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

For the Balmer series, ni = 2. Thus, the expression of wavenumber is given by,

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

For 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:

= 1.5236 × 106 m–1

  • 105
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