# Reduced Mass Proof

In a two-bodied system of hydrogen, an electron and a proton orbit each other about a shared center of mass. If I wanted to analyze the atomic motion of just the electron, the situation would become a one-body problem, and I would have to replace the mass of the electron with its corresponding reduced mass which is expressed in terms of the mass of the nucleus and the mass of the electron. In this proof I want to suppose that this reduced mass changes by a small amount $\Delta \mu$ when the electron jumps from the energy level with quantum number $n_{i}=3$ to the one with quantum number $n_{f} = 2$. From this I want to show that the wavelength changes by a corresponding amount $\Delta \lambda$ that approximately satisfies $\frac{\Delta \lambda}{\lambda} = -\frac{\Delta \mu}{\mu}$. In this situation, I am going to assume that $\Delta \mu$ is small in comparison with $\mu$.

I begin by stating that $\Delta \lambda \approx \frac{d\lambda}{d\mu} \Delta \mu$, which represents a change in wavelength. This is because in calculus $\Delta \mu$ and $d\mu$ are practically the same for small $\Delta$. I can now use the equation $\frac{1}{\lambda} = R(\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}})$ where $\lambda$ is the wavelength of an emitted photon during electronic transition from $E_{i}$ to $E_{f}$. $R$ is known as the Rydberg constant which is the physical constant relating to atomic spectra and each $n$ refers to a quantum number in a particular energy level. I can call everything on the right of this equation $b$ for the sake of simplification and write the formula as $\lambda = R^{-1}b^{-1}$, where $b$ represents everything that was on the left side fo the equation other than the Rydberg constant. Now, I can differentiate each side of the equation with respect to $\mu$, and because $R$ is a function of $\mu$ I must use the chain rule which looks as follows: $\frac{d\lambda}{d\mu} = -R^{-2}\frac{dR}{d\mu}b^{-1}$. I can now approximate $\frac{dR}{d\mu}$ as $\frac{R}{\mu}$. This will cause certain terms to cancel and I will be left with the formula $\frac{d\lambda}{d\mu} = -\frac{1}{R\mu b}$. Next, I can use the formula derived above, that $\lambda = \frac{1}{Rb}$, and plug this in for $R$ which will give me $\frac{d\lambda}{d\mu} = -\frac{\lambda}{\mu}$. I can then write the left side of the equation as $\frac{\Delta\lambda}{\Delta\mu}$ and rearrange terms to give me $\frac{\Delta\lambda}{\lambda} = -\frac{\Delta\mu}{\mu}$ which is what I wanted to show.