The document discusses the atomic arrangements in crystalline solids. It defines key concepts such as crystals, lattice, basis, unit cell, coordination number, Miller indices, and common crystal structures including simple cubic, body centered cubic, and face centered cubic. It provides examples of calculating the number of atoms in a unit cell, lattice parameter, atomic packing factor, atomic concentration, and drawing and identifying crystallographic planes from their Miller indices.

1. Atomic arrangements in Solids

2. Crystal: Definition
⚫A crystalline solid is a solid in which the
atoms bond with each other in a regular
pattern to form a periodic array of
atoms.
⚫Crystals are constructed by the infinite
repetition of identical structural units in
space.
⚫The most important property of a crystal is
periodicity, which leads to what is termed
long-range order.

3. Lattice and Basis
⚫All crystals can be described in terms of a
lattice and a basis.
⚫A lattice is an infinite periodic array of
geometric points in space, without any
atoms.
⚫When we place an identical group of
atoms or molecules, called a basis, at
each lattice point, we obtain the actual
crystal structure.

4. Lattice and Basis
⚫ The crystal is thus a lattice plus a basis at each lattice point.
⚫ Crystal = Lattice + Basis
⚫ Crystals with the same lattice but different basis create
different crystalline solids but have the same symmetry.
⚫ Here, we have a simple square 2D lattice, and the basis has 2
atoms.

5. Unit Cell
⚫ The unit cell is the most convenient building block unit in
the crystal structure that carries the properties of the
crystal. The repetition of the unit cell in three
dimensions generates the whole crystal structure.
⚫ In the previous example, the unit cell is a simple square
with two atoms.

6. Unit Cell
⚫ Crystals are made up of 3-dimensional arrays of
atoms. Since unit cell is the smallest repeating
unit having the full symmetry of a crystal
structure, crystal structure is described in terms
of the geometry of arrangement of particles in the
unit cell.
⚫ Unit cells can be visualized in the 2 following ways:
◦ Hard sphere representation: Atoms are denoted by
hard, touching spheres showing their realistic
arrangement.
◦ Reduced sphere representation: For clarity, it is
often more convenient to draw the unit cell with the
spheres reduced in size and the distance between
spheres highly exaggerated.

7. Common Crystal Structures
⚫Simple Cubic (SC)
⚫Body Centered Cubic (BCC)
⚫Face Centered Cubic (FCC)

8. Simple Cubic
⚫ The unit cell is cubic with one atom at each of the eight
corners.
⚫ Because atoms are shared with neighboring cells, only one-
eighth of the corner atoms belong to each unit cell. So the
number of atoms per unit cell is: 8 corners x 1/8 = 1
⚫ Rare due to low packing density (0.52), only Polonium (Po)
has this structure.
Close packed hard sphere unit cell Reduced sphere unit cell

9. BCC Crystals
⚫ The unit cell is cubic with one atom at each of the eight corners
and one atom at the center of the cell.
⚫ Because atoms are shared with neighboring cells, only once-
eighth of the corner atoms belong to each unit cell. So there are
effectively 2 atoms in the space of the unit cell (1 center + 8
corners x 1/8).
Close packed hard sphere unit cell Reduced sphere unit cell
Note: In a
true BCC
cell all
atoms are
identical,
the center
atom is
shaded
differently
only for
ease of
viewing.

10. BCC Crystals
⚫The body centered atom is in contact with all
the eight corner atoms. Each corner atom is
shared by eight unit cells and hence, each of
these atoms is in touch with eight body
centered atoms.
⚫Atoms touch each other along cube
diagonals.
⚫Examples: iron, chromium, tungsten, niobium,
vanadium, barium.
⚫Volume of the BCC unit cell is 68 percent
full of atoms, which is high but lower than the
maximum possible packing.

11. Generation of BCC structure
through unit cell repetition
Note: In the simulation some space is allowed between the atoms for better visibility, but
originally the atoms in a BCC structures touch each other along cube diagonals.

12. FCC Crystals
⚫ The unit cell is cubic with one atom at each of the eight corners
and one atom at the center of each of the six faces.
⚫ Because atoms are shared with neighboring cells, only one-eighth
of the corner atoms and half of the face centered atom belong to
each unit cell. There are effectively 4 atoms in space of the unit
cell (6 face x 1/2 + 8 corners x 1/8).
Close packed hard sphere unit cell Reduced sphere unit cell

13. FCC Crystals
⚫Atoms touch each other along face
diagonals.
⚫Examples: copper, nickel, gold, and silver.
⚫Volume of the FCC unit cell is 74
percent full of atoms, which is the
maximum possible packing possible
with identical spheres.
⚫Also known as Close Packed Cubic
(CPC) structure.

14. Generation of FCC structure
through unit cell repetition
Note: In the simulation some space is allowed between the atoms for better visibility, but
originally the atoms in a FCC structures touch each other along face diagonals.

15. ⮚ The number of nearest neighbours of which an
atom has in the unit cell of any crystal
structure.
Examples
⮚ Simple Cubic (SC) coordination number = 6
⮚ Body-Centered Cubic(BCC) coordination number = 8
⮚ Face-Centered Cubic(FCC) coordination number = 12
Coordination Number

16. • Coordination # = 6
(# nearest neighbors)
SIMPLE CUBIC STRUCTURE (SC)

17. Coordination # = 8
BODY CENTERED CUBIC STRUCTURE(BCC)

18. • Coordination # = 12
FACE CENTERED CUBIC STRUCTURE (FCC)

19. Number of atoms in unit cells

20. Atomic Radius
Simple Cubic (SC) Lattice

21. Atomic Radius of BCC latice

23. Atomic Radius of FCC lattice

24. Atomic Packing Factor
⚫Atomic packing factor (APF) or packing
efficiency indicates how closely atoms are
packed in a unit cell and is given by the
ratio of volume of atoms in the unit cell
and volume of the unit cell.

27. =74%

28. Unit cell geometry
⚫ To establish a reference frame
and to apply three-dimensional
geometry, we insert an xyz
coordinate system. The x, y, and
z axes follow the edges of the
parallelepiped and the origin is at
the lower-left rear corner of the
cell.
⚫ The unit cell extends along the x
axis from 0 to a, along y from 0
to b, and along z from 0 to c.
⚫ Conventionally the geometry of the unit cell is represented as
a parallelepiped with sides a, b, and c and angles α, β, and γ,
as depicted in figure. The sides a, b, and c and angles α, β, and
γ are referred to as lattice parameters.

29. Miller lndices
⚫ Miller indices of a plane refers to the general
convention for describing a particular plane in
a crystal based on three-dimensional geometry.
⚫ They are the reciprocals of the (three) axial
intercepts for a plane, cleared of fractions & common
multiples.
⚫ Algorithm
1) Choose a origin and unit cell of convenience (the plane
should not pass through the origin, if it does, shift the
origin or draw a plane parallel to it by an appropriate
translation).
2) Read off intercepts of plane with axes in terms of a, b, c
3) Take reciprocals of intercepts
4) Clear all fractions without reducing to smallest integer
values
5) Enclose in parentheses, no commas i.e., (hkl)

30. Miller lndices Example
⚫ Intercepts of the plane are a/2, b, and ∞ along x, y, z axis (the plane
is parallel to the z axis) respectively, so intercepts in terms of a, b,
c are ½, 1, and ∞.
⚫ Taking reciprocals of intercepts we get 1/(½) , 1/1, 1/ ∞ = 2,1,0.
⚫ This set of numbers does not have fractions, so it is not necessary
to clear fractions.
⚫ Hence, the Miller indices (hkl) are (210).

31. Miller lndices
⚫ If there is a negative integer due to a negative intercept, a
bar is placed across the top of the integer.
⚫ Also, if parallel planes differ only by a shift that involves a
multiple number of lattice parameters, then these planes may
be assigned the same Miller Indices. indices. For example, the
plane (010) is the xz plane that cuts the y axis at —b. If we
shift the plane along y by two lattice parameters (2b), it will
cut the y axis at b and the Miller indices will become (010).
¯

32. Miller lndices
⚫ Planes can have the same Miller indices only if they are
separated by an integer multiple of the lattice
parameter. For example, the (010) plane is not identical to
the (020) plane, even though they are geometrically parallel.
The (020) plane cannot be shifted by the lattice parameter b
to coincide with plane (010).
(010) (020)

33. Drawing Planes from Miller Indices
◦ Draw unit cell
◦ Shift origin if necessary. Must if the plane passes through origin,
advisable if negative integers present in Miller Indices. (Example: If
negative integer is along z axis, move origin along the positive
direction of z axis.)
◦ Invert the given indices to get intercepts
◦ Plot the intercepts along the x, y, z axis for non-∞ values.

34. Drawing Planes from Miller Indices
◦ For a single ∞ value, plot the
other 2 values and connect.
Then draw 2 lines along the
axis with the ∞ value, starting
from the intercept points and
ending at the unit cell surface.
Connect the points where
these lines end at the surface.
Example: (110), (101)
¯

35. Drawing Planes from Miller Indices
◦ For two ∞ values, plot the single non-∞
value and draw 2 lines along the axes
with the ∞ value, starting from the
intercept point and ending at the unit
cell surface. Draw another 2 lines the
same way, but starting from the points
where the previous lines ended.
Example: (100),(100)
¯

36. Example 1.13
⚫ Consider the FCC unit cell of the copper crystal
shown in the figure.
a. How many atoms are there per unit cell?
b. If R is the radius of the Cu atom, show that the
lattice parameter a is given by a = R 2√ 2
c. Calculate the APF.
d. Calculate the atomic concentration (number of
atoms per unit volume) and the density of the
Crystal given that the atomic mass of copper is
63.55 gmol-1 and the radius of the Cu atom is
0.128nm.

37. Example 1.13
a)
There are four atoms per
unit cell. The Cu atom at
each corner is shared with
eight other adjoining unit
cells. Each Cu atom at the
face center is shared with
the neighboring unit cell.
Thus, the number of atoms in the unit cell
= 8 corners (1/8 atom) + 6 faces (1/2 atom)
= 4 atoms.

38. b)
Consider one of the cubic
faces in above figure.
The face is a square of side a
and the diagonal
is or .
The diagonal has one atom at
the center of diameter 2R,
which touches two atoms at
the corners.
Example 1.13
The diagonal, going from corner to corner, is therefore R + 2R +R
= 4R.
⇒ 4R =
⇒

39. Example 1.13
c)
From part (b), we know
that

40. d) In general, atomic concentration is:
where x= number of atoms in unit cell.
From part (a), we know that x =4 for Cu.
Since 1 mole of matter weighs Mat grams and contains NA atoms,
each atom weighs Mat/NA grams.
Example 1.13

41. Example 1.14
a. Find the Miller Indices for the plane shown in
the figure which passes through one side of a
face and the center of an opposite face in the
Cu FCC lattice with a=0.3620 nm.
b. Draw (100) and (110) crystallographic planes
for the lattice.
c. Calculate the planar concentration for each of
these planes.

42. Example 1.14
Since the plane passes through the origin at the lower-left
rear corner, we place the origin to point 0’ at the lower-right
rear corner of the unit cell.
In terms of lattice parameter a:
x, y, and z axes intercepts: ∞, -1, 1/2
reciprocals: 0, -1, 2
Miller indices: (012) ¯

43. Example 1.14
⚫To calculate the planar concentration
n(hkl) on a given (hkl) plane, we consider a
bound area A of the plane within the unit
cell and treat it as the repeat unit in 2D.
Only atoms whose centers lie on A are
involved in the calculation.
⚫For each atom, we then evaluate what
portion of the atomic cross section (a
circle in 2D) cut by the plane (hkl) is
contained within A.

44. Example 1.14
The (100) plane corresponds to a cube face
and has an area A = a2. There is one full
atom at the center; that is, the (100) plane
cuts through one full atom, one full circle in
two dimensions, at the face center.
However, not all corner atoms are within A.
Only a quarter of a circle is within the bound
area.

45. Example 1.14
⚫ Number of atoms in A = (4 corners) x 1/4
atom) + 1 atom at face center = 2
⚫ Planar concentration n(100) of (100) is:

46. Example 1.14
⚫ Consider the (110) plane. The number of atoms in
the area A = (a)(a √ 2) defined by two face diagonals
and two cube sides is:
⚫ (4 corners) x 1/4 atom) + (2 face diagonals) x (1/2
atom at diagonal center) = 2
⚫ Planar concentration n(110) of (110) is: