The first four spectral lines in the Lyman series

Question:

The first four spectral lines in the Lyman series of an H-atom are λ = 1218 Å, 1028Å, 974.3 Å and 951.4Å. If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

Solution:

Let μH and μD be the reduced masses of electrons of hydrogen and deuterium respectively.

ni and nf are the fixed mass series for hydrogen and deuterium

Rh/Rd = μH/ μD

Reduced mass of hydrogen = μH = me/1+me/M = me (1-me/M)

Reduced mass of deuterium = μD = 2M.me/2M(1+me/2M) = me(1- me/2M)

μH/ μD = 0.99973

λD/ λH = 0.99973

λD = 0.99973 λH

Therefore, change in wavelength = λH – λD = 0.3 Å

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