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

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Numerical

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 (`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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Chapter 2: Structure of Atom - EXERCISES [Page 70]

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NCERT Chemistry Part 1 and 2 Class 11
Chapter 2 Structure of Atom
EXERCISES | Q 2.17 | Page 70
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