Talk:Rydberg constant

RH = 2.179 X 10^-18 J = 1.10 x 10^7 m^-1
My love khushbu Why in the world does this article only include the Rydberg constant expressed in the form 1.10 x 10^7 m^-1??? This is easily understood by intermediate chemistry students as the Rydberg cosntant after it is divided by the speed of light and Planck's constant, but it doesn't at all explain the original ACTUAL constant (2.179 x -18 J) and how it relates to ΔE=hν=hc/γ. Why would that not be included anywhere in an article that is to be seen by people who have never dealt with Rydberg before? If I come back in around 6 months and the article looks the same I guess I'll add a whole new section explaining this, unless someone else would rather go ahead. 2605:6000:FE40:A400:38CD:1C6A:D1E7:BF4F (talk) 20:58, 4 July 2014 (UTC)Devin 7/4/2014

A suggestion about the articles
I strongly suggest that the articles on the Rydberg Constant and the Rydberg Formula be merged into one article. These two articles overlap a lot, and the reason is that you can do one derivation in quantum mechanics, and out of that pops both the Rydberg constant and the Rydberg formula. voila! Then, all you have to say is that in the mid-1880s, Dr. Rydberg worked out the whole formula experimentally and empirically -- well-before quantum mechanics had been developed. Rydberg had a value for his constant, but he had no idea where thatn constant came from. He didn't know that it was a simple (mathematically) function of the charge on the electron, the mass of an electron, the permitivity of free space, and the speed of light - plus another constant that he had no idea about because it wasn't discovered until 1900 or a little later: Planck's Constant. 98.67.111.148 (talk) 06:51, 25 July 2010 (UTC)

Wrong units
The given formula for the Rydberg constant R_\infty would lead to a dimension of mass times velocity, whicjh is obviously not an inverse length. DumbBoy 12:09, 11 October 2005 (UTC)

The formula is correct, see eg http://scienceworld.wolfram.com/physics/RydbergConstant.html I don't like the line about being well-determined due to containing five other constants. This needs clarifying (if it's even right), it seems to me that the errors from all the other measurements add, making the Rydberg by far the most errored.

163.1.146.226 19:32, 6 November 2005 (UTC)

The above user refers to a paragraph that I'm now removing:


 * As the formula for the Rydberg constant contains no less than five other physical constants, namely the elementary charge $$e \ $$, the electron rest mass $$m_e \ $$, the permittivity of vacuum $$\epsilon_0 \ $$, reduced Planck's constant $$ \hbar \ $$, and the speed of light in vacuum $$c \ $$, it is one of the most well-determined physical constants. Measuring the Rydberg constant confirms the proportions of the values of the other five constants.

Melchoir 16:39, 16 January 2006 (UTC)

I think user 163.1.146.226 may not have understood that in high-precision measurement of fundamental constants, one normally determines combinations of constants from a single experiment. When one then wants to tease out a single constant, e.g., h, or e, or the mass of the electron, it typically becomes necessary to combine results of several experiments, so that it may very well be true that individual constants are more poorly determined that their combinations. The statement in the article that it tests QED is totally wrong, and I'm removing it. QED doesn't predict the value of, e.g., the electromagnetic coupling constant or the mass of the electron. No measurement of the Rydberg constant could ever, even in theory, disprove QED.--76.93.42.50 (talk) 03:55, 26 May 2008 (UTC)

Info from former page "Rydberg"
"The Rydberg is a unit of energy defined in terms of the ground state energy of the Hydrogen atom (see Rydberg formula). $$E=\frac{-m_e e^4}{2\hbar^2}\frac{Z^2}{n^2}$$ The Rydberg is abbreviated Ry. $$1 Ry = \frac{-m_e e^4}{2\hbar^2} = 13.6 eV $$"

I don't know how correct this info is, so i'm putting it here for now. Fresheneesz 23:16, 17 April 2006 (UTC)

Comments
I am quite used to seeing the formula expressed as $$\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2})$$. It would be more intuitive if the formula were expressed this way. —Preceding unsigned comment added by 69.91.126.99 (talk) 02:05, 1 March 2008 (UTC)
 * This is not the Wiki article for the formula, but the one for the constant. There is another wiki article on Rydberg formula where the equation you mention is given sooner. Take a look at that Wiki article and see what you think. S  B Harris 04:12, 1 March 2008 (UTC)

Rewriting
I did some rewriting. The lead contained a lot of errors, and was poorly written. The body of the article duplicated a lot of material from the lead, and also from Bohr model.--76.93.42.50 (talk) 03:49, 26 May 2008 (UTC)

Derivation section
User Razorflame reverted my edit in which I took out the derivation of the Bohr model, which IMO was a pointless duplication of material at Bohr model. I've invited Razorflame to talk about this here.--76.93.42.50 (talk) 03:59, 26 May 2008 (UTC)
 * That is fine. I realized my mistake, and I immediately self-reverted my reversion that I made to this article.  If I made a mistake, feel free to ask me again on my talk page. Cheers, Razorflame 04:00, 26 May 2008 (UTC)
 * Cool -- I'm glad we were able to work it out :-) --76.93.42.50 (talk) 04:03, 26 May 2008 (UTC)

Rydberg constant for hydrogen?
In the section on the derivation of the Rydberg constant, it states "We have therefore found the Rydberg constant for hydrogen to be"


 * $$ R_H = \frac{ m_e e^4}{8 \epsilon_0^2 h^3 c} $$

However, earlier in the article, R∞ is used to designate this quantity, where the infinity sign shows that the mass of the nucleus is assumed to be infinite. However, if the nucleus is assumed to be infinite, how can the constant be said to be specifically "for hydrogen"? Moreover, Eric Weisstein's World of Physics has the reduced mass of one proton and one electron in their formula for RH, $$ \frac{ m_e m_p}{m_e + m_p} $$ Thus, it seems to me that the symbol should be changed to R with an infinity sign and "for hydrogen" changed to "for an atom with one electron and a nucleus with infinite mass." Or, alternately, the reduced mass of hydrogen could be substituted for the mass of an electron. Let me know if I have misunderstood anything here. WilliamJenkins09 (talk) 13:59, 15 December 2009 (UTC)


 * The "assumption that the mass of the atomic nucleus is infinite compared to the mass" does NOT say that the mass of the atomic nucleus is infinite!
 * Somebody needs to learn something about the mathematics of approximations. What this DOES say is the the ratio (mass of the nucleus)/(mass of the electron) is so damn large that we are justified in treating this ratio as infinite. For hydrogen, given that the mass of a proton is ever-so-much-higher than the mass of an electon, then we are safe to approximate this ratio by infinity.
 * Then, in the physical sciences, the next question to ask is, "What if that ratio is not large enough to treat it as infinite? I can assure you that this usually leads to a lot more complicated math - but sometimes to some interesting results.   DAW - M.S. in electrical engineering and M.A. in mathematics, and who had taken approximations up to his gills!

98.67.111.148 (talk) 06:42, 25 July 2010 (UTC)
 * You are correct! I would keep R∞ and say "for an atom with one electron and one nuclear charge, but a nucleus of infinite mass". The reason for this is that R∞is usually this R constant which is given in texts. Then add "For the case of hydrogen-1 (protium), the reduced mass $$ \frac{ m_e m_p}{m_e + m_p} $$ must be substituted for $$m_e$$"-- or something like that. Perhaps adding how much this changes the "constant." Historically, a lot of people didn't believe Bohr's equation, and somebody or other (Rutherford, I think) noted that its prediction of the spectrum for He(+1) was just a little bit off (the He lines were not exactly 1/4 the H frequencies). Bohr pointed out that they should not be 1/4, but that factor multiplied by the ratios of the reduced masses for H-1 and He-4, and that got the ratio exactly. This made an instant believer out of Rutherford. Later, much the same thing was found to apply to the spectrum of deuterium. S  B Harris 23:45, 15 December 2009 (UTC)


 * SBHarris, thanks for your response. I made some changes to the derivation section to indicate that the equation is strictly true only for the imaginary system with an infinitely massive nucleus. I was going to add some remarks about adjusting for hydrogen with its finite mass, but then I noticed that the section "Value of the Rydberg constant" has an equation for atoms with nuclei of finite mass, so then I had second thoughts. I'm not sure whether this point is even worth bringing up, but I realized that the specification of a +1 nuclear charge is necessary in the equations for energy, but not in the final equation for the Rydberg constant (because there is a Z2 term right next to R in the Rydberg formula, right?) I'm not going to worry about this issue right now, though. If I got anything wrong or left something out, feel free to make the necessary corrections. WilliamJenkins09 (talk) 03:56, 18 December 2009 (UTC)

Circle reference?
In this article it says that the Rydberg constant can be calculated from fundamental constants using quantum mechanics. The equation to be used is


 * $$R_\infty = \frac{m_e e^4}{8 \varepsilon_0^2 h^3 c} = 1.097\;373\;156\;852\;5\;(73) \times 10^7 \ \mathrm{m}^{-1},$$

where me is the rest mass of the electron. However, in the article electron rest mass, it says that the rest mass of the electron is calculated from the definition of the Rydberg constant, namely


 * $$R_{\infty} = \frac{m_{\rm e}c\alpha^2}{2h} \Rightarrow m_{\rm e} = \frac{2R_{\infty}h}{c\alpha^2}\,.$$

So, the question is, which constant is calculated from the other one and which is not? The article says that, as of 2010, the Rydberg constant is the most accurately measured fundamental physical constant. But if it is calculated from the electron rest mass, wouldn't it be necessary that the electron rest mass is even more accurately measured? --Kri (talk) 22:13, 13 February 2011 (UTC)


 * I edited the article to make it clearer: $$R_\infty$$ is known by measuring atomic spectral lines. It is not obtained by multiplying e and m_e and h etc. (OK, well, it involves more than just directly measuring atomic spectral lines...there are also complicated theoretical calculations to correct for finite nuclear mass and various other effects.) --Steve (talk) 04:53, 14 February 2011 (UTC)

Hey Steve - your amendment is still not satisfactory. I came to the main article wanting to know what R is - i.e. how it is defined, and I found the article ambiguous. The article starts off saying that R is the limit of the H spectrum, but then goes on to define it (apparently) in terms of fundamental constants, but with the warning that this is only an approximation based on a nucleus with infinite mass relative to the mass of an electron. Then, in the section on Measurements we are back to spectra again, with yet another putative definition - this time of much greater complexity. Finally, in the 2nd paragraph of this section there is an statement that the constant is defined by a spectrum which does not exist in reality - to me a mind-boggling concept. This particular 'definition' seems to be an error, and I have substituted '..describes..' for '..is defined by...', which seems to make more sense, albeit with the logical meaning reversed. But we are still left with 2 or 3 possible definitions. It would help readers like me if a clear definition of what this constant is (in terms of how its value is found) were put right at the top of the page, and if the various other meanings of the term are put forward as explanations rather than as alternative definitions. I have left this site with a still unclear idea of what R actually is, which seems unsatisfactory - I must look elsewhere for what I want.


 * The most pedantic and logical (but not pedagogical) approach would be to say that $$R_\infty = m_e e^4/(8 \varepsilon_0^2 h^3 c)$$ by definition, and it just coincidentally happens to describe (in a good approximation) the spectrum of hydrogen. On the other hand, the most pedagogical approach -- the approach taken by intro-physics books and courses -- is to forget about fine structure and say (untruthfully) that the Rydberg formula is exact, and $$R_\infty$$ is defined in terms of the spectrum of hydrogen, and then it was the brilliance of Bohr to discover that $$R_\infty$$ just happens coincidentally to be a simple function of the electron mass etc. For this article, I don't know what the best approach is, i.e. how to keep things pedagogical and simple without saying anything inaccurate. I'm sure it can be done...maybe if I have more time later... --Steve (talk) 12:43, 18 March 2012 (UTC)
 * UPDATE: I had a go re-editing the intro and first three sections. Does it help? --Steve (talk) 00:03, 19 March 2012 (UTC)