Page 51 - Electrical Properties of Materials

34 The electron as a wave

Amplitude Amplitude
(a) (b)

Fig. 2.6
τ t l/τ ω
A rectangular pulse and its frequency
spectrum.

In the mathematical formulation, k and z of eqn (2.20) are to be replaced
by the frequency ω and time t. Hence, the relationship for communication
engineering takes the form,

ω t =2π. (2.37)
∗ Incidentally there is another reason
why radio telescopes must be bigger
than, say, radar aerials. They do a lot The analogy is close indeed.
of work at a wavelength of 210 mm, In the second analogue the size of an aerial and the sharpness of the radi-
which is seven times longer than the ation pattern are related. It is the same story. In order to obtain a sharp beam
wavelength used by most radars. Hence
for the same resolution an aerial seven one needs a big aerial. So if you have ever wondered why radio astronomers
times bigger is needed. use such giant aerials, here is the answer. They need narrow beams to be able
to distinguish between the various radio stars, and they must pay for them by
erecting (or excavating) big antennas. ∗
θ is the beamwidth, z is the lin- The mathematical relationship comes out as follows:
ear dimension of the aerial, and λ
is the wavelength of the electro- θ z = λ. (2.38)
magnetic radiation (transmitted or
received).

Exercises
2.1. Find the de Broglie wavelength of the following Find the spectral composition of the wave train at x =0 and
particles: find from that the spectral width ω (range within which the
amplitude drops to 63% of its maximum value).
(i) an electron in a semiconductor having average thermal
Prove the uncertainty relationship
velocity at T = 300 K and an effective mass of m = am 0 ,
∗
e
where a is a constant, p l = h.
(ii) a helium atom having thermal energy at T = 300 K, 2.3. A typical operating voltage of an electron microscope is
4
(iii) an α-particle (He nucleus) of kinetic energy 10 MeV. 50 kV.
(i) What is the smallest distance that it could possibly
2.2. A finite wave train moves with constant velocity v.Its resolve?
profile as a function of space and time is given as
(ii) What energy of neutrons could achieve the same
exp iu |u| < u 0 resolution?
f (a)= if
0 |u| > u 0 (iii) What are the main factors determining the actual resolu-
tion of an electron microscope?
where u = t – kx and k = /v.
At t = 0 the wave packet extends from x =–u 0 v –1 to 2.4. Electrons accelerated by a potential of 70 V are incident
–1
–1
x = u 0 v , that is, it has a length of l =2u 0 v . perpendicularly on the surface of a single crystal metal. The