In my quantum mechanics textbook, it claims that the Balmer series between hydrogen and deuterium is different. However, I was under the impression that the Balmer series
$$H_\alpha, H_\beta, H_\gamma$$ is related by the equation $$\lambda=C\frac{n^2}{n^2-4}$$ where $$ C=3646 \mathring{\text{A}} $$ $$n=3,4,5$$
Is there a equation relating the mass and the Balmer series?
Any hint would be appreciated
The full quantum analysis of the hydrogen atom is a quantum two body problem, however, one of those bodies is extremely massive compared to the other, so that this problem, as a first approximation, is analysed by solving either the first quantised (i.e. for one quantum particle in a classical environment) Schrödinger or Dirac equations for inverse square potential relative to a fixed central point. As you are likely aware, in reality the central point is not fixed and the full solution requires both charged nucleus and electron to be treated quantum mechanically. So clearly the energy eigenvalues for this two body system will be different from those derived from the first quantised treatment.
A first-order correction to the simplifying assumption is to use a reduced mass for the electron as described in Ben's comment:
$$\mu = \frac{m_e\,M_N}{m_e+M_N}$$
where $m_e$ is the true electron mass and $M_N$ the mass of the nucleus. As you can see, since the ratio of protons electron mass is 1836 plus a bit, $\mu$ is less than $m_e$ by the percentage of approximately $100 m_e / M_n$, or about 0.2%. If we replace $M_N$ by the mass of the deuteron, this difference shrinks by a half, i.e. the reduction is now 0.1%.
The paper "Reduced Dirac Equation And Lamb Shift As An Off-mass-shell Effect In Quantum Electrodynamics" by Ni Guang-jiong, Xu Jianjun and Lou Senyue discusses the validity of the reduced mass for the Dirac equation.
However, hopefully can now see what to do for a simple first approximation. Your constant $C = 4/R_H$, where $R_H$ is the Rydberg constant:
$$R_\infty = \frac{4}{C} = \frac{m_e q^4}{8 {\epsilon_0}^2 h^3 c}$$
but now you use the reduced mass $\mu$ instead of $m_e$. More generally, for any line, you do the same with the hydrogen atom Dirac equation general solution given either in the paper above or in the "Mathematical Summary of the eigenstates" section in the Hydrogen atom Wiki page.
Your problem is a very simple illustration of how the assertion that "chemistry is independent of all nuclear properties aside from the atomic number" begins to break down. Isotope masses all influence the electron eigenstates, and this influence is strongest with the lightest elements. In particular, a slightly kooky fact that I like keeping in mind is that heavy water will not support "higher" life. The hydrogen-oxygen bond energies depend on whether the hydrogen in the water is protium or deuterium, and this difference influences biochemistry ultimately lethally for eukaryotes and multicelled eukaryotic organisms. In particular, the workings of the mitotic spindle and other eukaryotic sexual reproductive structures that are strongly dependent on hydrogen bonding, in turn affected by the OH bond energy, can be shut down. See the Heavy Water Wiki Page: "Effect on biological systems" section. Many prokaryotes (asexual bacteria and archaeans), however, thrive in heavy water unhindered.
$$ \frac{1}{\lambda} = \frac{4}{C_M}\left(\frac{1}{4} - \frac{1}{n^2}\right) = R_M\left(\frac{1}{4} - \frac{1}{n^2}\right), $$ where $R_M = 4/C_M$ is the Rydberg constant for the particular atom: $$ R_M = R_\infty\left(1+\frac{m_\text{e}}{M}\right)^{-1}, $$ with $m_\text{e}$ the electron mass, $M$ the mass of the atomic nucleus and $$ R_\infty = 1.0973\,731\,568\,539\times 10^7\;\text{m}^{-1}. $$ For Hydrogen I get, $$ \begin{align} R_H &= R_\infty\left(1 + \frac{5.4857990943\times 10^{-4}\;\text{u}}{1.007276466812\;\text{u}}\right)^{-1},\\ &= 1.0967\,758\,341\times 10^7\;\text{m}^{-1},\\[2mm] C_H &= 4/R_H = 364.70534\;\text{nm}, \end{align} $$ and for Deuterium I get, $$ \begin{align} R_D &= R_\infty\left(1 + \frac{5.4857990943\times 10^{-4}\;\text{u}}{2.013553212724\;\text{u}}\right)^{-1},\\ &= 1.0970\,742\,659\times 10^7\;\text{m}^{-1},\\[2mm] C_D &= 4/R_D = 364.60613\;\text{nm}. \end{align} $$ Note: I haven't verified these numbers.
