Talk:Balmer series

Formerly untitled comment
Number change by vandal?: The Rydberg constant for an infinitely heavy nucleus is 10,973,731.534 m&minus;1.

Merged with Balmer Line and set up redirect on that page. I don't see what is wrong with this article that requires cleanup myself. --OmegaPaladin 16:48, 11 December 2005 (UTC)

The colors in the image of the Balmer series are too blue - I've looked at this series through a diffraction grating many times, and H-beta is greenish blue, and the two lines after that are flat-out violet. Is this a rendering problem? Lgrodnicki (talk) 14:08, 5 October 2016 (UTC)lgrodnicki

Balmer vs. Fraunhofer
Is there a difference--aside from the use to which they are put in physics theory--between "Balmer lines" and "Fraunhofer lines?" I take it that "Balmer series" would be a legit accreditation if B. was the first one to link those particular lines to hydrogen, for example, but since F. had noted and measured them, why "Balmer LINES?"

Terry J. Carter Terry J. Carter (talk) 23:12, 27 June 2008 (UTC)

Vacuum or Air
Following work conducted at the University of Oxford, it has come to my attention that the values quoted for the wavelength on this page are in fact air values. The page on Wavelength states that values are assumed to be taken in vacuo unless otherwise stated. This means the required edit is one of two possibilities - change the values to the vacuum values, or make it evident that these are air values. Jonathan Lipscombe [undergraduate physicist] 81.109.21.59 (talk) 11:43, 15 December 2009 (UTC)
 * Well they match those given at Hydrogen spectral series, which are directly derived from the vacuum formula. OrangeDog  (τ • ε) 18:51, 15 December 2009 (UTC)
 * You may be right, the data given in the article agrees with the observed wavelengths in air from NIST, but not with the observed wavelengths in a vacuum. I'll try to look more into this when I have time.  J kasd  03:33, 16 December 2009 (UTC)
 * Since the index of refraction of air is 1.0003 or so, the two values should be the same to the 4th decimal, right? S  B Harris 03:48, 16 December 2009 (UTC)
 * I originally assumed that too, but if I'm reading the data at NIST right, they differ by more than that. For example, they list the H-&alpha line to be at 656.3 nm for air but at 656.4 nm for a vacuum. (Actually they list several very close values for each line, I'm not quite sure why.)  J kasd  04:22, 16 December 2009 (UTC)
 * Well, I meant the fourth significant figure, not literally the fourth decimal (since where the decimal is, depends on the units you use). 656.4/656.3 = 1.00022 which is just about what you'd expect from the index of refraction for air, and in the right direction also (slower in air means longer wavelength in air, as λ = c(medium)/ν. I would have calculated 456.3 *1.0003 = 456.43 for air, which is close enough if you don't want to go past the 4th sig figure. S  B Harris 16:59, 7 January 2010 (UTC)
 * I am not certain which formula you are referring to, but the values do not agree with those derived from the Rydberg equation when using the Rydberg constant with the reduced mass - the wavelengths differ from those given by the equation when using the infinitely massive nucleus. Jonathan Lipscombe 81.109.21.59 (talk) 14:03, 30 December 2009 (UTC)
 * This one: Hydrogen spectral series. OrangeDog  (τ • ε) 15:03, 30 December 2009 (UTC)
 * I see - if one uses the reduced mass, as discussed immediately afterwards on the page, one comes up with the correct wavelengths, in vacuo. Using the infinitely heavy nucleus value for the constant does not come up with accurate vacuum values. This is just a case of applying a different formula, and doesn't affect the issue at hand. I was unsure whether the values should be unchanged, with a label, or whether the values should be changed to the vacuum values. Jonathan Lipscombe 81.109.21.59 (talk) 11:55, 7 January 2010 (UTC)
 * I would have though it was obvious that Hydrogen isn't infinitely heavy... OrangeDog  (τ • ε) 13:02, 7 January 2010 (UTC)

Vanishing m
The equation commonly used to calculate the Balmer series is a specific example of the Rydberg formula and follows as a simple reciprocal mathematical rearrangement of the formula above (conventionally using a notation of m for n as the single integral constant needed):

[ Display equation with m replaced by the integer 2 ]

[ Text computing some units in which m occurs to designate meters ]

That's not good. The phrase "notation of m for n", when caught by the corner of the eye, leads one to scour the display equation for m (m was clearly fundamental in preceding formulas, when it occurred). And you do find m, a little further down, but by then it means something else entirely.

Any practicing physicist would barely be able to experience this. But as someone who once took a year of undergraduate physics, thirty years ago, I had to stare at this in mild bafflement for 30 s.

Please make it more clear than the formula is back in Balmer land, and that m=2 is now assumed. Note that once the brain has seen the simple Rydberg generalization, the brain resists going back to the wacky subcase. (That might just be the brain of a computer scientist, that has spent three decades seeking and destroying manifest constants that never should have been manifest in the first place.)

f(n) = 1/square(BALMER_SPECIAL_M) - 1/square(n) /* something like this */
 * 1) DEFINE BALMER_SPECIAL_M (2)

&mdash; MaxEnt 16:51, 16 October 2018 (UTC)

Visible lines in the Balmer series
According to the article, "lines five and six can be seen with the naked eye" (counting from the right). Is there a source for this? From experiments with a diffraction grating, four (or sometimes three) lines are reported to be visible, see for example https://physics.stackexchange.com/q/219218/220957.


 * I suspect this is a poorly-worded way of pointing out that a human retina is sensitive to wavelengths somewhat shorter than 400 nm. Perception in this range varies between individuals but is generally very weak, mostly because a (natural) cornea is progressively more opaque to shorter wavelengths, so such "marginal violet" colours will appear very dim. Give the other constraints, such "visibility" is only theoretical, and of little practical effect.
 * Martin Kealey (talk) 03:59, 16 March 2024 (UTC)

Various Problems
There are a few issues that need to be resolved here. First of all one cannot just pull up the NIST value of the Rydberg constant and directly "reverse engineer" it to determine B in the empirical Balmer series formula. The B in this formula was an empirically determined constant and will differ from that calculated from the NIST value firstly on account of the wavelengths being measured in air rather than in vacuum and secondly because the NIST value is based on an infinitely heavy nucleus. Since the Balmer series formula (and B) is historical, a more realistic value would be that obtained from regression:

x = n^2/(n^2-4) vs y (measured Balmer series wavelengths - in air).

Keeping the historical 'flavour' a good place to look for measured values of these wavelengths would be the Rowland Solar Spectrum:

3-2 Balmer Alpha 6562.808 A

4-2 Balmer Beta 4861.342 A

5-2 Balmer Gamma 4340.475 A

6-2 Balmer Delta 4101.748 A

7-2 Balmer Epsilon 3970.076 A

An excellent regression fit is obtained leading to a value of B: 3646.022 A.

Alternatively determine 4/R(H) and then correct for vacuum/air typically at 15C 1atm pressure.

Most certainly the Balmer Alpha value in the table needs to be corrected since it is a vacuum wavelength whereas all the rest are air wavelengths. The reason why the NIST database shows several values for Balmer Alpha is because it's not just a single line - it consists of a number of fine structure components which have actually been resolved and separately measured via laser absorption spectroscopy. The Balmer Alpha 'line' (singular) is essentially an averaged value of these components weighted according to their intensities. NIST do give this value: it's labelled simply 3 2 without any indication of 3p, 3d, 3s and 2p, 2s etc. You could also look up similarly 4-2, 5-2 etc all of which have associated fine structure "forests".

Neil Parker (talk) 12:16, 29 July 2019 (UTC)

John law
Is John Dalton law still valid 105.112.16.182 (talk) 19:36, 16 January 2022 (UTC)