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Question
Determine the shortest wavelengths of Balmer and Pasch en series. Given the limit for the Lyman series is 912 Å.
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Solution
The wavelength of lines are given by,
`1/lambda = "R"_"H" (1/"n"^2 - 1/"m"^2)`
For lyman series limit. n = 1. m = ∞
`therefore 1/lambda_"L" = "R"_"H" (1/1^2 - 1/oo) = "R"_"H"`
`therefore lambda_"L" = 1/"R"_"H" = 912`Å
For Balmer series limit, n = 2, m = ∞
`therefore 1/lambda_"B" = "R"_"H" (1/2^2 - 1/oo) = "R"_"H"/4`
`therefore lambda_"B" = 4/"R"_"H"`
= 4 × 912
= 3648 Å
For Paschen series limit. n = 3, m = ∞
`1/lambda_"P" = "R"_"H" (1/3^2 - 1/oo) = "R"_"H"/9`
`therefore lambda_"P" = 9/"R"_"H" = 9 xx 912`
= 8208 Å
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