> If the 'mean free path' between collisions is approximately 400 angstroms
> (in copper) how far apart are the individual atoms in a copper lattice?
As far as I know Cu has the face-centered cubic crystal structure
and if I look more closely to my text books I find a distance of
about 1.3 Angstroms to the nearest neighbour...
> As an aside, I had forgotten how much "empty space" there actually is in an
> atom until I recently read the following analogy (which may not be correct
> at all). The allegation was that if a typical atom (defined by the size of
> the electron rings) were the size of the earth, the nucleus would be the
> size of a football field! Is that even close to the truth?
There's a famous number in physics - the so called Bohr radius for the
hydrogen atom -- which is about 0.529 Angstroms. On the other side you
find the approximate proton radius to about 1.3 fm according to
Rutherford. That gives you a ratio atom radius/nucleus radius of about
40'700
for the hydrogen atom. Hence, if you take the radius of the earth to
about 6'370 km you end up with about 156.5 m ...
Nevertheless let me tell you that this "planet model" for atoms
is a very simple one. The restrictions of this model lead to the
development of quantum mechanics in the beginning of this century
where everything is described by ubiquitous wave functions
(must say this as a physicist :-).
> Thanks again for the thoughts.
Hope this helps!
Kind regrads,
Ch. Schuster
--------------------------------------------------------------------- Christian Schuster, Dipl.-Phys. IIS, ETH Zurich Integrated Systems Laboratory Swiss Federal School of Technology Fax: +41 1 63 21194 Gloriastrasse 35 Phone: +41 1 63 27595 CH-8092 Zurich E-Mail: [email protected] Switzerland WWW: http://www.iis.ee.ethz.ch/~schuster/ ------------------------------------------------------------------- Disclaimer: This mail reflects only my personal opinion. I do not guarantee for any results of using it.
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