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Question
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Solution
For the Balmer series, ni = 2. Thus, the expression of wavenumber (`bar "v"`)is given by,
`bar "v" = [1/(2)^2 - 1/"n"_"f"^2] (1.097 xx 10^7 "m"^(-1))`
Wave number (`bar "v"`) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, `bar "v"` has to be the smallest.
For (`bar "v"`) to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3, we get:
`bar "v" = (1.097xx10^7)[1/2^2 - 1/3^2]`
`bar "v" = (1.097 xx 10^7)[1/4 - 1/9]`
`= (1.097 xx 10^7) ((9-4)/36)`
`=(1.097 xx 10^7)(5/36)`
`bar "v"` = 1.5236 × 106 m–1
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