For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

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Lyman λ 1 ​ 1 ​ = R [ 1 1 ​ − 4 1 ​ ] = 4 3 R ​ λ 1 ​ = 3 R 4 ​ ________...(1) and Balmer λ 2 ​ 1 ​ = R [ 4 1 ​ − 9 1 ​ ] = 36 5 R ​ λ 2 ​ = 5 R 36 ​ Then, λ 2 ​ λ 1 ​ ​ = 27 5 ​

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