Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in 1H and 2H. This is because the wavelength of transition depends to a certain extent on the nuclear mass. If the nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre of mass.

Such a system is equivalent to a single particle with a reduced mass $\mathrm{\mu }$, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here $\mathrm{\mu }={\mathrm{m}}_{\mathrm{e}}\mathrm{M}/\left({\mathrm{m}}_{\mathrm{e}}+\mathrm{M}\right)$, where M is the nuclear mass and me is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in 1H and 2H. (mass of 1H nucleus is $1.6725×{10}^{-27}$ kg, mass of 2H nucleus is $3.3374×{10}^{-27}$ kg, Mass of electron.)

Hint: The wavelength of the transition depends on the reduced mass.
Step 1: Find the wavelength of the transition for Hydrogen and Deuterium.
The total energy of the electron in the nth states of the hydrogen-like atom of atomic number Z is given by
where signs are usual and the $\mathrm{\mu }$ that occurs in the Bohr formula is the reduced mass of electron and proton.
Let ${\mathrm{\mu }}_{\mathrm{H}}$ be the reduced mass of hydrogen and ${\mathrm{\mu }}_{\mathrm{D}}$ that of Deuterium. Then, the frequency of the 1st Lyman line in hydrogen is:
${\mathrm{h\nu }}_{\mathrm{H}}=\frac{{\mathrm{\mu }}_{\mathrm{H}}{\mathrm{e}}^{4}}{8{\mathrm{\epsilon }}_{0}^{2}{\mathrm{h}}^{2}}\left(1-\frac{1}{4}\right)=\frac{{\mathrm{\mu }}_{\mathrm{H}}{\mathrm{e}}^{4}}{8{\mathrm{\epsilon }}_{0}^{2}{\mathrm{h}}^{2}}×\frac{3}{4}$
Thus, the wavelength of the transition is ${\mathrm{\lambda }}_{\mathrm{H}}=\frac{3}{4}\frac{{\mathrm{\mu }}_{\mathrm{H}}{\mathrm{e}}^{4}}{8{\mathrm{\epsilon }}_{0}^{2}{\mathrm{h}}^{3}\mathrm{c}}.$ The wavelength of the transition of the same line in Deuterium is ${\mathrm{\lambda }}_{\mathrm{D}}=\frac{3}{4}\frac{{\mathrm{\mu }}_{\mathrm{D}}{\mathrm{e}}^{4}}{8{\mathrm{\epsilon }}_{0}^{2}{\mathrm{h}}^{3}\mathrm{c}}$.
Step 2: Find the percentage difference.
Hence, the percentage difference is:
$100×\frac{∆\mathrm{\lambda }}{{\mathrm{\lambda }}_{\mathrm{H}}}=\frac{{\mathrm{\lambda }}_{\mathrm{D}}-{\mathrm{\lambda }}_{\mathrm{H}}}{{\mathrm{\lambda }}_{\mathrm{H}}}×100=\frac{{\mathrm{\mu }}_{\mathrm{D}}-{\mathrm{\mu }}_{\mathrm{H}}}{{\mathrm{\mu }}_{\mathrm{H}}}×100\phantom{\rule{0ex}{0ex}}=\frac{\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{M}}_{\mathrm{D}}}{\left({\mathrm{m}}_{\mathrm{e}}+{\mathrm{M}}_{\mathrm{D}}\right)}-\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{M}}_{\mathrm{H}}}{\left({\mathrm{m}}_{\mathrm{e}}-\mathrm{M}\right)}}{\frac{{\mathrm{m}}_{\mathrm{e}}{\mathrm{M}}_{\mathrm{H}}}{\left({\mathrm{m}}_{\mathrm{e}}+{\mathrm{M}}_{\mathrm{H}}\right)}}×100\phantom{\rule{0ex}{0ex}}=\left[\left(\frac{{\mathrm{m}}_{\mathrm{e}}+{\mathrm{M}}_{\mathrm{H}}}{{\mathrm{m}}_{\mathrm{e}}+{\mathrm{M}}_{\mathrm{D}}}\right)\frac{{\mathrm{M}}_{\mathrm{D}}}{{\mathrm{M}}_{\mathrm{H}}}-1\right]×100$