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Nuclear Radius and Diffraction

  1. Mar 20, 2012 #1
    Hi there,

    My textbook says "When a beam of high energy electrons is directed at a thin solid sample of an element, the incident electrons are diffracted by the nuclei of the atoms in the foil."

    What does this mean exactly? Are the atoms diffracted by the gap between different nuclei?

    It goes on to say they are diffracted because the de broglie wavelength of the electrons is similar to the diameter of the nuclei - but I thought the wavelength had to be similar to the gap width, i.e the distance between two nuclei. This gap width wouldn't necessary by the same as the diameter of the nucleus would it?

    Thanks in advance!
     
  2. jcsd
  3. Mar 20, 2012 #2
    It is true that waves can be diffracted around obstacles as well as through apertures.
    This partly explains how you can hear sound behind an obstacle.
    The effects of diffraction are apparent when the wavelength of the wave is of the same order as the dimensions of the obstacle.
     
  4. Mar 22, 2012 #3
    Thanks for your reply but I'm still confused.

    I understand the principle of diffraction. What I don't understand is what is actually diffracting the electrons in this case? Is it the gap between the nuclei or the nucleus itself? If it is the gap between the nuclei then how does that give an approximation for the diameter of a nucleus?

    If the beam are diffracted by travelling through the nucleus itself then what is that all about?!

    Thanks
     
  5. Mar 22, 2012 #4
    From the wording I would say that the electrons are diffracted by the nucleus. The nucleus is behaving as an obstacle and the details of the diffraction pattern will indicate the size of the nucleus.
    I would not say the beams are "travelling through the nucleus itself"
     
  6. Mar 22, 2012 #5

    Drakkith

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    Staff: Mentor

    The distance between nuclei of atoms is far larger than the nuclei themselves. If the de broglie wavelength is approximately the size of the nucleus then it is being diffracted by the nucleus itself.
     
  7. Mar 26, 2012 #6
    Thanks Drakkith - That's what I think it is suggesting but what does that mean to say the electron is being diffracted by travelling through the nucleus?

    Aren't nuclei relatively dense? 6x10 to the 16 kg per metre cubed. (more dense than a neutron star).

    If electrons are firing into this dense object and diffracting then, well, how is this happening!? This doesn't sound like diffraction to me. The electrons are not travelling through a gap or passing an edge - and if they are passing an edge then it will be of something inside the nucleus (a quark possibly)- how would that than tell us the diameter of the nucleus?

    Help!
     
  8. Mar 26, 2012 #7

    Drakkith

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    Staff: Mentor

    The nucleus has an "outer edge" if you want to call it that. Since the electron has wave-like properties, it diffracts off of it just like a sound wave does when it passes over an object. Higher energy electrons have a smaller wavelength and can actually tell us what's in the nucleus through deep inelastic scattering, but longer wavelength ones simply diffract off of it as if it were one object. As to how that tells you the diameter of the nucleus, that is beyond my knowledge and ability to explain. I only know that it does.
     
  9. Mar 26, 2012 #8
    See the differential cross section plots of electron-nuclear scattering in

    http://www.jpoffline.com/physics_docs/y3s6/nuclearphysics_ln.pdf

    See the obvious "diffraction" shape in Fig, 6 on page 8. This is due to the electron deBroglie wavelength and the nuclear size, See the discussion in Section 1.4. For a E = 100 MeV electron, the deBroglie wavelength is λ = hc/(2πE) = 197 MeV-fermi / 100 MeV = 1.97 fermi (1.97 x 10-13 cm). So it is similar to nuclear sizes.

    This article shows rings fron optical diffraction around a circular dark spot

    http://en.wikipedia.org/wiki/Arago_spot
     
    Last edited: Mar 26, 2012
  10. Mar 26, 2012 #9

    sophiecentaur

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    The term diffraction refers to the whole thing. The individual diffracting centres and the pattern in which they are arranged (as in the 'array factor' of antenna arrays). The same happens with the 'two slits' experiment, you have the broad beam pattern from the slit width and the banded pattern from the 'array' of two, spaces, slits.
    The overall pattern is given by the product of the two patterns but one will often dominate.

    In the case of the array of nuclei and fast, diffracting electrons, the pattern is dominated by the individual nuclei. For a longer wavelength (X rays or slow electrons) the array factor will dominate and you get an array of spots for a single crystal or rings (as with the School Electron diffraction through a graphite target).

    Btw. The shape of diffraction patterns for a circular obstacle is the same as for a circular hole - the dark areas being replaced by light areas etc.
     
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