Capítulo 5: Contrastes de estilo
5.3 Potencialidades infinitas
Consider the diffraction of x-rays from a single crystal sample as shown in Fig.2.21. Let XX’ and YY’ be two parallel planes with an interplanar spacing d.
A D F C B E M N θ d x x′ y y′ θ
Figure 2.21 Diffraction of x-rays by crystal planes.
Let AB represent an x-ray incident at an angle θ to the plane which is scattered along BC. DE is a parallel ray scattered along EF. The two rays BC and EF should interfere constructively to produce a resultant diffracted beam. The path difference
between the two rays can be found by drawing perpendiculars BM and BN to the rays DE and EF respectively.
The total path difference = ME + EN = d sin θ + d sin θ
= 2d sin θ (2.10)
The two reflected rays will be in phase if this path difference is equal to an integral multiple of λ, the wavelength of x-rays.
Hence, the condition for producing a maximum becomes
nλ = 2d sin θ (2.11) where n is an integer called order of diffraction. This is known as Bragg’s law.
2.3.2 Bragg’s spectrometer
Bragg’s spectrometer (more commonly referred to as x-ray diffractometer) is an in- strument used to study the angles and intensities of diffracted beams for a given crystal sample. A conventional Bragg spectrometer is shown in Fig.2.22.
S F S T C D R S 1 2
Figure 2.22 Bragg spectrometer.
It consists of an x-ray source S producing characteristic x-rays of required wave- length. Unwanted continuous x-rays as well as characteristic x-rays of other wave- lengths are removed from the beam using a suitable absorption filter. The monochro- matic x-ray beam is then passed through a pair of slits S1 and S2 which will define the divergence of the beam. The x-rays will then be incident on the crystal sample C
mounted at the centre of a turn table T. The turn table also carries a side arm on which a detector D for x-rays (usually an ionization counter like G.M. counter) is mounted. The sample holder and the detector can be rotated about an axis perpendicular to the direction of the incident x-rays. The rates of rotation of the crystal holder and the detector arm are such that the detector always receives the x-rays reflected from the crystal. When x-rays are incident on any particular plane at an angle θ, the diffracted beam which will be at an angle 2θ will be received by the detector. The output of the detector is measured or recorded on a x-y recorder R to obtain a chart showing the variation of diffracted intensity of x-rays as a function of diffraction angle.
The verification of Bragg’s law is carried out by mounting a standard sample on the turn table of the spectrometer. The x-rays are allowed to fall on the sample at an angle close to the glancing angle (θ = 0) and the angle is gradually increased. The detector output observed at angle 2θ in each case is measured and the angles corresponding to peak intensities are noted. For the standard sample, knowing the crystal structure and the lattice parameters, the interplanar spacings may be calculated. For example, for a cubic crystal, the permitted d-spacing can be calculated as
d = a
(h2+ k2+ l2)1/2 (2.12)
Hence, Bragg’s law can be rewritten as
sin2θ = n 2λ2 4a2 (h
2+ k2+ l2) (2.13)
Knowing, λ and a, by assigning different integer values for h, k, l and order n, allowed values of θ can be computed. These values of θ are compared with those obtained from the Bragg spectrometer recordings. A perfect match between the computed values and the observed values of Bragg angles will be obtained thereby establishing the validity of Bragg’s law.
2.3.3 Structure determination
The Bragg spectrometer analysis is a widely used experimental technique for the rou- tine determination of crystal structure. It is very much suitable for identification and structure determination of crystals possessing high symmetry (eg. cubic crystals). It is also possible to distinguish between the lattices of the cubic system using extinction rules.
The θ values for different planes can be obtained from the spectrometer recording. For a given cubic lattice, it is possible to list of all combinations of h, k and l. It is seen
that sin2θ values will increase with (h2+ k2+ l2). Since λ is a constant and known, the lattice parameter ‘a′ can be found out for which sin2 values will be in the same ratio as (h2+ k2+ l2).
The distinction between lattices of cubic system (i.e s.c., b.c.c. and f.c.c.) is possi- ble by using the fact that not all combinations of (h2+ k2+ l2) lead to diffraction for a given lattice. Some planes may not give diffraction peaks if the net diffracted intensity add up to be zero.
It is possible to derive the extinction rules for different cubic lattices. These rules are listed below:
Lattice type Allowed reflections 1. Simple cubic All values of (h2+ k2+ l2) 2. Body centred cubic Even values of (h2+ k2+ l2) 3. Face centred cubic h, k, l, all odd or all even.
From the extinction rules, we can derive the ratio of (h2+k2+l2) values and compare with the observed sin2θ value to identify the lattice type. The ratio of (h2 + k2 + l2) values (or sin2θ values) will be as follows:
Lattice type Ratio
1. Simple cubic 1:2:3:4:5:6:8:9:.... 2. Body centred cubic (2:4:6:8:10:12:14:16: ...)
1:2:3:4:5:6:7:8:... 3. Face centred cubic 3:4:8:11:12:16:19:20:...
It may be mentioned that (h2 + k2+ l2) = 7, 15, . . . etc., correspond to forbidden reflections as the sum of the squares of three integers can not be 7, or 15, etc. Thus, if 7 appears in the ratio of (h2+ k2+ l2) values, it must correspond to 14 and the lattice type will be b.c.c. If 7 is absent, the lattice is simple cubic. Face centred lattice can be easily identified by the distinct ratio.
2.4
Electron diffraction
De Broglie’s hypothesis suggests that an electron of mass m travelling with a velocity vis associated with a de Broglie wavelength λ given by
λ = h p =
h mv
It is interesting to know that Davisson and Germer’s experiment on the scattering of electrons shows that electrons can be diffracted and the diffraction obeys Bragg’s law.
An electron diffraction apparatus consists of a long evacuated column with a fila- ment suitably heated to emit electrons (Fig.2.23).
filament Anode sample
Screen
Figure 2.23 Experimental study of electron diffraction.
These electrons are accelerated by a high positive potential applied to an anode. The accelerated electrons passing through the small aperture in the anode are made to fall on a crystalline sample. The diffraction pattern may be observed on a fluorescent screen or may be photographed. The measurements made on the diffraction pattern indicate that the diffraction of electrons is in accordance with the Bragg’s law.
The advantage of electron diffraction is that the wavelength of the electron beam can be conveniently modified to suit our requirement by altering the accelerating po- tential. But the electron beam has a much lower penetration power compared to that of x-rays and hence restricts its use to very thin samples.
2.5
Neutron diffraction
Neutrons produced in a nuclear reactor, after undergoing a large number of collisions will assume an average energy equal to the thermal energy kT which is of the order of 0.025eV. Such neutrons are called thermal neutrons and the de Broglie wavelength associated with them is about 1.8A◦. Hence, it can be used as an useful probe to study crystalline materials. The advantage of using neutrons for diffraction studies is due to the fact that they are neutral particles and are not influenced by the electron cloud or the nuclear charge of the atoms. Further, because of the magnetic moment carried by the neutrons, they have been used to probe the magnetic properties of crystals.
The main difficulty with neutron diffraction is to obtain mono energetic neutron beams. Neutron monochromators, which are modified versions of Bragg spectrometer, can be used to obtain neutron beams of thermal as well as higher energies. The detec- tion and recording of diffracted beams is also complicated since ionization counters do not detect neutrons directly.
Numerical Examples
2.1 A cubic crystal has a lattice parameter of 5.6A◦. Calculate the interplanar spacing
for (110) and (111) set of planes. Solution:
d = a (h2+ k2+ l2)1/2
where h,k,l are the Miller indices of the set of planes under consideration. For (110) planes, d110= 5.6 (12+ 12+ 02)1/2 = 5.6 √ 2 = 3.96A ◦(Ans.) For (111) planes, d111= 5.6 (12+ 12+ 12)1/2 = 5.6 √ 3 = 3.23A ◦(Ans.)
2.2 A cubic crystal of lattice parameter 4.31A◦ has a set of planes with interplanar
spacing of 1.76A◦. Find the Miller indices of the set of planes. Solution: (h2+ k2+ l2)1/2 = a d = 4.31 1.76 = 2.45 (h2+ k2+ l2) = (2.45)2 = 6
Sum of the squares of three integers is 6. Hence, the Miller indices of the planes are (211). Since it is cubic system, (211), (121), and (112) are identi- cal.
2.3 Find the Miller indices of a set of parallel planes which makes intercepts in the ratio 3a : 4b on the x and y axis , and are parallel to the z axis where a, b and c are the primitive translation vectors of the lattice.
Solution:
Intercepts as multiples of 3a 4b ∞ lattice parameter
Devide by lattice parameter 3 4 ∞
Take reciprocals 1/3 1/4 1/∞
Reduce to smallest set 4 3 0 of integers
Miller indices of the given set of planes are(430) (Ans.).
2.4 Show that in a cubic crystal, the interplanar spacing of the (300) set of planes is equal to the inter planar spacing of the (221) set of planes.
Solution:
Interplanar spacing is given by
d = a/(h2+ k2+ l2)1/2 d300= a/(32+ 0 + 0)1/2 = a/3
d221= a/(22+ 22+ 12)1/2 = a/3
Hence the interplanar spacing of the set of planes (300) and (221) are equal. 2.5 A cubic crystal of lattice parameter 5.60A◦ has a set of planes with interplanar
spacing of 1.77A◦. Find the Miller indices of the set of planes.
Solution: We have
d = a/(h2+ k2+ l2)1/2 Or (h2+ k2+ l2)1/2 = a/d
= 5.60/1.77 = 3.16 ≈ 3
Hence, (h2+ k2 + l2) is equal to 9. The possible values of miller indices are either (300) or (221).
2.6 In an x-ray tube, an accelerating voltage of 35 kV is applied between the filament and the target. Calculate the minimum wavelength of continuous x-ray emitted.
Solution: The minimum wavelength of x-rays emitted is given by
λmin = hc eV 6.62 × 10−34× 3 × 108 1.6 × 10−19× 35 × 103 = 3.55 × 10−11m (Ans.).
2.7 First order Bragg diffraction of x-rays of wavelength 1.54 A◦was observed at an
angle of 12◦. Calculate the interplanar spacing for the set of planes responsible for the diffraction. If the crystal has a lattice parameter of 6.41 A◦, find the planes giving rise to diffraction.
Solution: d = nλ 2 sin θ = 1 × 1.54 2 × sin 12◦ d = a/(h2+ k2+ l2)1/2 ∴(h2+ k2+ l2)1/2= a/d = 6.41/3.70 = 1.73 ∴(h2+ k2+ l2) = (1.73)2 = 3.
Hence, the set of planes are (111) (Ans.).
2.8 An x-ray diffraction pattern for cubic lead crystal is obtained with x-rays of wave- length 1.54 A◦. If the first order reflection from (110) planes is observed at 12.7◦,
determine the interplanar spacing for (110) planes and also the lattice parameter of lead. Solution: Interplanar spacing d = nλ 2 sin θ = 1 × 1.54 2 × sin 12.7 = 3.50A◦(Ans.) Lattice parameter a = d(h2+ k2+ l2)1/2 = 3.50 × √2 = 4.95A◦. (Ans.).
2.9 The Kα line of x-ray spectrum from a molybdenum target (z = 42) has a wave-
length of 0.71A◦. Calculate the wavelength of Kα line when a copper target
(z = 29) is used. Solution:
ν = 3cR
4(z − 1)2 where R is the Rydberg constant. λ(cu) λ(Mo) = (42 − 1)2 (29 − 1)2 ∴λ(Cu) = 0.71(41) 2 (28)2 = 1.52A◦(Ans.).
2.10 Calculate the minimum accelerating potential to be applied to an x-ray tube in order to produce continuous x-ray emission down to a wavelength of 1 Acirc.
Solution: We have λmin = hc/eV V = hc/eλmin = 6.62 × 10 −34× 3 × 108 1.6 × 10−19 × 1 × 10−10 = 12412.5V (Ans.).
2.11 In an x-ray diffraction experiment, the first order diffraction from a particular set of planes was observed at 12◦. Find the angle at which the second order diffraction occurs for the same set of planes.
Solution: nλ = 2d sin θ
For n = 1, sin θ1 =λ/2d = sin 120 For n = 2, sin θ2 = 2λ/2d = 2 sin 120
= 2 × 0.2079 = 0.4158
θ2 = sin−1(0.4158) = 24.570 (Ans.).
2.12 The Bragg diffracted beam of x-rays from a set of planes is observed at an angle of 15◦when a molybdenum target (λ = 0.71A◦) is used. Find the position of the diffracted beam for the same set of planes when a copper target (λ = 1.54A◦) is
used.
Solution: We have
nλ = 2d sin θ
Hence, nλ1 = 2d sin θ1 and nλ2 = 2d sin θ2 sin θ2 =λ2sin θ1/λ1
= 1.54 × sin 150/0.71 = 0.5614
θ2 = sin−1(0.5614) = 34.150 (Ans.).
2.13 The wavelength of Kαlines from a copper target is found to be 1.54 A◦. Find the
wavelength of Lα line from the same target.
Solution: We know that 1 α = R(z − b) 2 1 n21 − 1 n22 !
For Lαline, b = 7.4, n1= 2 and n2= 3
Substituting and taking the ratio of λKαand λLα λLα/λKα= 9.074
Hence, λLα = 9.074 × 1.54 = 13.97 × 10−10m (Ans.).
Exercise
2.1 State and explain Moseley’s law, and explain on of its important applications. (March 1999). 2.2 What is a space lattice ? Describe briefly the seven systems of crystals.
(March 1999). 2.3 Describe Bragg’s law for x-ray diffraction in crystals. Describe how Bragg spec- trometer is used for determination of wavelength of x-rays. (August 1999). 2.4 Explain Miller indices and indicate the following planes in a cubic unit cell:
(110), (101) and (111). (August 2000).
2.5 Explain Moseley’s law and mention its applications. (August 2000). 2.6 Derive Bragg’s law for x-ray diffraction. Describe the determination of wave- length of x-rays using Bragg spectrometer. (August 2000). 2.7 Explain Moseley’s law and its importance. (March 2001). 2.8 If the lattice parameter of a cubic crystal is 3 A.U.,find the interplanar spacing
between (111) planes. (March 2001)
2.9 Derive Bragg?s law of x-ray diffraction. Explain how Bragg spectrometer is used
in the verification of the law. (March 2001).
2.10 Define unit cell and Bravais lattice. Derive the expression for the interplanar spacing in terms of Miller indices. The interplanar spacing of (110) planes is 2 A.U. for a cubic crystal. Find out the atomic radius. (August 2001). 2.11 State Moseley’s law. Discuss its applications. Calculate the energy of electrons that produce Bragg diffraction of first order at an angle of 220 when incident on crystal planes with interplanar spacing of 1.8 A.U. (August 2001).
2.12 What are Miller indices? How do you find the Miller indices of a given plane? (March 2002).
2.13 Derive Bragg’s law. (July 2005)
2.14 Explain the structure of NaCl and calculate the packing factor for a bcc structure. (July 2005) 2.15 Obtain the expression for interplanar distance in terms of Miller indices.
(July 2005) 2.16 Sketch the following planes in a cubic unit cell : (101), (121), (010). (July2005)
Electrical Conductivity In Metals
3.1
Introduction
Metals have been known for their excellent ability to conduct heat and electricity. Such a high conductivity is attributed to the presence of free electrons in metals and attempts have been made to explain the conductivity of metals quantitatively. Earlier theories were classical in nature. In the year 1900, Paul Drude proposed a theory of conductiv- ity of metals based on the free electron model which was later developed and refined by Hendrik Lorentz. According to the classical free electron theory, a metal can be considered to be made up of ion cores comprising of nucleus and inner electrons ex- cluding the valence electrons. These ion cores are immobile and do not take part in the conduction process. The valence electrons are considered to be free and almost independent of the ions to which they belong. These valence electrons can move under the influence of an applied electric field and hence contribute to the conductivity.
Before discussing the classical free electron theory, let us examine the Ohm’s law which is the most fundamental of the laws governing the charge transport in materials. Consider a metal sample to which a voltage source is connected. This results in the flow of an electric current through the sample. The magnitude of the current I is found to be proportional to the applied voltage V and the relation is expressed as
I ∝ V or I = GV (3.1) where G is a constant of proportionality called the conductance of the sample. The relation may also be expressed in terms of resistance R of the sample as
V = IR where R = 1/G (3.2) This relation is known as Ohm’s law. The quantities involved in this relation are sample specific and depend on the dimensions of the sample. A more general form of Ohm’s law is
J = σE (3.3)
where J is the current density representing the charge moving past a reference point per unit area of cross section of the sample per unit time, E is the Electric field equivalent to the voltage per unit length of the sample and σ is the electrical conductivity of the
material. The electrical resistivity is the reciprocal of electrical conductivity and is defined as the electrical resistance per unit area per unit length of the sample and is given by the relation
ρ = 1 σ =
RA
l (3.4)
where A is the area of cross section and l is the length of the sample. The advantage of using Ohm’s law in its general form is that the quantities involved are not specific to the sample and are independent of sample dimensions. The electrical resistivity is a constant for the material and may be used to specify the relative ease with which an electric current passes in metals.