Page 52 - Electrical Properties of Materials

Exercises 35

crystal planes are parallel to the metal surface and have a (cu- 1966) it was found that a certain fraction of the electrons
bic) lattice spacing of 0.352 nm. Sketch how the intensity of passing through the metal had a loss of energy of 14.97 eV.
the scattered electron beam would vary with angle. We could explain this loss as being the creation of a particle of
that much energy. But what particle? It cannot be a photon (a
2.5. A beam of electrons of 10 keV energy passes perpendicu-
transverse electromagnetic wave in the wave picture) because
larly through a very thin (of the order of a few nanometres) foil
an electron in motion sets up no transverse waves. It must be
of our previous single crystal metal. Determine the diffraction
a particle that responds to a longitudinal electric field. So it
pattern obtained on a photographic plate placed 0.1 m behind
might be a plasma wave of frequency ω p whichwecouldcall
the specimen. How will the diffraction pattern be modified
a ‘plasmon’ in the particle picture. The energy of this particle
for a polycrystalline specimen? [Hint: Treat the lattice as a
would be ω p .
two-dimensional array.]
Calculate the value of ω p for aluminium assuming three
2.6. Consider again an electron beam incident upon a thin free electrons per atom. Compare it with the characteristic
metal foil but look upon the electrons as particles having a energy loss found.
certain kinetic energy. In experiments with aluminium foils The density of aluminium is 2700 kg m –3 and its atomic
(J. Geiger and K. Wittmaack, Zeitschrift für Physik, 195, 44, weight is 27.