The first four spectral lines in the Lyman series of a $H$ atom are $\lambda = 1218 \, \mathring{A}, 1028 \, \mathring{A}, 974.3 \, \mathring{A}, 951.4 \, \mathring{A}$. If instead of Hydrogen,we consider Deuterium,calculate the shift in the wavelength of these lines.

(A) The Rydberg constant is given by $R = \frac{\mu e^4}{8 \epsilon_0^2 h^3 c}$,where $\mu$ is the reduced mass of the electron-nucleus system,$\mu = \frac{M m_e}{M + m_e}$.
For Hydrogen $(H)$,$M_H \approx 1836 m_e$,so $\mu_H = \frac{1836 m_e^2}{1837 m_e} \approx 0.99945 m_e$.
For Deuterium $(D)$,$M_D \approx 3670 m_e$,so $\mu_D = \frac{3670 m_e^2}{3671 m_e} \approx 0.99973 m_e$.
Since $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,we have $\lambda \propto \frac{1}{R} \propto \frac{1}{\mu}$.
Thus,$\frac{\lambda_D}{\lambda_H} = \frac{\mu_H}{\mu_D} \approx \frac{0.99945}{0.99973} \approx 0.99972$.
The shift is $\Delta \lambda = \lambda_H - \lambda_D = \lambda_H (1 - 0.99972) = 0.00028 \lambda_H$.