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Most Detailed Photos of an Atom Yet 229

BuzzSkyline writes "Ukrainian researchers have managed to take pictures of atoms that reveal structure of the electron clouds surrounding carbon nuclei in unprecedented detail. Although the images offer no surprises (they look much like the sketches of electron orbitals included in high school science texts), this is the first time that anyone has directly imaged atoms at this level, rather than inferring the structure of the orbitals from indirect measurements such as electron or X-ray interferometry."
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Most Detailed Photos of an Atom Yet

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  • by Genda ( 560240 ) <> on Tuesday September 15, 2009 @04:41AM (#29423887) Journal

    The ability to directly measure electron density is quite an old technique. STMs and AFMs have been doing this since the very beginning.. I agree with the researcher's quote in the article that it's good to develop a complementary technique(FEEM) abd at best that's its contribution. I'd be happy to hear what else it contributes. though I don't quite agree with his or the editors spelling! ;) "it's always good to have complimentary approaches,"

    In this particular application, its simply a very cool thing to be able to prove theory with direct measurement. In the future I can imagine viewing electron orbitals for test samples of high temperature superconductors or producing high resolution images of the electron cloud density for a protein (get a better idea of the quantum component for protein folding) might prove extremely useful and interesting.

    In my experience, no sooner does someone come up with a better device for viewing, then someone comes up with a exquisite need for that device.

  • by PatrickThomson ( 712694 ) on Tuesday September 15, 2009 @04:56AM (#29423979)

    We can only "approximate" orbitals in atoms with more than one electron, true. That's the same as saying that numerical methods won't "exactly" solve a function. We can still get really accurate results, even if it's computationally expensive. FT-IR of heteronuclear diatomics, anyone?. Orbitals still retain the basic shapes in multiple-electron atoms.

  • by PeterBrett ( 780946 ) on Tuesday September 15, 2009 @05:32AM (#29424121) Homepage

    Here's a picture of a dupe [], complete with comments.

    It's not a dupe -- that was a different story, as you would know if you had compared TFA from each story.

  • by PatrickThomson ( 712694 ) on Tuesday September 15, 2009 @05:35AM (#29424129)

    Basically, a chemistry education is very much like fast-forwarding through 300+ years of science history. Some dead-ends are skipped, but by and large, the simpler and more self-contained a theory was, the older it is and the earlier it's taught in school. The university-taught molecular orbital theory is (debatably) too rich and complex to be taught any earlier.

    The moons-orbiting theory fit with all the available evidence at the time it was developed. Think of orbitals as clouds of probability where, if you tried to pin down the electron, it might be. A moons-orbiting theory would give this probability cloud as a thin donut around the atomic waist. The shapes of orbitals as depicted in wikipedia etc. are consequences of the maths of quantum mechanics. It's annoyingly non-intuitive.

  • by locofungus ( 179280 ) on Tuesday September 15, 2009 @05:55AM (#29424207)

    It's wrong to think of the electron as a particle when it's "orbiting" in an atom. Instead you should think of it as a probability density. This is Schroedinger's cat all over again, the electron is "smeared out" all over its "orbit" but instead of being "half dead, half alive" it's x% here, y% there.

    This is also like the two slit experiment. The electron doesn't go through one slit or the other, it goes through both slits (not 50% dead and 50% alive; 50% went through that slit and 50% went through this slit) but when it hits the phosphor screen it's a particle as its "where is it" probability function collapses to a point.

    The "wavefunction" is (as far as we can tell) a mathematical curiosity that when squared gives us the observed probability function. The probability distribution is real, the wavefunction gives us a convenient handle to calculate probabilities and how they evolve.

    But now that I've said that the wavefunction is an imaginary curiosity, imagine a sine wave on a string and then join the two ends of the string together. There will only be a few discrete lengths of string where the sine wave will "join up" correctly (ok there's an infinite number but the length of the string is limited). It turns out that, with rather a lot of unpleasant maths (see the wikipedia page for spherical harmonics), our wavefunction works like that sine wave and we find that there are only certain orbitals where the wavefunction is well behaved.


  • by fastest fascist ( 1086001 ) on Tuesday September 15, 2009 @06:27AM (#29424347)
    Not to pick nits, but is a picture that is the result of electrons striking a surface actually a photograph?
  • by vikhyat ( 1593841 ) on Tuesday September 15, 2009 @06:35AM (#29424385)
    It's probably like one of those long exposure photographs.
  • by master_p ( 608214 ) on Tuesday September 15, 2009 @06:37AM (#29424397)

    Could it be that we can't pinpoint the exact position and velocity of an electron at the same time because they are interlinked with all the surrounding particles? i.e. the act of measurement affects the outcome.

  • by Anonymous Coward on Tuesday September 15, 2009 @08:56AM (#29425711)
    Based on your previous posts, your lack of education was never in doubt.
  • by Barterer ( 878209 ) on Tuesday September 15, 2009 @09:19AM (#29426017)

    Very informative, thanks. When you say the electrons have a definite momentum about the Z-axis, do you mean just one chosen axis (depending on your perspective and axes you assign) or does it have something to do with which way is "up" or gravity?

  • by Anonymous Coward on Tuesday September 15, 2009 @10:10AM (#29426689)

    ...Think of orbitals as clouds of probability where, if you tried to pin down the electron, it might be. ...


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