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.

Question

Philoid · Accepted Answer

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 m_p and that of the electron be m_e

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