The wavelengths corresponding to the first four spectral lines of the Lyman series of the $H$-atom are $\lambda = 1218 \, \mathring{A}, 1028 \, \mathring{A}, 974.3 \, \mathring{A}$,and $951.4 \, \mathring{A}$. Now,consider a deuterium atom instead of a hydrogen atom. Given the mass of the hydrogen atom is $1.6725 \times 10^{-27} \, kg$,the mass of the deuterium atom is $3.3374 \times 10^{-27} \, kg$,and the mass of the electron is $9.109 \times 10^{-31} \, kg$,calculate the percentage change in the wavelength of the first spectral line of the Lyman series for the deuterium atom relative to the hydrogen atom.

(A) The wavelength of a spectral line is inversely proportional to the Rydberg constant $R$,which depends on the reduced mass $\mu = \frac{m_e M}{m_e + M}$.
Thus,$\lambda \propto \frac{1}{\mu} = \frac{m_e + M}{m_e M} = \frac{1}{M} + \frac{1}{m_e}$.
For hydrogen $(H)$ and deuterium $(D)$:
$\frac{\lambda_D}{\lambda_H} = \frac{\frac{1}{M_H} + \frac{1}{m_e}}{\frac{1}{M_D} + \frac{1}{m_e}} = \frac{M_D(m_e + M_H)}{M_H(m_e + M_D)}$.
Using the approximation $\lambda_D \approx \lambda_H \left(1 - \frac{m_e}{M_H} + \frac{m_e}{M_D}\right)$ or the exact ratio:
$\lambda_D = \lambda_H \left( \frac{1 + \frac{m_e}{M_H}}{1 + \frac{m_e}{M_D}} \right) \approx 1218 \left( 1 + \frac{9.109 \times 10^{-31}}{1.6725 \times 10^{-27}} - \frac{9.109 \times 10^{-31}}{3.3374 \times 10^{-27}} \right)$.
$\lambda_D \approx 1218 \left( 1 + 5.446 \times 10^{-4} - 2.729 \times 10^{-4} \right) = 1218 \left( 1 + 2.717 \times 10^{-4} \right) = 1218.33 \, \mathring{A}$.
Wait,the standard formula for reduced mass correction is $\lambda_D = \lambda_H \frac{\mu_H}{\mu_D}$.
$\frac{\mu_H}{\mu_D} = \frac{M_H(m_e + M_D)}{M_D(m_e + M_H)} \approx 1 - \frac{m_e}{M_H} + \frac{m_e}{M_D} \approx 1 - 2.717 \times 10^{-4}$.
$\lambda_D = 1218 \times (1 - 0.0002717) = 1217.669 \, \mathring{A}$.
Percentage change $= \frac{\lambda_H - \lambda_D}{\lambda_H} \times 100\% = \frac{1218 - 1217.669}{1218} \times 100\% \approx 0.0272\%$.