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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:$$mu=fracm_e,m_Nm_e+m_N$$
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 $rmH$ and deuterium $rmD$.

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 $frac11836$, and the fractional change of wavelength for deuterium is $frac13672$. The fractional change between these two is therefore also $frac13672=0.00027$. For the Balmer-$alpha$ line at $656$nm this becomes a shift of $0.18$nm.

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 $frac11836$"

Here is my attempt:

First calculating the reduced mass for $rmH$, noting that $m_p=1836,m_e$ then
$$mu=fracm_e ,m_pm_e+m_p=frac1836 ,m_e^2m_e+1836,m_e=frac18361837m_e$$

Now the Rydberg formula is $$frac1lambda=R_inftyleft(frac1n_1^2-frac1n_2^2right)tag1$$

Where $$R_infty=fracme^42hbar^2(4pi epsilon_0)^2frac1hctag2$$ is the Rydberg constant and is approximately $1.097373times 10^7rmm^-1$

Substituting $mto mu$ in $(2)$ gives a new value of the Rydberg constant, which I will call $R_H=1.096776times 10^7rmm^-1$

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

$$lambda=fracn_1^2 times n_2^2n_2^2-n_1^2timesfrac1R_infty=frac2^2 times 3^23^2-2^2timesfrac11.097373times 10^7=frac365timesfrac11.097373times 10^7$$
$$approx 656.1123701785992....,rmnm$$ as expected.

Now the shifted Balmer-$alpha$ line is given by $$frac365timesfrac1rmR_H=frac365timesfrac11.096776times 10^7$$
$$approx 656.4695069914002....,rmnm$$

The problem is that I unable to (or don't know how) show that the fractional change between the shifted and unshifted wavelength is $frac11836$ 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 $frac11836$?

Any hints or tips are greatly appreciated.

From equations (1) and (2), and substituting $m$ with $mu$, you can write $$lambda_mu=frac Cmu$$
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:$$fraclambda_H-lambda_Dlambda_H=fracfrac 1mu_H-frac 1mu_Dfrac 1mu_H=fracfracm_e+m_pm_em_p-fracm_e+2m_p2m_em_pfracm_e+m_pm_em_p=frac2(m_e+m_p)-(m_e+2m_p)2(m_e+m_p)=fracm_e2(m_e+m_p)approxfracm_e2m_p$$

Setting $m_N=km_e$, we have $mu=frac kk+1 m_e$. 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 $$k+1over klambda_0 = left(1+frac1kright)lambda_0.$$ That is, the fractional change in the wavelength due to the mass adjustment is equal to $1/k$.