/2
b
912Å
c
912×2Å
d
912×4Å

detailed solution

Correct option is D

For Lyman series, the series limit wavelength is given by1λ=R112−1∞2=R  or  λ=1RFor Balmer series, the series limit wavelength is given by1λ′=R122−1∞2=R4  or  λ′=4RClearly,          λ′=41R  or  λ′=4λ=4×912Å

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