Download m - Egloos

Chapter 10 The Problem Solutions
1. The ion spacings and melting points of the sodium halides are as follows:
NaF
NaCl
NaBr
Nal
Ion spacing, nm
0.23
0.28
0.29
0.32
Melting point, oC
988
801
740
660
Explain the regular variation in these quantities with halogen atomic number.
【Sol】
The halogenic atomic numbers are Z = 9 for fluorine (F), Z = 17 for chlorine (Cl), Z =35 for
bromine (Br) and Z = 53 for iodine (I). The greater the atomic number of a halogen atom,
the larger the atom is, hence the increase in the interatomic spacing with Z. The larger the
ion spacing, the smaller the cohesive energy, hence the lower the melting point.
3.
(a) The ionization energy of potassium is 4.34 eV and the electron affinity of chlorine is
3.61 eV. The Madelung constant for the KCl structure is 1.748 and the distance between
ions of opposite sign is 0.314 nm. On the basis of these data only, compute the cohesive
energy of KCl. (b) The observed cohesive energy of KCl is 6.42 eV per ion pair. On the
assumption that the difference between this figure and that obtained in a is due to the
exclusion-principle repulsion, find the exponent n in the formula Br-n for the potential
energy arising from this source.
Inha University
Department of Physics
【Sol】
(a) The cohesive energy will be the negative of the Coulombic energy as given in Equation
(10.1), minus the difference between the ionization energy of potassium and the electron
affinity of chlorine;
e2
α
− (E i − Ea )
4πεor
= (1.748)
(8.988 × 109 N ⋅ m 2 /C 2 )(1.602 × 10−19 C)2
−9
( 0.314 × 10 m )(1.602 × 10
−19
J/eV )
− ( 4.34 eV − 3.61 eV ) = 7.29 eV
(b) The difference between the observed binding energy and that found in part (a) must be
due to the repulsive energy as given in Equation (10. 1). From the observed binding energy,
UO must be given by
-UO = 6.42 eV + (4.34 eV - 3.61 eV) = 7.15 eV.
The Couloumbic energy, an intermediate calculation in part (a) , is Ucoulomb -8.0156 eV, and
so solving Equation (10.5) for n,

Uo 
n = 1 −

 U coulomb 
Inha University
−1

7.15 eV 
= 1 −

 8.0156 eV 
−1
= 9.26
Department of Physics
5.
The joule-Thomson effect refers to the drop in temperature a gas undergoes when it
passes slowly from a full container to an empty one through a porous plug. Since the
expansion is into a rigid container, no mechanical work is done. Explain the JouleThomson effect in terms of the van der Waals attraction between molecules.
【Sol】
The heat lost by the expanding gas is equal to the work done against the attractive van der
Waals forces between the gas molecules.
7.
What is the effect on the cohesive energy of ionic and covalent crystals of (a) van der
Waals forces and (b) zero-point oscillations of the ions and atoms about their
equilibrium positions?
【Sol】
(a) Van der Waals forces increase the cohesive energy because they are attractive, and the
ions in the crystals tend more to cohere. (b) Zero-point oscillations decrease the cohesive
energy because these oscillations represent a mode of energy that is present in a solid but not
in individual atoms or ions.
Inha University
Department of Physics
9.
Does the "gas" of freely moving electrons in a metal include all the electrons present? If
not, which electrons are members of the "gas"?
【Sol】
The electrons that constitute the "gas" of freely moving electrons are only those that are
loosely bound to the nuclei, specifically those electrons in the outer shells. As has been seen,
the innermost electrons have binding energies that give rise to x-ray spectra, and will not be
members of the free-electron gas.
11. Silver has an atomic mass of 108 u, a density of 10.5 x 103 kg/m3 , and a Fermi energy of
5.51 eV. On the assumptions that each silver atom contributes one electron to the
electron gas and that the mean free path of the electrons is 200 atom spacings, estimate
the resistivity of silver. (The actual resistivity of silver at 200C is 1.6 x 10-8 Ω ⋅ m.)
【Sol】
The number density n is the mass density ρAg divided by the mass mAg of each atom,
ρ Ag
(10.5 × 103 kg/m 3 )
n=
=
= 5.855 × 10− 28 atoms/m 3
27
m Ag (108u)(1.66054 × 10 kg/u )
keeping extra significant figure. With the assumption stated in the problem, this is the same
as the electron density.
2m e εF
mev F
2( 9.11 ×10 −31 kg )(5.51eV )(1.60 ×10 −19 J/eV )
ρ=
=
=
ne 2 (200n −1 / 3 ) 200e 2n 2 / 3
(200)(1.60 ×10-19 C) 2 (5.86 ×10 28 m − 3 ) 2 / 3
= 1.64 ×10 −8 Ω ⋅ m
Inha University
Department of Physics
13. How are the band structures of insulators and semiconductors similar? How are they different?
【Sol】
In both insulators and semiconductors, a forbidden band separates a filled valence band from
the conduction band above it. In semiconductors, the band gap is smaller than in insulators,
and the property of the gap that makes a semiconductor a semiconductor is that in the
semiconductor some valence electrons have enough thermal energy to jump across the gap to
the conduction band.
15. (a) Why are some solids transparent to visible light and others opaque? (b) The
forbidden band is 1.1 eV in silicon and 6 eV in diamond. To what wavelengths of light
are these substances transparent?
【Sol】
(a) Photons of visible light have energies of ~ 1 – 3 eV, which can be absorbed by free
electrons in a metal without leaving the electrons' valence band. Hence metals are opaque.
The forbidden bands in insulators and semiconductors are too wide for valence electrons to
jump across the gaps by absorbing 1 – 3 eV. Hence such solids are transparent.
(b) The minimum wavelengths are given by λ = hc/Eg, where E. is the gap energy; light with
shorter wavelengths would have energies larger than the gap energy. For silicon,
λmin
Inha University
hc 1.24 × 10− 6 eV ⋅ m
=
=
= 1.13 × 10−6 m = 1.13 ìm
Eg
1.1 eV
Department of Physics
keeping an extra significant figure; elemental silicon is not transparent to visible light.
For diamond,
hc 1.24 × 10− 6 eV ⋅ m
λmin =
=
= 2.07 × 10− 7 m = 207 nm
Eg
6 eV
17. (a) When germanium is doped with aluminum, is the result an n-type or a p-type
semiconductor? (b) Why?
【Sol】
Aluminum atoms have 3 electrons in their outer shell, germanium atoms have 4 (see Table
7.4). Replacing a germanium atom with an aluminum atom leaves a hole, and the the result
is a p-type semiconductor.
19. Draw the third Brillouin zone of the two-dimensional square lattice whose first two
Brillouin zones are shown in Fig. 10.41.
【Sol】
The figure on the next page shows the third Brillouin zone. The wavevectors in this zone will be
those that do not fit into the first two zones but are not diffracted by the diagonal sets of atomic
planes in Figure 10.40 that make angles of ± arctan(l/2) = ± 26.6 0 or ± arctan(2) = ± 63.4 0 with
the x- or y-axes. These wavenumbers correspond to |kx|+|ky| > 2π/a (the condition that k not be
in the first or second zones) and
Inha University
Department of Physics
2π
π
π π
2π 
π

,0 < ky >  or 0 < kx < ,
< ky >
 < kx <

a
a
a a
a 
a

Figure for Solution to Problem 9-19: More text on previous page
In the figure, the darker-shaded square is the first Brillouin zone and the lighter-shaded square
is the second zone, as in Figure 10.41 (the axes and axis scales are not shown, but are the same
as in the text figure). The unshaded area is the third Billouin zone.
Inha University
Department of Physics
21. Phosphorus is present in a germanium sample. Assume that one of its five valence
electrons revolves in a Bohr orbit around each P + ion in the germanium lattice. (a) If the
effective mass of the electron is 0.17me, and the dielectric constant of germanium is 16,
find the radius of the first Bohr orbit of the electron. (b) The energy gap between the
valence and conduction bands in germanium is 0.65 eV, How does the ionization energy
of the above election compare with this energy and with kT at room temperature?
【Sol】
(a) The radius of a Bohr orbit for a given energy level n is proportional to the relative
permittivity (the dielectric constant) and the mass, as given in Equation (4.13). The radius of
the first Bohr orbit, in terms of ao, would then be
ε /ε
16
r1 = a 0 * o = ( 5.29 × 10−11 m )
= 5.0 nm.
0.17
m /m
(b) The energy of an electron in a given Bohr orbit is porportional to the effective mass and
inversely proportional to the square of the relative permittivity (see Equation (4.15)). The
ionization energy is then
m * /m
0.17
E = ( −E1)
=
(
13
.
6
eV
)
= 9.0 × 10− 3 eV.
2
2
(ε /ε o )
(16)
This is much smaller than the energy gap of 0.65 eV but comparable to the product kT ≈ 0.025 eV.
Inha University
Department of Physics
23. The effective mass m* of a current carrier in a semiconductor can be directly determined
by means of a cyclotron resonance experiment in which the carriers (whether electrons
or holes) move in helical orbits about the direction of an externally applied magnetic
field B. An alternating electric field is applied perpendicular to B, and resonant
absorption of energy from this field occurs when its frequency ν, is equal to the
frequency of revolution νc, of the carrier. (a) Derive an equation for νc, in terms of m*,
e, and B. (b) In a certain experiment, B = 0.1 T and maximum absorption is found to
occur at ν = 1.4 x 1010 Hz. Find m*. (c) Find the maximum orbital radius of a charge
carrier in this experiment whose speed is 3 x 104 m/s.
【Sol】
(a) In a uniform magnetic field of magnitude B, an electron moving with nonrelativistic
speed v at right angles to the field will experience a force of magnitude evB. The inward
acceleration of a particle moving with constant speed v in a circle of radius R has magnitude
v2/R. From F = m*a,
2
*v
evB = m
, or eB = m * ( 2πνc ),
R
where νc = v/(2πR) has been used. Solving for the cyclotron frequency νc,
eB
νc =
.
2πm *
Note that the cyclotron frequency is independent of the electron's speed, and hence its orbit
radius.
Inha University
Department of Physics
(b) From the above expression for νc solving for m* gives
eB
(1.60 × 10−19 C)( 0.1 T )
m =
=
= 1.82 × 10−31 kg = 0.2me .
10
2πνc
2π (1.4 × 10 Hz )
*
(c) Solving for R as a function of v and ν,
v
3 × 104 m/s
−7
R=
=
=
3
.
4
×
10
m.
2πν 2π (1.4 × 1010 Hz )
25. A voltage of 5.0 V is applied across a Josephson junction. What is the frequency of
radiation emitted by the junction?
【Sol】
From Equation (10.28), the frequency is
2(Ve ) 2(5.0 × 10−6 eV)
9
ν=
=
=
2
.
4
×
10
Hz = 2.4 GHz.
h
4.14 × 10-15 eV ⋅ s
Inha University
Department of Physics