The first four spectral lines in the Lyman series of a 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.

We will first look at energy in the nth state of the electron of the atom of z atomic number,


where n is nth orbit


h is Planck’s constant,


m is reduced mass of electron and nucleus system


e is the charge of the electron


is the energy of the electron in nth orbit


Now consider the mass of proton to be mp and that of the electron be me


Reduced mass for hydrogen atom is given as follows,


------(1)


Reduced mass for deuterium atom is given as follows,


------(2)


Now energy difference,


∴ for same transition in hydrogen and deuterium atom we have,







On putting the value of mass of proton and electron in the above formula,


We get


∴ putting the values of wavelength in the above formula gives the corresponding values of wavelength in deuterium, which are nearly,


1217.7 Å, 1027.7 Å, 974.0223 Å, 951.130 Å


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