Page 40 - Electrical Properties of Materials

Exercises 23

an insulator having the same number of lattice atoms because of the electronic
contribution. These expectations are, however, wrong. It turns out that metals Metals behave as if the free elec-
and insulators have about the same specific heat. Our model fails again to ex- trons make practically no contri-
plain the experimentally observed value. What shall we do? Modify our model. bution to the specific heat.
But how? Up to now the modifications have been fairly obvious. The ‘wrong
sign’ of the Hall voltage could be explained by introducing positive carriers,
and when cyclotron resonance measurements showed that the mass of an elec-
tron in a solid was different from the ‘free’ electron mass, we simply said: ‘all
right, the electron’s mass is not a constant. How should we modify our model
now?’ There seems to be no simple way of doing so. An entirely new start is
needed.
There is no quick fix for this real dilemma. We have to go quite deeply into
wave theory and quantum mechanics. Finally, all is revealed in Section 6.3
when we find that electrons do make a quantifiable contribution to specific
heat, which turns out to be very small.

Exercises
1.1. A 10 mm cube of germanium passes a current of 6.4 mA 1.4. Measurements on sodium have provided the following
–8
when 10 mV is applied between two of its parallel faces. data: resistivity 4.7 × 10 ohm m, Hall coefficient –2.5 ×
3
–1
Assuming that the charge carriers are electrons that have a 10 –10 m C , critical wavelength of transparency 210 nm, and
2
–3
–1 –1
mobility of 0.39 m V s , calculate the density of carriers. density 971 kg m .
What is their collision time if the electron’s effective mass in Calculate (i) the density of electrons, (ii) the mobility,
germanium is 0.12 m 0 where m 0 is the free electron mass? (iii) the effective mass, (iv) the collision time, (v) the number
of electrons per atom available for conduction.
1.2. An electromagnetic wave of free space wavelength Electric conduction in sodium is caused by electrons. The
0.5 mm propagates through a piece of indium antimonide number of atoms in a kg mole is 6.02 × 10 26 and the atomic
that is placed in an axial magnetic field. There is resonant weight of sodium is 23.
absorption of the electromagnetic wave at a magnetic field,
–2
B = 0.323 wb m . 1.5. For an electromagnetic wave propagating in sodium plot
the real and imaginary part of the wavenumber k as a function
(i) What is the effective mass of the particle in question? 6 16
of frequency (use a logarithmic scale) from 10 to 10 Hz.
(ii) Assume that the collision time is 15 times longer (true 6 15
Determine the penetration depth for 10 ,10 ,and 2 ×
for electrons around liquid nitrogen temperatures) than 15
10 Hz.
in germanium in the previous example. Calculate the
Use the conductivity and the collision time as obtained from
mobility.
Exercise 1.4.
(iii) Is the resonance sharp? What is your criterion?
1.6. A cuboid of Ge has contacts over all of its 2 mm × 1mm
1.3. If both electrons and holes are present the conductivit- ends and point contacts approximately half way along its 5 mm
ies, add. This is because under the effect of an applied electric length, at the centre of the 5 mm × 1 mm faces. A magnetic
field the holes and electrons flow in opposite directions, and a field can be applied parallel to this face. A current of 5 mA
negative charge moving in the (say) +z-direction is equivalent is passed between the end contacts when a voltage of 310 mV
to a positive charge moving in the –z-direction. is applied. This generates a voltage across the point contacts
Assume that in a certain semiconductor the ratio of elec- of 3.2 mV with no magnetic field and 8.0 mV when a field of
tronic mobility, μ e , to hole mobility, μ h , is equal to 10, 0.16 T is applied.
20
–3
the density of holes is N h =10 m , and the density of
–3
19
electrons is N e =10 m . The measured conductivity is (i) Suggest why an apparent Hall voltage is observed with
–1
–1
0.455 ohm m . Calculate the mobilities. no magnetic field.