Deutrium was discovered in 1931by Harold Urey by measuring the small change in wavelength for a particular transition in ¹H and ² This is because, the wavelength of transition depends to a certain extent on the nuclear mass. If 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 μ , revolving around the nucleus at a distance equal to the electron-nucleus separation. Here μ = m_eM / ( m_e + M ) where M is the nuclear mass and m_e 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^-27kg , mass of  ¹H nucleus is 3.3374 * 10^-27kg 3.3374 * 10^-27 ?g , Mass of electron = 9.109 * 10^-31 kg ).

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- total energy of electron

E= $\frac{\mu Z^2{e}^4}{8{\epsilon }_o^2{h}^2}\frac{1}{n^2}$

The frequency hv= $\frac{\mu e^4}{8{\epsilon }_o^2{h}^2}\left(1-\frac{1}{4}\right)=\frac{\mu e^4}{8{\epsilon }_o^2{h}^2}\frac{3}{4}$

$?\lambda$ = ${\lambda }_D-{\lambda }_H$

$100*\frac{?\lambda }{{\lambda }_H}$ = $\frac{{\lambda }_D-{\lambda }_H}{{\lambda }_H}*100$ = $\frac{{\mu }_D-{\mu }_H}{{\mu }_H}*100$

= $\frac{\frac{m_{e{M}_D}}{m_e+M_H}-\frac{m_{e{M}_D}}{m_e-M_H}}{\frac{m_{e{M}_D}}{m_e+M_H}}*100$

When m_e<H<

after solving we get 2.714 $*{10}^{-2}\%$