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Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

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#### Solution

For the Balmer series, n_{i} = 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, n_{f} should be minimum. For the Balmer series, a transition from n_{i} = 2 to n_{f} = 3 is allowed. Hence, taking n_{f} = 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 × 10^{6} m^{–1}

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[R_{H} = 1 × 10^{5} cm^{−1}, h = 6.6 × 10^{−34} Js, c = 3 × 10^{8} ms^{−1}]