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.
In hydrogen atom, one electron (of mass meme) revolves around one proton (of mass M) Reduced mass for hydrogen, μH=me×Mme+M=me(1+meM)μH=me×Mme+M=me(1+meM) In deuterium, .1D2.1D2, one electron ( of mass meme) revolves around nucleus containing one proton and one neutron (of mass 2M). ∴∴ Reduced mass for deuterium, μD=me×2M2M+me=2M.me2M(1+me2M)=me(1−me2M)μD=me×2M2M+me=2M.me2M(1+me2M)=me(1-me2M) When an electron jumps form orbit j to orbit i, the frequency of radiation emitted hvji=(Ej−Ei)∝μhvji=(Ej-Ei)∝μ (reduced mass) ∴λji∝1μ∴λji∝1μ If λDλD is wavelength emitted in case of deuterium, and λHλH is wavelength emitted in case of hydrogen, then λDλH=μHμD=me(1−meM)me(1−me2M)=(1−meM)(1−me2M)−1=(1−meM)(1+me2M)λDλH=μHμD=me(1-meM)me(1-me2M)=(1-meM)(1-me2M)-1=(1-meM)(1+me2M) =(1−meM+me2M)=(1−me2M)=(1-meM+me2M)=(1-me2M) As meM=11840meM=11840, therefore λDλH=(1−12×1840)=0.99973λDλH=(1-12×1840)=0.99973 λD=(0.99973)λHλD=(0.99973)λH Using λH=1218Å,1028Å,974.3Åand951.4ÅλH=1218Å,1028Å,974.3Åand951.4Å, we get λD=1217.7Å,1027.7Å,974.04Å,951.1ÅλD=1217.7Å,1027.7Å,974.04Å,951.1Å.
