Page 28 - Electrical Properties of Materials

Electromagnetic waves in solids 11

but that is because radio waves have not got high enough frequencies; let us
try light waves. Can they penetrate a metal? No, they can not. It is another
empirical fact that metals are not transparent. So we should try even higher
frequencies. How high? Well, there is no need to go on guessing, we can work
out the threshold frequency from eqn (1.52). Taking the electron density in a
3
typical metal as 6 × 10 28 per m , we then get
1 N e e 2 1/2
f p =
π m 0
28
)
1 6 × 10 (1.6 × 10 –19 2
1/2
=
2π 9.11 × 10 –31 × 8.85 × 10 –12
15
=2.2 × 10 Hz, (1.54)
where ε 0 is the free-space permittivity.
At this frequency range you are probably more familiar with the
wavelengths of electromagnetic waves. Converting the above calculated fre-
quency into wavelength, we get

c 3 × 10 8
λ = = 15 = 136 nm, (1.55)
f p 2.2 × 10
where c is the velocity of light.
Thus, the threshold wavelength is well below the edge of the visible re-
gion (400 nm). It is gratifying to note that our theory is in agreement with our
everyday experience; metals are not transparent.
There is one more thing we need to check. Is the condition ωτ 1 satis-
fied? For a typical metal at room temperature, the value of τ is usually above
10 –14 s, making ωτ of the order of hundreds at the threshold frequency.
By making transmission experiments through a thin sheet of metal, the crit-
ical wavelength can be determined. The measured and calculated values are
compared in Table 1.1. The agreement is not too bad, considering how simple
our model is.
Before going further I would like to say a little about the relationship of
transmission, reflection, and absorption to each other. The concepts are simple
and one can always invoke the principle of conservation of energy if in trouble.

Table 1.1 Threshold wavelengths for alkali
metals

Metal Observed Calculated
wavelength wavelength
(nm) (nm)
Cs 440 360
Rb 360 320
K 315 290
Na 210 210
Li 205 150