Re:IONS/metal pedagogy.

[Joseph Bellina <jbellina@saintmarys.edu>]

I think you have just explained how thermionic emission works...that is 
what happens to the comparison between the thermal and the electrostatic 
terms when the temperature is 2000 K or more...are there enough electrons 
in the Boltzmann tail that their thermal energy is sufficient to swamp 
the electostatic term?
Having said that I guess I would add that for your consideration you need 
to remember that most of these thermal processes involve the few highly 
energetic particles on the high energy end of the Boltzmann distribution, 
so the mean thermal energy might not be the thing to look at.

Fun calculation.

joe

On Tue, 13 Oct 
1998, Ludwik Kowalski wrote:

> Some numerology:
>
> Suppose that the radius of the metallic sphere is 1 cm and that the
> excess charge is 1.6 nC (10^10 e). This means that a typical distance
> between adjacent excess electrons is about 0.35 micrometers. The
> corresponding electric potential energy,  0.004 eV, is six times less
> than the average thermal energy at room temperature (0.025 eV).
> The rms speed of electrons at room temperature is about 10^5 m/s.
> And collisions between them (in a vacuum) are not very frequent.
>
> The distance between "electrons in the vacuum" and the "layer
> of exposed positive charges", which binds them, must be much
> smaller that the mean distance between the adjacent electrons in
> the outer layer.  Otherwise the effect of electrostatic interaction
> would be negligible at room temperatures. Working backward (and
> demanding V=0.25 eV; factor of ten above thermal) one finds that
> the distance between the outer layer of electrons and "the exposed
> layer of induced positive charges" is about 0.005 microns or 50
> Angstroms. From a distance of 10 A, which is reasonable (?), the
> electrostatic interaction would be 50 times stronger than thermal.
>
> What speculative suggestions can be made on the basis of these
> trivial considerations?