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Question
The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 Å. Calculate the short wavelength limit for the Balmer series of the hydrogen spectrum.
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Solution
Given short wavelength limit of the Lyman series
`1/λ_"L" = "R"(1/1^2 - 1/∞)`
`1/(913.4 Å) = "R"(1/1^2 - 1/∞)`
λL = `1/"R"` = 913.4 Å
For the short wavelength limit of Balmer series
n1 = 2, n2 = ∞
`1/λ_"B" = "R"(1/2^2 - 1/∞)`
λB = `4/"R"`
= 4 × 913.4 Å
= 3653.6 Å
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