Calculate the difference in wavelength of the Balmer-$\alpha$ line ($n = 3$ to $n = 2$) in hydrogen and deuterium

Question

In order to predict correctly the wavelengths of the hydrogen lines it is necessary to use in the expression for $R_{\infty}$ the reduced mass of the electron:$$\mu=\frac{m_e\,m_N}{m_e+m_N}$$ where $m_N$ is the mass of the nucleus.

Deuterium has a nuclear mass of approximately $2m_p$.

Calculate the difference in wavelength of the Balmer-$\alpha$ line ($n = 3$ to $n = 2$) in hydrogen $\rm{H}$ and deuterium $\rm{D}$.

The answer given to this question is

Since the mass of the proton is $1836$ times the mass of the electron, the fractional change in wavelength for hydrogen due to the use of the reduced mass is $\frac{1}{1836}$, and the fractional change of wavelength for deuterium is $\frac{1}{3672}$. The fractional change between these two is therefore also $\frac{1}{3672}=0.00027$. For the Balmer-$\alpha$ line at $656$nm this becomes a shift of $0.18$nm.

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Basically, I am struggling to understand how the author deduced that "the fractional change in wavelength for hydrogen due to the use of the reduced mass is $\frac{1}{1836}$"

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Here is my attempt:

First calculating the reduced mass for $\rm{H}$, noting that $m_p=1836\,m_e$ then $$\mu=\frac{m_e \,m_p}{m_e+m_p}=\frac{1836 \,{m_e}^2}{m_e+1836\,m_e}=\frac{1836}{1837}m_e$$

Now the Rydberg formula is $$\frac{1}{\lambda}=R_{\infty}\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)\tag{1}$$

Where $$R_{\infty}=\frac{me^4}{2\hbar^2(4\pi \epsilon_0)^2}\frac{1}{hc}\tag{2}$$ is the Rydberg constant and is approximately $1.097373\times 10^7\rm{m}^{-1}$

Substituting $m\to \mu$ in $(2)$ gives a new value of the Rydberg constant, which I will call $R_H=1.096776\times 10^7\rm{m}^{-1}$

From $(1)$ with $n_1=2$ and $n_2=3$, the unshifted Balmer-$\alpha$ line wavelength is given by

$$\lambda=\frac{{n_1}^2 \times {n_2}^2}{{n_2}^2-{n_1}^2}\times\frac{1}{R_{\infty}}=\frac{{2}^2 \times {3}^2}{{3}^2-{2}^2}\times\frac{1}{1.097373\times 10^7}=\frac{36}{5}\times\frac{1}{1.097373\times 10^7}$$ $$\approx 656.1123701785992....\,\rm{nm}$$ as expected.

Now the shifted Balmer-$\alpha$ line is given by $$\frac{36}{5}\times\frac{1}{\rm{R_H}}=\frac{36}{5}\times\frac{1}{1.096776\times 10^7}$$ $$\approx 656.4695069914002....\,\rm{nm}$$

The problem is that I unable to (or don't know how) show that the fractional change between the shifted and unshifted wavelength is $\frac{1}{1836}$ and because of this I feel that I am missing a much simpler way of doing this than what I tried above.

Could someone please explain how the fractional change was determined to be $\frac{1}{1836}$?

Any hints or tips are greatly appreciated.

Answer 1

From equations (1) and (2), and substituting $m$ with $\mu$, you can write $$\lambda_\mu=\frac C\mu$$ Here $C$ is a factor that contains $h$, $c$, $\epsilon_0$, and some numerical constants for the Balmer-$\alpha$ line. Then the fractional change is:$$\frac{\lambda_H-\lambda_D}{\lambda_H}=\frac{\frac 1{\mu_H}-\frac 1{\mu_D}}{\frac 1{\mu_H}}=\frac{\frac{m_e+m_p}{m_em_p}-\frac{m_e+2m_p}{2m_em_p}}{\frac{m_e+m_p}{m_em_p}}=\frac{2(m_e+m_p)-(m_e+2m_p)}{2(m_e+m_p)}=\frac{m_e}{2(m_e+m_p)}\approx\frac{m_e}{2m_p}$$

Answer 2

Setting $m_N=km_e$, we have $\mu=\frac k{k+1} m_e$. After suppressing irrelevant detail, we can see that $R$ is proportional to adjusted mass, hence the wavelength that corresponds to a particular transition is inversely proportional to the adjusted mass, so if we let $\lambda_0$ be the unadjusted wavelength, then the adjusted wavelength is simply $${k+1\over k}\lambda_0 = \left(1+\frac1k\right)\lambda_0.$$ That is, the fractional change in the wavelength due to the mass adjustment is equal to $1/k$.