[This document contains special characters/formulas that cannot be displayed in the webpage. Kindly download !]
Transcript:
Q 1) Prove:
(a)
than an infinite point lattice is only capable of showing 2, 3, 4, or 6fold type rotational symmetry;
(b)
the Weiss zone law, i.e. if [uvw] is a zone axis and (hkl) is a face in the zone, then
hu + kv + lw = 0;
(c)
that in the cubic system the direction [hkl] is parallel to the facenormal (hkl);
(d)
that in the cubic system the angle f between the facenormals (h1k1l1) and (h2k2l2) is given by
(e)
that, when using MillerBravais indices (hkil), h + k + i = 0
Ans 1)
(a) Let’s consider a regular polygon with n sides. Each vertex acts like a rotation point. According to trigonometry of a regular polygon, interior angle subtended by two successive arms of a regular polygon is always
radians. If m faces meet at vertex, then equation that satisfies this condition is:
=>
=> m and n can only be 3, 4 or 6. It implies that a plane can only be filled with convex or regular polygons which are either equilateral triangles, squares, or hexagons (i.e. n = 3, 4, or 6)
Note: n cannot be 2. It means there’s no plane and polygon is actually a straight line. And a straight line cannot fill any plane. It needs to have at least 3 points to form a closed structure and fill the plane.
(b) Unit vector passing through the face plane (hkl) is nothing but
and the Weiss zone axis vector for its direction [uvw] is
But these are actually perpendicular, and must satisfy the equation .W= 0 (i.e. dot product)
i.e.
=> hu + kv + lw = 0
(c) In the cubic system, equation of unit vector of the plane becomes
, where a is the lattice parameter
The direction vector for (hkl) is
Their cross product is
=
, which is equal to 0. It implies that they are parallel to each other.
(d) Let us consider the unit vector passing through h1k1l1 and h2k2l2 be
and respectively, where d
1 and d2 are the interplanar distances and a1 and a2 are the lattice parameters. Their cross product is, but their mod is equal to 1 as they are unit vectors. Hence, the equation reduces to
Also, interplanar distance is related to lattice parameter and position vector as
Hence,
Replacing a1 and a2, and the equation reduces to
….. proved
(e)
In the aforedrawn figure, Area of triangle OAB is equal to sum of areas of triangles of OAC and OCB. From trigonometry we know that area of triangle is ½ ab sinC (where a and b are the lengths of the sides of triangle and C is angle subtended between those sides. Hence,
, sin AOC = sin BOC = sin AOB *
=>
=> ………….. proved
* (Note = AOC = BOC = 60o and AOB = 120o, also sin 120o = sin 60o)
O
B
A
C
Y
X
U
a/h
a/k
a/i
Q 2) Derive an expression for the Fermi energy of a free electron metal at zero temperature. Using the data given, and any other constants, evaluate the Fermi energy of the alkali metals.
Li
Na
K
Rb
Cs
Density, gcm3
0.534
0.971
0.86
1.53
1.87
Atomic weight
6.939
22.99
39.102
85.47
132.905
How would you measure the Fermi energy experimentally for these metals?
Ans 2) Fermi energy is defined as the highest energy occupied by an electron at absolute zero.
According to FermiDirac statistics, the probability that a particle (fermion) will have energy (E) is given by
, and at absolute zero its value is 1  (1)
Suppose the volume (V) contains Ne electrons. Then,
, but =1 from equation 1. The equation reduces to:
Q 3) Derive an expression for the ratio
of the thermal and electrical conductivities of a freeelectron metal. Calculate the value of the Lorentz number
Where T is the absolute temperature. Explain the discrepancy between the calculated value and the following measured values of L for sodium at low temperatures.
T, oK
10
20
30
60
L, W ohm deg2
1.37 x 108
0.7 x 108
1.0 x 108
1.2 x 108
Ans 3)
Thermal conductivity (.) =
Electrical conductivity (s) =
Using these two equations we can derive an expression for the ratio
, which is
=>
 (1)
, it implies that
or
, where
, which is called as Lorentz number. Its value can be easily calculated by substitution the values of and e. (is 1.3807 x 1023 and e is 1.6 x 1019)
=>
=> Wohm/K2
Lorentz number equation is
, which doesn’t depend on Temperature
i.e. Lorentz number is temperatureindependent and its value depends on the values of Boltzmann constant and electronic charge. But in reality Lorentz number depends on the relaxation processes for electrical and thermal conductivity being the same, which is not true for all the temperatures, and because of various assumptions taken (according to Drude’s theory), there is a discrepancy between the calculated values and the measured values.
Q 4) Assuming that silver is a monovalent metal with a spherical Fermi surface, calculate the following quantities: (i) Fermi energy and Fermi temperature
(ii) Radius of the Fermi surface
(iii) Fermi velocity
(iv) Crosssectional area of the Fermi surface
(v) Cyclotron frequency in a field of 5000 oersted
(vi) Mean free path of electrons at room temperature and near absolute zero
(vii) Orbital radius in a field of 5000 oersted
(viii) Length of the side of the cubic unit cell
(ix) Lengths of the first two sets of reciprocal lattice vectors in kspace
(x) Volume of the Brillouin zone
Density of silver = 10.5 g cm3
Atomic weight = 107.87 g
Resistivity = 1.61 x 106 ohm cm at 295oK and 0.0038 x 106 ohm cm at 20oK
Ans 4) For silver
(referred page 36, Ashcroft/Mermin)
(i)
Fermi energy )21.50areV and Fermi temperature
=> Fermi energy
is 5.493 eV and Fermi temperature is 6.381 x 104 K
(ii)
Radius of the Fermi sphere Å1
=> Radius of the Fermi sphere
is 1.201 Å1
(iii)
Fermi velocity
=> Fermi velocity is 1.39 x 108 cm/sec
(iv)
Crosssectional area of the Fermi surface 24F Å2
=> Crosssectional area of the Fermi surface A is 18.125 Å2
(v)
Cyclotron frequency in a field of 5000 oersted
=> .c = 2.8H x 106 Hz = 14 x 109 Hz or 14 TeraHz
(vi)
Mean free path of electrons at room temperature and near absolute temperature
=> Mean free path is
=> lRT = 521.17 Å and lNAZ = 22.08 µ
(RT is Room temperature and NAZ is Near Absolute Zero)
(vii)
Orbital radius in a field of 5000 oersted
, and
=>
= 2.9 x 106 s1 and r = 103.44 m
(viii)
Length of the side of the cubic cell
Given the density of Silver (10.5 gcm3) and Atomic weight (107.87 amu)
Atomic weight = 107.87 x 1.66 x 1024 g = 1.79 x 1022 g
Density of silver = 10.5 = mass of one atom/volume of one atom = 1.79 x 1022/(4pr3/3)
=> r3 = 4 x 1024, and r = 1.588 .
As Silver is a monovalent element, lattice parameter is
times of radius
=> lattice parameter (a) = 4.46 .
(ix)
Lengths of the first two sets of reciprocal lattice vectors in kspace
It’s nothing but
=> lengths = 1.409 .1 each
(x)
Volume of the Brillouin zone
Volume of Brillouin zone is nothing but the volume of reciprocal lattice in kspace
i.e. VB = , where V is the volume of the unit cell i.e. a3
=> VB = , and V is 88.72 .3
=> VB = 2.796 .3
Q 5) Show that the mean energy per particle at 0oK for electrons obeying FermiDirac statistics is
,
Where EF(0) is the Fermi energy at T = 0. Assuming the value of this quantity at a finite temperature is
Find the value of the ratio (Cv)FD/(Cv)Cl for an electron gas with Fermi energy 7 eV, where (Cv)FD is the specific heat of a gas of particles obeying FermiDirac statistics and (Cv)Cl is the specific heat of a gas obeying classical statistics.
Ans 5)
Total energy of electrons at 0oK can be expressed as:
… equation (1)
We know that
Thus equation (1) reduces to
 (equation 2)
But we know that at absolute zero Fermi energy expression is:
(proved earlier, refer solution to problem number 2)
 equation (3)
Equation 2 can rewritten as
This equation finally reduces to
, where Ne is total number of electrons
mean energy per particle is nothing but Total energy divided by total number of electrons
i.e.
, where EF is at absolute zero i.e. EF(0)
Hence
As given in the problem the expression for mean energy per particle at a finite temperature T is
We know that (Cv)FD is nothing partial derivative of Emean at constant volume, the equation for (Cv)FD is
We know that specific heat for gas obeying classical statistics is given by
The ratio
, given EF(0) = 7 eV and = 0.8617 x 104 eV/K
, where T is the temperature in absolute scale
Hence the ratio of (Cv)FD/(Cv)Cl for an electron gas with Fermi energy 7 eV is 4.05 x 105 T
