eee 315 - electrical properties of materials lecture 2

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EEE 315 - Electrical Properties of Materials Lecture 2

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Page 1: EEE 315 - Electrical Properties of Materials Lecture 2

EEE 315 - Electrical Properties of Materials

Lecture 2

Page 2: EEE 315 - Electrical Properties of Materials Lecture 2

Bravais Lattice

The unit cell of a general 3D lattice is described by 6 numbers

3 distances (a, b, c)3 angles (, , )

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Page 3: EEE 315 - Electrical Properties of Materials Lecture 2

Bravais Lattice

There are 14 distinct 3D lattices which come under 7 Crystal SystemsThe BRAVAIS LATTICES (with shapes of unit cells as) : Cube (a = b = c, = = = 90) Square Prism (Tetragonal) (a = b c, = = = 90) Rectangular Prism (Orthorhombic) (a b c, = = = 90) 120 Rhombic Prism (Hexagonal) (a = b c, = = 90, = 120) Parallelepiped (Equilateral, Equiangular)(Trigonal) (a = b = c, = = 90) Parallelogram Prism (Monoclinic) (a b c, = = 90 ) Parallelepiped (general) (Triclinic) (a b c, )

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Page 4: EEE 315 - Electrical Properties of Materials Lecture 2

Bravais Lattice

The lattice centering are:Primitive (P): lattice points on the cell corners

only.Body (I): one additional lattice point at the center

of the cell.Face (F): one additional lattice point at the center

of each of the faces of the cell.Base (A, B or C): one additional lattice point at

the center of each of one pair of the cell faces.

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Page 5: EEE 315 - Electrical Properties of Materials Lecture 2

Bravais Lattice

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Page 6: EEE 315 - Electrical Properties of Materials Lecture 2

Miller Indices

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The planes passing through lattice points are called ‘lattice planes’.

The orientation of planes or faces in a crystal can be described in terms of their intercepts on the three axes

Miller introduced a system to designate a plane in a crystal.

He introduced a set of three numbers to specify a plane in a crystal.

This set of three numbers is known as ‘Miller Indices’ of the concerned plane.

Page 7: EEE 315 - Electrical Properties of Materials Lecture 2

Miller Indices

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Miller indices is defined as the reciprocals of

the intercepts made by the plane on the three

axes.

Page 8: EEE 315 - Electrical Properties of Materials Lecture 2

Miller Indices

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Procedure for finding Miller Indices

Step 1: Determine the intercepts of the plane along the axes X,Y and Z in terms of the lattice constants a,b and c.

Step 2: Determine the reciprocals of these numbers.

Page 9: EEE 315 - Electrical Properties of Materials Lecture 2

Miller Indices

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Step 3: Find the least common denominator (lcd) and multiply each by this lcd.

Step 4:The result is written in parenthesis.

This is called the `Miller Indices’ of the plane in the form (h k l).

Page 10: EEE 315 - Electrical Properties of Materials Lecture 2

Miller Indices

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Plane ABC has intercepts of 2 units along X-axis, 3

units along Y-axis and 2 units along Z-axis.

PLANES IN A CRYSTAL

ILLUSTRATION

Page 11: EEE 315 - Electrical Properties of Materials Lecture 2

Miller Indices

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DETERMINATION OF ‘MILLER INDICES’

Step 1:The intercepts are 2,3 and 2 on the three axes.

Step 2:The reciprocals are 1/2, 1/3 and 1/2.

Step 3:The least common denominator is ‘6’. Multiplying each reciprocal by lcd, we get, 3,2 and 3.

Step 4:Hence Miller indices for the plane ABC is (3 2 3)

ILLUSTRATION

Page 12: EEE 315 - Electrical Properties of Materials Lecture 2

Miller Indices

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Example

• In this plane, the intercept along X axis is 1 unit.

• The plane is parallel to Y and Z axes. So, the intercepts along Y and Z axes are ‘’.

• Now the intercepts are 1, and .

• The reciprocals of the intercepts are = 1/1, 1/ and 1/.

• Therefore the Miller indices for the above plane is (1 0 0).

Page 13: EEE 315 - Electrical Properties of Materials Lecture 2

Miller Indices

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SOME IMPORTANT PLANES

Page 14: EEE 315 - Electrical Properties of Materials Lecture 2

Miller Indices

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Problem # 1

A certain crystal has lattice parameters of 4.24, 10 and 3.66 Å on X, Y, Z axes respectively. Determine the Miller indices of a plane having intercepts of 2.12, 10 and 1.83 Å on the X, Y and Z axes.

Lattice parameters are = 4.24, 10 and 3.66 ÅThe intercepts of the given plane = 2.12, 10 and 1.83 Åi.e. The intercepts are, 0.5, 1 and 0.5.Step 1: The Intercepts are 1/2, 1 and 1/2.Step 2: The reciprocals are 2, 1 and 2.Step 3: The least common denominator is 2.Step 4: Multiplying the lcd by each reciprocal we get, 4, 2 and 4.Step 5: By writing them in parenthesis we get (4 2 4)

Therefore the Miller indices of the given plane is (4 2 4) or (2 1 2).

Page 15: EEE 315 - Electrical Properties of Materials Lecture 2

Miller Indices

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Problem # 2

Calculate the miller indices for the plane with intercepts 2a, - 3b and 4c the along the crystallographic axes.

The intercepts are 2, - 3 and 4

Step 1: The intercepts are 2, -3 and 4 along the 3 axes

Step 2: The reciprocals are 1/2, -1/3, 1/4

Step 3: The least common denominator is 12.

Multiplying each reciprocal by lcd, we get 6 -4 and 3

Step 4: Hence the Miller indices for the plane is

Page 16: EEE 315 - Electrical Properties of Materials Lecture 2

Classical Theory of Electrical and Thermal Conduction

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Page 17: EEE 315 - Electrical Properties of Materials Lecture 2

Electrical & Thermal Conduction

Electrical conduction: Motion of charges (conduction electrons) in a material under the influence of an applied electric field

Thermal conduction: Conduction of thermal energy from higher to lower temperature regions in a material

Thermal conduction involves carrying of energy by conduction electrons.

Good electrical conductors are also good thermal conductors!

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Page 18: EEE 315 - Electrical Properties of Materials Lecture 2

Next Week!

QUIZ!

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