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If $λ_{1} and λ_{2}$ are the wavelengths of the first members of the Lyman and Paschen series, respectively, then $λ_{1} : λ_{2}$ is

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Correct option is D

1λ∝1n12-1n22 ; ∴λ1λ2=132-142paschen112-122Lymen =7/1443/4=7108

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If λ1 and λ2 are the wavelengths of the first members of the Lyman and Paschen series, respectively, then λ1 : λ2 is

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If $λ_{1} and λ_{2}$ are the wavelengths of the first members of the Lyman and Paschen series, respectively, then $λ_{1} : λ_{2}$ is