# I Relevance of the Rydberg Constant?

#### neilparker62

Homework Helper
Summary
Summary: Question about the Rydberg Constant.

The Rydberg constant is essentially a 'relic' of a now obsolete model (the Bohr model) of the Hydrogen atom. So my question is why do scientists keep trying to measure it to greater and greater degrees of precision?

Also I thought that the Rydberg constant / Rydberg energy would relate directly to accurately obtained values of atomic hydrogen ionization energy. But this does not seem to be the case. So what 'correction' needs to be applied to the Rydberg constant (adjusted for Hydrogen) such that it agrees with an accurately obtained value of ionization energy for the same atom ? Why do we keep it , if it is no longer directly reflective of the physical quantity it was designed to determine (via calculation of 1 - infinity transition) ?

RH = 109677.576 cm-1.
Ei(H) = 109678.7717426 (10) cm−1.

I know my RH is showing way less significant figures than for Ei(H) but the difference is already evident in the units column.

Formulations of the Dirac energy equation I have seen appear to depend only upon mc^2 and the fine structure constant although elsewhere it appears there are more complex versions which somehow 'reincorporate' the Rydberg constant. Why ?

Disclaimer: I have no idea what's going on in the Maths per above link - was just interested in the formula at the end of it!

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#### Anand Sivaram

• neilparker62

#### neilparker62

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Thanks for the refs. I had a look.

I know about the use of the Rydberg formula in calculating spectral lines. Point is there is now a more accurate formula for calculating spectral lines. This formula doesn't seem to depend upon the Rydberg constant .

Don't know if small corrections such as for Lamb shift and/ or recoil energy rely on Rydberg constant ?

#### neilparker62

Homework Helper
Explicitly, here is a calculation of the ionization energy (as frequency) of atomic hydrogen. The +- 8GHz correction is for so called "Ground State Lamb shift" although there appears to be some debate in the literature as to the exact composition of "Ground State Lamb shift". I have omitted a smaller correction for recoil energy.

As can be seen, the Rydberg constant is not used at all (unless perhaps in a determination of the Lamb shift correction ?). So my question remains "Quo vadis Rydberg Constant" ?? What does the 'super accurate' value of this constant represent if 'divorced' from it's original physical meaning as the limit of the Hydrogen Lyman series transitions?

• Anand Sivaram

#### ZapperZ

Staff Emeritus
2018 Award
The use of Rydberg constant and the modelling of Rydberg atom are extremely useful in dealing with excitons.

Zz.

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#### Cthugha

Explicitly, here is a calculation of the ionization energy (as frequency) of atomic hydrogen. The +- 8GHz correction is for so called "Ground State Lamb shift" although there appears to be some debate in the literature as to the exact composition of "Ground State Lamb shift". I have omitted a smaller correction for recoil energy.

As can be seen, the Rydberg constant is not used at all (unless perhaps in a determination of the Lamb shift correction ?).
Besides the excellent remarks by ZapperZ: Your statement is incorrect. The calculation you quote contains the Rydberg constant modified by the correction for the finite proton mass.

Neglecting the 8 GHz shift, the calculation you posted yields:
$$f_{Ryd,c}=\frac{m_p m_e}{m_p+m_e} c^2 \frac{1-\sqrt{1-\alpha^2}}{h},$$
while the standard Rydberg frequency is given by the Rydberg constant times c, which yields
$$f_{Ryd}=m_e c^2 \frac{\alpha^2}{2 h},$$

The change in the mass section is just the difference between a system considering a single electron and a positive charge of infinitely heavy mass (which is what the Rydberg constant is for) and a system, where the positive charge has finite mass. If you let the proton mass go to infinity, the first term tends towards $m_e$ again. For the second term containing the fine structure constant, you can simply calculate the two values and will see that they are pretty much the same. The above equation just introduces a tiny correction to the initial value.

The Rydberg constant is present in the equation you posted. The equation just adds some small corrections.

#### neilparker62

Homework Helper
The Rydberg constant is present in the equation you posted. The equation just adds some small corrections.
The devil is in the detail: $$E_\text{k} = m \gamma c^2 - m c^2 = \frac{m c^2}{\sqrt{1 - v^2/c^2}} - m c^2$$
At a low speed v<<c relativistic kinetic energy is approximated well by the classical kinetic energy. This is done by binomial approximation or by taking the first two terms of the Taylor expansion for the reciprocal square root: $$E_\text{k} \approx m c^2 \left(1 + \frac{1}{2} v^2/c^2\right) - m c^2 = \frac{1}{2} m v^2$$ (Hope Wikipedia will forgive the copy/paste from their article on kinetic energy!)

Similarly here we may write: $$f= μc^2 \frac{1-\sqrt{1-\alpha^2}}{h} \approx μc^2 \frac{1-(1-\alpha^2/2)}{h} = \frac{½μ\alpha^2c^2}{h}$$

In both cases I think fair to say (for the binomial approximations) "close but no cigar". And of course there is a world of difference as to where the relativistic equations are coming from. As a final note Bohr himself introduced the correction for reduced mass which applies equally to both equations.

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#### Cthugha

I do not see your point. The "world of difference" is between considering the relativistic corrections and neglecting them. The approach using the Dirac equation of course automatically includes the relativistic equations. However, it also involves the "standard" Rydberg constant (which only contains other natural constants, which makes it extremely interesting) times a correction function.
See e.g. section 1.2. of the following thesis:
https://edoc.ub.uni-muenchen.de/21102/1/Beyer_Axel.pdf

In that section you also find the reason why the precision of measuring the Rydberg constant and its relativistic corrections is pushed to its limits nowadays: The correction depends on the size of the proton, which is a not-so-well-known natural constant. Accordingly, these measurements also open up a way to determine this size, which has been subject of lots of discussions with respect to the so-called proton size puzzle recently (https://www.nature.com/articles/nature09250 , https://science.sciencemag.org/content/358/6359/79). The articles are behind a paywall, but mostly covered by the thesis I posted, which is also a good summary on the recent renewed interest in the Rydberg constant.

#### neilparker62

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I do not see your point. The "world of difference" is between considering the relativistic corrections and neglecting them. The approach using the Dirac equation of course automatically includes the relativistic equations. However, it also involves the "standard" Rydberg constant (which only contains other natural constants, which makes it extremely interesting) times a correction function.
See e.g. section 1.2. of the following thesis:
https://edoc.ub.uni-muenchen.de/21102/1/Beyer_Axel.pdf
Thanks for the various references in your post - was aware of the 'proton radius puzzle' albeit only in a peripheral way from reading the literature since I have neither the Maths nor Physics to understand it in detail.

However solving for r and v as is done in the Bohr model, we obtain v=αc for the electron and therefore the simplest interpretation of Rydberg energy is just:
$$E_k=½mv^2=½m_e{(\alpha c)}^2$$
ie: Kinetic energy of a planetary model electron revolving around the nucleus - infinite mass assumed.

It strikes me as a little incongruous that despite the sophistication of relativity and quantum electrodynamics, all we seem to be doing is tacking on various 'corrections' to this neo-classical model. Somewhat akin to saying that actually ½mv^2 is "fundamental" and all we need do is tack on various "corrections" accordingly ?!

The "world of difference" I referred to is in regard to the theoretical underpinnings of relativity as compared to classical/Newtonian. The actual differences may indeed be very small for v<<c but significant enough in the Hydrogen atom to improve accuracy (as compared to the Bohr model) by orders of magnitude.

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#### ZapperZ

Staff Emeritus
2018 Award
But in your first post, your main question was not asking about the accuracy of the Rydberg constant, but "... Why do we keep it , if it is no longer directly reflective of the physical quantity it was designed to determine (via calculation of 1 - infinity transition) ? ..."

I thought I have given you sufficient references on its use in excitons. Are you still questioning its usefulness, or has that question been sufficiently addressed? If it has, then I do not know why you are still questioning its existence.

Zz.

#### neilparker62

Homework Helper
Many thanks for the very interesting introduction to 'excitons' - not something I had come across previously I have to admit.

I don't question the usefulness of the Rydberg constant any more than I do the general usefulness of the Rydberg formula for the Hydrogen spectrum - and atomic spectra generally. Or for that matter the various formulae we get from Newtonian mechanics. They don't suddenly 'fail' on account of more accurate formulae being discovered. And we do expect - indeed demand - of such formulae that they somehow encompass older theories. In this case as can be seen via the use of binomial approximations.

Apologies for the use of the word 'obsolete' in respect of older theory - not at all the right word to use!

#### Cthugha

Well, yes the Rydberg constant indeed is pretty universal. If you consider the field of Rydberg physics, you will find that hydrogen itself is of little interest and people study other elements. However, in every atom beyond hydrogen, spherical symmetry is absent. You have some screening due to inner electrons and other stuff. Taking the full relativistic approach for anything beyond hydrogen-like atoms is pretty hopeless.

Still, what people do is to introduce a quantum defect. A simple angular momentum-dependent correction to the standard Rydberg formula that just shifts the effective quantum numbers by a constant amount and effectively maps the problem to the hydrogen problem using the bare Rydberg constant and gives you pretty correct results. It is just much more efficient to look up this value in a table than to have someone experienced in theory dig through the whole problem for several years just to find out that it cannot be solved exactly. And the nice thing is that this strategy works pretty much universally. If you have a look at the first Rydberg exciton paper linked by ZapperZ above, you will find that the authors also introduce a quantum defect for excitons. The authors do so for completely different physical reasons. Instead of screening the authors need to consider central cell corrections and non-parabolicity of the involved bands that cause the deviations, but still the formalism based on the standard Rydberg constant (modified by the different values taking the dielectric surroundings into account) works well for us.

• SpinFlop

#### neilparker62

Homework Helper
However, it also involves the "standard" Rydberg constant (which only contains other natural constants, which makes it extremely interesting) times a correction function.
Seems to depend upon what thesis and/or research paper you are reading. See for example Equation 2.3 on page 6 of Parthey's thesis on the 1s-2s transition.

Deploying this equation (more or less) for calculation of ionization energy (atomic hydrogen) gives a result which differs from the value in the NIST database (3.2880868571276 E15 Hz) by about 8.172 GHz , a value in good agreement with the last known value of "Ground State Lamb Shift" (8.172874 GHz) before the latter seemingly disappeared off the theoretical landscape altogether. I have no idea why but I am told it was something to do with the fact that scientists couldn't quite agree on it's definition as anything that was not covered by Dirac energy (with reduced mass corrections) and hyperfine splitting.

#### neilparker62

Homework Helper
But in your first post, your main question was not asking about the accuracy of the Rydberg constant, but "... Why do we keep it , if it is no longer directly reflective of the physical quantity it was designed to determine (via calculation of 1 - infinity transition) ? ..."

I thought I have given you sufficient references on its use in excitons. Are you still questioning its usefulness, or has that question been sufficiently addressed? If it has, then I do not know why you are still questioning its existence.

Zz.
Well I read the exciton paper with interest and played around quite a bit with the data set provided - the absorption peaks recorded do indeed fit a 'Rydberg type' model with some degree of precision. As indeed is the case with the Hydrogen spectrum itself. I actually did a linear regression between the exciton data and the Hydrogen Lymen series - it fitted with a correlation of 0.99989.

However the model used does not employ "the" Rydberg constant - it simply fits the data to an equation of the form: $$E_n=E_g-\frac{Ry}{n^2}$$ and from the fit obtains statistical values for $E_g$ and $R_y$. Then refines the model via the use of 'quantum defects' (which I won't pretend to understand at all).

At some level of precision in the hydrogen spectrum, the Rydberg model proved insufficient and was superceded by Dirac's equations (if I understand the history correctly). Naturally we do expect the Rydberg formula to constitute an accurate approximation of Dirac energy just as we do Newtonian kinetic energy vs relativistic kinetic energy.

#### Cthugha

To be honest, your last few posts were a bit obscure and I am not sure I understood your points adequately. But let me get at least the obvious things right.

However the model used does not employ "the" Rydberg constant - it simply fits the data to an equation of the form: $$E_n=E_g-\frac{Ry}{n^2}$$ and from the fit obtains statistical values for $E_g$ and $R_y$. Then refines the model via the use of 'quantum defects' (which I won't pretend to understand at all).
Of course the model employs "the" Rydberg constant. We are discussing semiconductor physics here, so the Coulomb problem must be considered with respect to the semiconductor surroundings. First, when deriving the energies of bound states, one must consider that the effective mass of excitons is different from the mass of an electron. The energy of the bound states scales linearly with the mass ratio. Also, of course Cuprous Oxide is a dielectric. Accordingly, you need to take screening into account. The Rydberg energy scales with the inverse dielectric constant squared of the material. For Cu2O the reduced mass is about 0.365 times the bare electron mass and the dielectric constant is about 7.5 for the wavelength range of interest. You can calculate yourself, what to expect for excitons in Cu2O starting from the 13.6 eV binding energy known from the hydrogen atom.

Quantum defects are used to account for problems that arise in problems, which do not have the same symmetry as the hydrogen problem. Which is...every problem besides hydrogen. For atoms, you get screening due to the other charges present. For excitons you get non-parabolicity of the bands and the presence of the lattice, which breaks rotational symmetry. The latter for example results in a mixing of P- and F-excitons in this material system for principal quantum numbers of 4 or higher.

At some level of precision in the hydrogen spectrum, the Rydberg model proved insufficient and was superceded by Dirac's equations (if I understand the history correctly). Naturally we do expect the Rydberg formula to constitute an accurate approximation of Dirac energy just as we do Newtonian kinetic energy vs relativistic kinetic energy.
I still have no idea what your point is. So let me state it again in simple words. The Rydberg constant is not the Rydberg model. There are two reasons, why people are interested in the Rydberg energy.

First: precision measurements of natural constants:
The Rydberg constant is simply$\frac{m_e c \alpha^2}{2 h}$. That is it. It is just a composition of other natural constants. And as the Rydberg energy can be measured quite precisely, it is also a reasonable way to measure the value of the fine structure constant as well. And of course people want to measure this value precisely. The easiest way to measure this constant is to measure several hydrogen-like systems and several hydrogen transitions. And yes: the relativistic corrections are important. However, in order to get to know the magnitude of these corrections, one must know the value of the Rydberg constant as well. Then, one for example may also derive a rather precise value for the proton radius. However, in order to be able to do that, both the Rydberg energy and the corrections need to be known. Knowing just the energy levels of the hydrogen atom is not sufficient here. And accordingly, pretty much nobody doing precision measurements of the hydrogen spectrum is doing that for the sake of hydrogen. The spectrum itself is somewhat irrelevant in this respect. Therefore, also the Rydberg model is not really of interest is. But the Rydberg energy is.

Second: Hydrogen-like spectra:
Of course people may also be interested inhydrogen-like spectra. For example this is of interest in the exciton systems mentioned above. However, here any relativistic corrections are of little interest. For the excitons mentioned above, they are so sensitive to external influences that already the presence of about one free charge per cubic micrometer (which is very little - usually the number of defects or surface charges is higher) will wreak havoc compared to the relativistic corrections. Here, the bare Rydberg energy adapted to the dielectric is fully sufficient. Potential relativistic corrections are tiny compared to other perturbations, anyway. So nobody would bother to do the full math in this case.

Of course the aim of precision measurements is always point 1. And here your assumption that the meaning of the Rydberg energy is to deduce the transition energies of hydrogen is simply inappropriate.

• DrClaude

#### ZapperZ

Staff Emeritus
2018 Award
Well I read the exciton paper with interest and played around quite a bit with the data set provided - the absorption peaks recorded do indeed fit a 'Rydberg type' model with some degree of precision. As indeed is the case with the Hydrogen spectrum itself. I actually did a linear regression between the exciton data and the Hydrogen Lymen series - it fitted with a correlation of 0.99989.

However the model used does not employ "the" Rydberg constant - it simply fits the data to an equation of the form: $$E_n=E_g-\frac{Ry}{n^2}$$ and from the fit obtains statistical values for $E_g$ and $R_y$. Then refines the model via the use of 'quantum defects' (which I won't pretend to understand at all).

At some level of precision in the hydrogen spectrum, the Rydberg model proved insufficient and was superceded by Dirac's equations (if I understand the history correctly). Naturally we do expect the Rydberg formula to constitute an accurate approximation of Dirac energy just as we do Newtonian kinetic energy vs relativistic kinetic energy.
But your initial question was on the Rydberg CONSTANT, not the Rydberg model. You asked for its purpose and why we continue to refine the Rydberg constant.

Zz.

#### neilparker62

Homework Helper
First: precision measurements of natural constants:
The Rydberg constant is simply$\frac{m_e c \alpha^2}{2 h}$. That is it. It is just a composition of other natural constants. And as the Rydberg energy can be measured quite precisely, it is also a reasonable way to measure the value of the fine structure constant as well.
For the sake of clarity , what exactly do we measure when we measure "Rydberg energy"? Originally it was meant to be ionization energy of Hydrogen. But then it was realized that the nucleus has a finite mass - hence we now need to introduce Rh. Then there are relativistic issues not to mention Lamb shift. So the Rydberg energy (if indeed it is still supposed to be ionization energy of hydrogen) is a completely different 'animal' from what it started out as. So for that matter is Rh.

You have said that the Rydberg energy is a compilation of constants. For me each of those constants makes perfect sense in its own right especially the fine structure constant. I think that in essence the fine structure constant is the 'relativistic successor' to the Rydberg constant and our efforts should centre on finding techniques of measuring it independently rather than 'extracting' from a somewhat undefined "Rydberg energy". My Physics text book eulogises the Rydberg constant as this wonderful combination of so many different natural constants. Well big deal - does it actually offer anything substantive that all the others in combination do not ??

Thankyou very much all the same for your very comprehensive responses and insights into the exciton model. And many thanks to Zapper for the same although I am a bit puzzled by his last response: yes my question is indeed to do with the Rydberg constant (and why we would want to keep improving its accuracy) but the exciton model is an exemplary illustration of the Rydberg model which doesn't really address the issue of the Rydberg constant per se.

#### Cthugha

For the sake of clarity , what exactly do we measure when we measure "Rydberg energy"? Originally it was meant to be ionization energy of Hydrogen. But then it was realized that the nucleus has a finite mass - hence we now need to introduce Rh. Then there are relativistic issues not to mention Lamb shift. So the Rydberg energy (if indeed it is still supposed to be ionization energy of hydrogen) is a completely different 'animal' from what it started out as. So for that matter is Rh.
I do not want to repeat, but: the Rydberg energy is the binding energy of an electron bound to an infinitely heavy mass in the absence of QED effects or $\frac{m_e c\alpha^2}{2h}$, which is the same. It has never been and will never be the binding energy of hydrogen. It is sometimes a good approximation for that, but it is much more important than the binding energy of hydrogen will ever be. People do not measure the Rydberg constant in order to know the hydrogen lines more precisely. People measure the hydrogen lines because they want to know the Rydberg constant.

What you measure in real experiments in order to determine it is a combination of several transitions of hydrogen-like systems. For example 1S-2S and 2S-4P transitions of hydrogen. Or similar transitions for muonic hydrogen. Or Uranium 91+. Now you measure several dozens of these lines for different orbital angular momenta, masses or principal quantum numbers and compare this to the the predictions of the Dirac equation. However, just as the standard model, you need several experimentally determined parameters as an input to the equation. These are the electron mass and the electromagnetic interaction constant (or fine structure constant). Now you will find that several transitions show a scaling that depend in a very systematic manner on a combination of these constants - which is the Rydberg energy. So from these transitions you can deduce the Rydberg energy quite precisely. Now the Rydberg constant only contains exactly the two variables mentioned before and defined constants (h and c have fixed values by definition). So, when trying to fix the input parameters for the Dirac equation, this is now like a joker. If you are in addition able to measure the fine structure constant precisely, you do not have to measure the electron mass anymore as you can calculate it from Rydberg energy and the fine structure constant. If you are in addition able to measure the electron mass precisely, you do not have to do a precision measurement of the fine structure constant anymore as you can calculate it from the Rydberg energy and the electron mass. And as already pointed out before several times: You can measure the Rydberg energy VERY precisely.

As it is known so well, it also allows you to determine the quantities that determine the corrections quite precisely. There is a correction due to the spatial extension of the nucleus. Measuring the energies of the associated transition gives you the size of the nucleus because the Rydberg energy is so much better known than these quantities.

You have said that the Rydberg energy is a compilation of constants. For me each of those constants makes perfect sense in its own right especially the fine structure constant. I think that in essence the fine structure constant is the 'relativistic successor' to the Rydberg constant and our efforts should centre on finding techniques of measuring it independently rather than 'extracting' from a somewhat undefined "Rydberg energy".
That is a rather questionable understanding of how metrology works. For the determination of each important constant, you need several others. For example, the electron mass was not measured directly, but the mass-to-charge ratio was measured first, followed by the charge. Nowadays, you get it from the Rydberg energy and the fine structure constant.

The Rydberg constant is known to about 0.006-ppb accuracy and is thus one of the best-known constants out there. Of course people use constants they can measure precisely to determine the constants that are known less precisely. What else should they do?

My Physics text book eulogises the Rydberg constant as this wonderful combination of so many different natural constants. Well big deal - does it actually offer anything substantive that all the others in combination do not ??
Yes, it can be measured. Very precisely.

• #### neilparker62

Homework Helper
I do not want to repeat, but: the Rydberg energy is the binding energy of an electron bound to an infinitely heavy mass in the absence of QED effects or mecα22hmecα22h\frac{m_e c\alpha^2}{2h}, which is the same. It has never been and will never be the binding energy of hydrogen. It is sometimes a good approximation for that, but it is much more important than the binding energy of hydrogen will ever be. People do not measure the Rydberg constant in order to know the hydrogen lines more precisely. People measure the hydrogen lines because they want to know the Rydberg constant.
As I understand the scientific method, we measure some or other aspect of physical reality and then compare the result to whatever theory we advance to describe that particular phenomenon. It therefore makes perfect sense to try and measure the binding energy of hydrogen which at least has the merit of having some basis in physical reality. The quantity described above - by its very definition - does not. That's why I am asking what do we actually measure when we measure the Rydberg constant ?

It is an obvious - and I believe at the time intended - result of the Rydberg formula that the Lymen series limit (ionization energy of hydrogen) corresponds to simply "R" in the formula. When - due to scientific progression - it became apparent this was not the case I don't see how this constant can subsequently assume a meaning far beyond what was intended (or even foreseen) in the formula with which it is associated.

It appears that the way these various progressions are viewed are as if they were layers of onion peel on top of the non relativistic Rydberg model. And if we can somehow quantify and hence strip away QED corrections, relativistic 'corrections' , finite nuclear size corrections etc etc, then at the heart of everything we will find the 'pristine' Rydberg model and hence extract the Rydberg constant.

I would guess that what we actually measure is in fact just electron mass wrapped up in various other constants - this would be in line with your description of the Rydberg constant as the 'joker' which can substitute for either electron mass or fine structure constant as required. Actually I tried it out - made a tiny difference (in the right direction!) to the calculation I posted previously.

#### Cthugha

As I understand the scientific method, we measure some or other aspect of physical reality and then compare the result to whatever theory we advance to describe that particular phenomenon. It therefore makes perfect sense to try and measure the binding energy of hydrogen which at least has the merit of having some basis in physical reality. The quantity described above - by its very definition - does not. That's why I am asking what do we actually measure when we measure the Rydberg constant ?
You can measure it, but it depends on plenty of quantities. You need to distribute the whole measured energy somehow between the bare Rydberg energy, the relativistic corrections, QED correction, finite nucleus corrections and so on and so forth. Also I strongly oppose to calling something THE binding energy of hydrogen. There are an enormous number of states in the hydrogen system. All of them have their own binding energy.

It appears that the way these various progressions are viewed are as if they were layers of onion peel on top of the non relativistic Rydberg model. And if we can somehow quantify and hence strip away QED corrections, relativistic 'corrections' , finite nuclear size corrections etc etc, then at the heart of everything we will find the 'pristine' Rydberg model and hence extract the Rydberg constant.
Well, of course you can do that. This is what Rydberg physics does. Some of the most precise measurements of the Rydberg constant for example may be done using circular Rydberg states (https://journals.aps.org/pra/abstract/10.1103/PhysRevA.96.032513). You can get rid of QED corrections and the like simply by going to states, where they are negligible. If you go to principal quantum numbers on the order of above 100 and states of the highest OAM possible, you end up with states where the corrections are as close to 0 as it gets. These states can be realized well in the lab.

I would guess that what we actually measure is in fact just electron mass wrapped up in various other constants - this would be in line with your description of the Rydberg constant as the 'joker' which can substitute for either electron mass or fine structure constant as required. Actually I tried it out - made a tiny difference (in the right direction!) to the calculation I posted previously.
One of the important things in understanding metrology is that there is no more fundamental wrapped up constant. If $\alpha$ and $m_e$ are your quantities of interest, it does not matter whether you measure them directly or whether you measure $\alpha$ and $m_e^4 \alpha^{17}$. This is not different from picking a suitable basis to describe 2D-space. You may pick $\pmatrix{1\\0}$ and $\pmatrix{0\\1}$ as a set of basis vectors or $\pmatrix{1\\1}$ and $\pmatrix{1\\-1}$ or any similar combination that suits the problem. Here, it is similar. You choose the basic quantities to measure according to whatever can be measured precisely.

Staff Emeritus
If you look at the hydrogen spectrum, it tells you what 0.9997R is. Singly-ionized helium tells you what 3.9995R is, and so on. Why does the fact that no atom has energy levels lined up exactly at 1.0000R bother you?

#### neilparker62

Homework Helper
At just 4 decimal places, doesn't bother me at all. The Bohr model (Rydberg formula) is accurate to about 6 decimal places I believe.

Staff Emeritus
But why does it matter that no energy level is an integer multiple of the Rydberg? They aren't integer numbers of eV either.

#### neilparker62

Homework Helper
The constant in front of the Rydberg formula is the Rydberg constant (with an adjustment for finite nuclear mass). The constant in front of the Dirac energy expression is (quite naturally I would say) $m_ec^2$, also with a correction for finite nuclear mass. So in principle I do not see any particular need for the Rydberg constant other than as the "joker" which one can substitute for either electron mass or fine structure constant. If we're saying the "joker" provides the most accurate representation of electron mass , that would make sense (it appeared to marginally improve the accuracy of my ionization energy calculation - as compared to NIST value - when I tried it) Since in essence the Rydberg constant - by its very definition - is a non-existent physical entity , one wonders exactly what we are in fact 'measuring' to such enormous degrees of precision.

Staff Emeritus
Cthuga's response is good. The Rydberg constant is not a property of any single atom. You seem to think it should be, but obviously the community disagrees.

"Relevance of the Rydberg Constant?"

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