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

Numerical

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

#### APPEARS IN

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