Reason for spectral lines in hydrogen-like atoms

For a hydrogen-like atom, the energy levels are given more accurately by using the reduced mass $\mu$ of the electron-nucleus system:

$E_n = -\frac{\mu e^4}{8\varepsilon_0^2 h^2 n^2} \propto -\frac{\mu}{n^2}$

Hence the Rydberg constant for the atom becomes

$R = R_\infty \frac{\mu}{m_e}$

where

$\mu = \frac{m_e M}{m_e + M}$

and $M$ is the nuclear mass.

Compare hydrogen and deuterium

• For hydrogen, the nucleus is a proton of mass $M_p$.
• For deuterium, the nucleus is a deuteron of mass approximately $2M_p$.

Since the nuclear masses are different, the reduced mass $\mu$ is different in the two cases.

Effect on wavelengths

Since the spectral wavelengths depend on the Rydberg constant, and the Rydberg constant depends on reduced mass,

$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

a different nuclear mass $\Rightarrow$ different reduced mass $\Rightarrow$ slightly different $R$ $\Rightarrow$ slightly different wavelengths.

Check the options

• A: Size of nuclei are different — this is not the main reason for the shift in ordinary spectral lines. ❌
• B: Nuclear forces are different — irrelevant to electronic spectral lines. ❌
• C: Masses of the two nuclei are different — this changes reduced mass and hence wavelengths. ✅
• D: Attraction between electron and nucleus is different — both have nuclear charge $+e$, so Coulomb attraction is essentially the same. ❌

Conclusion

The spectral lines of deuterium differ slightly from hydrogen because the nuclear masses are different, which changes the reduced mass of the electron-nucleus system.