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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