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Question 45 of 57

Q.Determine the shortest wavelengths of Balmer and Paschen series. Given the limit for Lyman series is 912 Å.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022Subjective· 3mImportance★★★★★
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Rydberg formula 1/λ = R(1/n1² − 1/n2²); use the Lyman limit to fix R, then evaluate the series limits for Balmer (n1=2) and Paschen (n1=3).

The series limit (shortest wavelength, n2→∞n_2\to\infty) for a series starting at level n1n_1 is 1λ=Rn12\dfrac{1}{\lambda} = \dfrac{R}{n_1^2}, i.e. λ=n12/R\lambda = n_1^2/R.

Given the Lyman limit (n1=1n_1=1) is 912912 Å: 1912=R  ⟹  R=1912 A˚−1\dfrac{1}{912} = R \implies R = \dfrac{1}{912}\ \text{Å}^{-1}.

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