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Question
Calculate the shortest wavelength of electromagnetic radiation present in Balmer series of hydrogen spectrum.
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Solution
For shortest wavelength in Balmer series nf = 2, n1 = ∞
`1/lambda = "R" [1/"n"_"f"^2 - 1/"n"_"i"^2]`
`1/lambda = "R" [1/4]`
`therefore lambda = 4/"R"`
`= 4/(1.097 xx 10^7) "m"`
`= 3.646 xx 10^-7"m"`
= 3646 Å
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