Talk:Mössbauer effect

Miscellaneous

Is there a reason someone removed the umlaut from Mössbauer? -- Matt Stoker

Umlauts are not Wikifiable. Meaning you can't put brackets around it and make it a link. Same goes for most special characters. --MemoryHole.com

You can make it display, just not actually link: [[Mossbauer Spectroscopy|Mössbauer Spectroscopy]] links to the first term and displays the second, like so: Mössbauer Spectroscopy. The linking is supposed to be resolved I think with .92, but I'm not sure. --KQ

Fine, but why not leave the umlauts in the unlinked text? -- Matt Stoker

With MediaWiki 1.5 we don't have such a problem anymore (as I've understood), so I just was bold moved the page a moment ago. I hope it's fine.—Pt_(T) 23:27, 30 December 2005 (UTC)[reply]

- Nice articles on Mossbauer effect and spectroscopy. Upon reading that the Mars Rover uses a Mossbauer spectrometer, I was curious how elements were identified. From my college days I vaguely remembered the basic physics. Your article is a clear, concise explanation - better than other Web pages I found using a Yahoo search. - A diagram, such as http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/mossb.html, could be helpful. - Who is the author? Are you a university professor? -- PWL, 1-23-04

The article says X-rays and gamma rays behave differently, with the former able to undergo resonant emission and absorption in gasses while the latter cannot, but the article on gamma rays says they're effectively the same thing. Can anyone clarify? Specific wavelengths would be nice ^_^ 192.171.162.101 09:38, 2 August 2005 (UTC)[reply]

Note that it's not the rays which are behaving in a certain way. It's due to the properties of the gas/metal target. X- and gamma rays are both forms of electromagnetic radiation, and they do overlap somewhat, but generally speaking gamma rays have higher energy, which is why the effect happens only for gamma rays. I can't say I understand the Mossbauer effect enough to give you specific wavelengths, but I'll put it on my to-do list. You may want to check out what HyperPhysics has on this (the link in the comment above yours). - mako 02:45, 3 August 2005 (UTC)[reply]

In the description section the line "In the case of a gas the emitting and absorbing bodies are atoms, so the mass is quite small," is used. This line is confusing and should be more specific. It is not the atomic mass which is "quite small" (since the atomic mass is essentially the nuclear mass with some electrons) but that of the atomic electrons which are the absorbing bodies in this case. I have changed the line to read "In the case of a gas the emitting and absorbing bodies are atomic electrons, so the mass is relatively small,". — Preceding unsigned comment added by 130.209.45.212 (talk) 10:11, 17 May 2013 (UTC)[reply]

The above comment is right in the sense that the first description phrasing was confusing, however the new sentence is physically wrong. "In the case of a gas the emitting and absorbing bodies are atomic electrons, so the mass is relatively small, resulting in a large recoil energy, which prevents resonance." It is not the electrons that emit the gamma ray and the mass of the whole atom has to be used in the formula. So I changed atomic electrons to atoms. — Preceding unsigned comment added by 195.184.106.238 (talk) 15:48, 31 May 2013 (UTC)[reply]

Graph

Could someone give it English subtitles?

I'll second that request. I assume the Chinese characters say something like "Photon counts" and "Energy [MeV]" for the vertical and horizontal axes, respectively. -- JeffreyRMiles 21:20, 23 April 2007 (UTC)[reply]

Using translate.google.com yielded "Relative count rate" and "Road Address" for the vertical and horizontal axes, respectively. "Road Address" may actually be "Channel Address". Galium111 (talk) 05:18, 11 October 2010 (UTC) The horizontal axis may be showing relative speed in centimeters per second between the turntable-mounted radiation source to the absorber. Reading with translate.google.com, the caption on zh.wikipedia.org's graph shows that "The horizontal axis is the channel address, on behalf of the relative velocity". The article describes using the Doppler effect to "regulate" the energy of the gamma ray between the source and the absorber with zero relative speed having the maximal resonance absorption. It says that "a few centimeters per second" of speed was "enough to break the resonance". Galium111 (talk) 13:49, 14 October 2010 (UTC)[reply]

I hope I will be able to upload a figure with a legend in English in a couple of days.

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/ironze.html#c1 has a graph of the same six absorption peaks with units on the horizontal axis. Please take a look before changing the graph. Thanks. Galium111 (talk) 19:43, 14 October 2010 (UTC)[reply]

Missing Stuff

There should be a notion as to why the Mößbauer Effect is so awesome that its used even though it has some very inconvienient requirements (like very low gamma decay energies for room-temperature meassurements). It would be visible in the graph, if it had english text.

The reason is: The energy resolution of the method is positively awesome. You can get resolutions in the sub- ueV range, for total energies in the 10keV range. There is no other spectroscopic methods that accurate.

seconded. I was also hoping to find a description of the use of the ME on Cobalt-60? gamma rays that showed the energy gain(=frequency increase) as the gamma rays fell some tens of metres. Rod57 (talk) 14:53, 19 February 2010 (UTC)[reply]

His name is Rudolf Mößbauer

Its Mößbauer and not Mössbauer! —Preceding unsigned comment added by 194.8.209.3 (talk) 14:08, 14 December 2007 (UTC)[reply]

And what is the text "Meß-Bauer" supposed to achieve, aside from confusion?

—DIV (128.250.80.15 (talk) 05:05, 27 March 2008 (UTC))[reply]

Reverse Mossbauer Effect?

I would like to see a section explaining what the Reverse (or Inverted) Mossbauer Effect is. — Preceding unsigned comment added by 71.191.244.247 (talk) 19:02, 12 July 2015 (UTC)[reply]

What does Admiral Poindexter have to do with this?

US Navy rear admiral John Poindexter is listed in one of the links at the bottom of the page. How is he relevant to zero-phonon nuclear transitions? His own Wikipedia page does not mention any involvement in the subject.

Stockpiling references for later use on inline citations

Patel, S.B. (1991). Nuclear Physics - An Introduction. New York: Wiley Eastern Limited (John Wiley & Sons). pp. 312–321. ISBN 0-470-211320-X. {{cite book}}: Check |isbn= value: length (help); Cite has empty unknown parameter: |coauthors= (help)
Semat, Henry (1972). Introduction to Atomic and Nuclear Physics. New York: Holt, Rinehart & Winston. pp. 470–478.
Cheers - Williamborg (Bill) 20:40, 8 June 2010 (UTC)[reply]

Relativity

Mossbauer Effect experiments helped to prove Einstein's theory that time is relative, and I expected to read about that here, but it's not covered. I can't add it because I don't know enough. Can someone who knows more please add something? Richard75 (talk) 19:45, 13 June 2020 (UTC)[reply]

You probably have in mind the Mössbauer rotor experiments. Tercer (talk) 14:45, 8 April 2021 (UTC)[reply]