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

Explanation

In hydrogen atom, one electron (of mass `m_e`) revolves around one proton (of mass M) Reduced mass for hydrogen, `mu_(H)=(m_exxM)/(m_e+M)=(m_e)/((1+(m_e)/M)`
In deuterium, `._1D^2`, one electron ( of mass `m_e`) revolves around nucleus containing one proton and one neutron (of mass 2M).
`:.` Reduced mass for deuterium, `mu_(D)=(m_exx2M)/(2M+m_e)=(2M.m_e)/(2M(1+(m_e)/(2M)))=m_e(1-(m_e)/(2M))`
When an electron jumps form orbit j to orbit i, the frequency of radiation emitted
`hv_(ji)=(E_(j)-E_(i)) propmu` (reduced mass) `:. lambda_(ji) prop 1/mu`
If `lambda_(D)` is wavelength emitted in case of deuterium, and `lambda_(H)` is wavelength emitted in case of hydrogen,
then ` (lambda_(D))/(lambda_(H))=(mu_(H))/(mu_(D))=(m_e(1-(m_e)/M))/(m_e(1-(m_e)/(2M)))=(1-(m_e)/M)(1-(m_e)/(2M))^(-1)=(1-(m_e)/M)(1+(m_e)/(2M))`
`=(1-(m_e)/M+(m_e)/(2M))=(1-(m_e)/(2M))`
As `(m_e)/M=1/1840`, therefore `(lambda_(D))/(lambda_(H))=(1-1/(2xx1840))=0.99973`
`lambda_(D)=(0.99973)lambda_(H)`
Using `lambda_(H)=1218Å,1028Å,974.3Å and 951.4Å`, we get `lambda_(D)=1217.7Å,1027.7Å,974.04Å,951.1Å`.