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Calculate the Wave Number for the Longest Wavelength Transition in the Balmer Series of Atomic Hydrogen. - Chemistry

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, ni = 2. Thus, the expression of wavenumber (`barv`)is given by,

`barv = [1/(2)^2 - 1/n_f^2] (1.097 xx 10^7 m^(-1))`

Wave number (`barv`) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, `barv` has to be the smallest.

For (`barv`) 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:

`barv = (1.097xx10^7)[1/2^2 - 1/3^2]`

`barv = (1.097 xx 10^7)[1/4 - 1/9]`

`= (1.097 xx 10^7) ((9-4)/36)`

`=(1.097 xx 10^7)(5/36)`

`barv` =  1.5236 × 106 m–1

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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 2 Structure of Atom
Q 17 | Page 66
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