Prof.P. Ravindran, - folk.uio.nofolk.uio.no/ravi/CMP2013/7. problem_rf.pdf · · 2013-10-17P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Problems & Solutions -I Geometry

P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Problems & Solutions -I

http://folk.uio.no/ravi/CMP2013

Prof.P. Ravindran, Department of Physics, Central University of Tamil

Nadu, India

Problems & Solutions -I

1


Geometry of a Cube

2

Diagonal Face

222

af

aaf

3

3

DiagonalBody

2222

ab

afab

2


Fundamental Properties of Matter

States of Matter

1. Solids – Definite volume, definite shape.

2. Liquids – Definite volume, no fixed shape. Flows.

3. Gases – No definite volume, no definite shape. Takes the

volume and shape of its container.

Matter: - Has mass, occupies space

Mass – measure of inertia - from Newton’s first law of

motion. It is one of the fundamental physical properties.

3


Regarded as fourth state of matter. No definite

volume, no definite shape. Composed of electrically

charged particles.

Fully ionized gas at low density with equal amount

of positive and negative charges – net electrically

neutral.

Affected by electric and magnetic fields.

Plasma is the main state of matter in planetary

objects such as stars.

Plasma: 4


Regarded as fifth state of matter obtained when

atoms/molecules are at very low temperature and

their motion is halted.

They lose their individual identity and become a

different entity.

Bose-Einstein condensates – Formed by bosons.

Fermionic condensates – By fermions.

Condensate: 5


Spin Occupancy Examples

Fermions Half

integral

Only one per

state

electrons,

protons,

neutrons,

quarks,

neutrinos

Bosons Integral

spin

Many

allowed

photons, 4He

atoms, gluons

6


7

The energy required to change one gram of

a solid at its melting point into a liquid is

called the heat of fusion.


8

The heat of fusion of ice at 0oC is 80. cal/g.

How many calories of energy are needed to change

10.0 g of ice at 0.00oC to water 20.0oC?

Determine the calories necessary to melt 10.0 g of ice.

800. cal 10.0 g80. cal

=1 g

P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Problems & Solutions -I 9

How many calories of energy are needed to change 10.0 g of

ice at 0.00oC to water 20.0oC?

Determine the calories necessary to heat 10.0 g of water from

0.00oC to 20.0oC.

The total heat absorbed by the system is the heat required to melt the ice

plus the heat required to raise the water temperature from 0.00oC to

20oC.

800 cal + 200 cal = 1000 cal

The specific heat of water is 1.00 cal/goC.

200 cal 10.0 go

1.00 cal

1 g C

o20.0 C =


10

The energy required to change one gram of a

liquid at its boiling point into a vapor is called

the heat of vaporization.


11


20.0 g of water at 20.0oC to steam at 100.0oC?

Determine the calories necessary to heat 20.0 g from 20oC to

100oC.

1600 cal 20.0 go

1.00 cal

g C

The specific heat of water is 1.00 cal/goC.

(mass) (sp.ht) (Δt) = energy

o o100. C - 20.0 C



20.0 g of water at 20.0oC to steam at 100.0oC?

Determine the calories necessary to change 20.0 g of water at 100.oC

to steam at 100oC.

10800 cal 20.0 g540 cal

1 g

The total heat absorbed by the system is the heat required to raise the

water temperature from 20oC to 100oC plus the heat required to change

the water to steam.

1600 cal + 10800 cal = 12400 cal

The heat of vaporization of water at 100oC is 540 cal/g.


Heating (cooling curve)

13


How much heat (in kJ) is needed to convert 80.0 g of ethanol at -130oC

to vapor at 96oC? The specific heat of solid, liquid, and vapor ethanol

are 0.97, 2.4, and 1.2 J/g.oC, respectively. DHfus= 5.02 kJ/mol, DHvap=

39.3 kJ/mol. The mp is -114oC and bp is 78oC.

Step 1: heat solid up to mp (-114oC), remember q=smDT

q1=0.97 J/g.oC x 80.0 g x (-114oC -(-130oC))x 1 kJ/1000J=1.24 kJ

Step 2: melt solid to liquid

q2= 80.0g x 1 mol C2H5OH/46.07 g x 5.02 kJ/mol=8.72 KJ

Step 4: convert liquid to vapor

q4= 80.0g x 1 mol C2H5OH/46.07 g x 39.3 kJ/mol=68.2 KJ

Step 3: heat liquid from mp (-114oC) to bp (78oC)

q3=2.4 J/g.oC x 80.0 g x (78oC -(-114oC))x 1 kJ/1000J=36.9 kJ

Step 5: heat vapor from 78oC to 96oC

q5=1.2 J/g.oC x 80.0 g x (96oC -78oC)x 1 kJ/1000J=1.73 kJ

Total heat=q1 +q2 + q3+ q4 + q5 =117 kJ

14


15

The Liquid State

How many joules of energy must be absorbed by 5.00 x 102 g of

H2O at 50.0oC to convert it to steam at 120oC? The molar heat

of vaporization of water is 40.7 kJ/mol and the molar heat

capacities of liquid water and steam are 75.3 J/mol oC and 36.4

J/mol oC, respectively.


16

The Liquid State

? J = 27.8 mol J

mol J

40 7 101131 10

35.

.

? .

.. . .

mol = 500 g H O1 mol H O

g H O mol H O

1st let's calculate the heat required to warm water from 50 to 100 C

? J = 27.8 mol J

mol CC J

22

2

2

o

o

o

1827 8

753100 0 50 0 105 105

Next, let’s calculate the energy required to boil the water.

Finally, let’s calculate the heat required to heat steam from

100 to 120oC.

? J = 27.8 mol J

mol C120.0 -100.0 C J

o

o36 40 20 105..


17

The Liquid State

The total amount of energy for this process is the sum of

the 3 pieces we have calculated

105 10 1131 10 0 20 10

12 56 10

5 5 5

5

. . .

.

J J J

J or 1.26 10 kJ3


18

The Liquid State

If 45.0 g of steam at 140oC is slowly bubbled into 450 g of

water at 50.0oC in an insulated container, can all the steam

be condensed?


19

The Liquid State

condensed. becannot steam theof all Thus

kJ. 94.1 is absorbcan water liquid theheat that ofAmount

kJ. 105 is steam theof all condense heat to ofAmount

kJ 94.1 C)50.0 -(100.0 75.3 mol 25.0

water.liquid in the availableheat ofamount theCalculate (2)

kJ .105 40.7 mol 2.50 C100.0 -140.0 36.4mol 2.50

steam. thecondense torequiredheat ofamount theCalculate (1)

mol .025g 18

mol 1 waterg 450 mol 2.50

steam g 18

mol 1 steam g 0.45

o

Cmol J

molkJo

Cmol J

o

o


The heat of vaporization of ammonia is 23.4 kJ/mol. How much

heat is required to vaporize 1.00 kg of ammonia?

– First, we must determine the number of moles of ammonia in

1.00 kg (1000 g).

– Then we can determine the heat required for vaporization.


15

Element Aluminum Argon Barium Beryllium Boron Bromine Cadmium Calcium Carbon Cesium Chlorine Chromium Cobalt Copper Flourine Gallium Germanium Gold Helium Hydrogen

Symbol Al Ar Ba Be B Br Cd Ca C Cs Cl Cr Co Cu F Ga Ge Au He H

At. Weight (amu) 26.98 39.95 137.33 9.012 10.81 79.90 112.41 40.08 12.011 132.91 35.45 52.00 58.93 63.55 19.00 69.72 72.59 196.97 4.003 1.008

Atomic radius (nm) 0.143 ------ 0.217 0.114 ------ ------ 0.149 0.197 0.071 0.265 ------ 0.125 0.125 0.128 ------ 0.122 0.122 0.144 ------ ------

Density

(g/cm3) 2.71 ------ 3.5 1.85 2.34 ------ 8.65 1.55 2.25 1.87 ------ 7.19 8.9 8.94 ------ 5.90 5.32 19.32 ------ ------

Adapted from

Table, "Charac-

teristics of

Selected

Elements",

inside front

cover,

Callister 6e.

Characteristics of Selected Elements at 20C 21


22

Solids - Unit Cell & Density

A metal X (atomic mass M) crystallizes in a simple cubic

crystal structure with one atom per unit cell with the length

of the unit cell a in cm, then its density d g cm-1 is.

M / NA

d = ------------

a 3

Mass of an atom / unit cell

Volume of a unit cell

If the cubic unit cell has n atoms,

then density

n M

d =

(NA a3).

NA is the Avogadro’s number


23

Solids - Unit Cell & Radius of atom

Simple cubic:

Cubic edge a = 2 time radius of

atoms,

a = 2 r

Body centre cubic:

face diagonal df2 = 2 a2

body diagonal, db2 = df2 + a2 = 3 a2

= (4r)2

Face centre cubic:

face diagonal df2 = 2 a2 = (4r)2

a = 22 r

http://www.webelements.com/webelements/elements/text/Na/xtal.html

http://www.webelements.com/webelements/elements/text/Po/xtal.html


24

We can calculate the density in a unit cell.

volume

massDensity

Mass is the mass of the number of atoms in the unit cell.

Mass of one atom =atomic mass/6.022x1023

Avogadro’s Number!

DENSITY

N0 = 6.022 x 1023

atoms per mole


25 Volume of a copper unit cell

r= 128pm = 1.28x10-10m = 1.28x10-8cm

Volume of unit cell is given by:

38( )1028.18 cmV

cell

Cu crystalizes as a fcc

3 8rV cell

3231075.4 cmV

cell


Copper Density Calculation

63.54 g Cu

mole Cu

mole Cu

6.022 X 1023

atoms

4 atoms Cu

unit cell 4.75X10-23

cm3

unit cell

= 8.89 g/cm3

Laboratory measured density: 8.92 g/cm3

26


27

Solids – Density and Radii of Atoms

Copper crystallize in ccp type structure with a density of 8.92 g mL-1.

Calculate its atomic radius based on the hard-sphere packing model.

Let the cell edge be a and radius be r. Data NA = 6.023e23, atomic mass of Cu

63.5 and a = 22 r

4 * 63.5 g

d = ------------------ = 8.92 g cm-3

6.02e23 * a3

Thus, a3 = 4*63.5 / (6.02e23 * 8.92) cm3

= 4.73e-23 cm3

And a = 3.617e-8 cm = 22 r

Thus, r = 1.279e-8 cm or 0.127 nm

volume

http://www.webelements.com/webelements/elements/text/Cu/xtal.html


Geometry of a Cube

2

Diagonal Face

222

af

aaf

3

3

DiagonalBody

2222

ab

afab

28


Atomic Radius and Cell Dimensions

Simple Cubic

r = a/2

1 atom/unit cell (in metals)

Body-Centered Cubic

b = 4r = a(3)1/2

r = a(3)1/2/4

2 atoms/unit cell (in metals)

Face-Centered Cubic

f = 4r = a(2)1/2

r = a(2)1/2/4

4 atoms/unit cell (in metals)

29


Problem:

Copper has an atomic radius of 0.128 nm, an FCC crystal structure, and an atomic

weight of 63.5 g/mol. Compute its theoretical density and compare the answer with

its measured density.

Given:

Atomic radius = 0.128 nm (1.28 Ǻ)

Atomic weight = 63.5 g/mole

n = 4 ACu = 63.5 g/mol


Solution:

Unit cell volume = 16 R3

√2

R = Atomic Radius

= 8.89 g/cm3

Close to 8.94 g/cm3

in the literature

)atoms103cell](6.02/unit cm)10(1.282[16

63.5g/mole4ρ

2338


32

When silver crystallizes, it forms face-centered cubic cells.

The unit cell edge length is 4.087Å . Calculate the density

of silver.

o

g107.167amu106.022

g 1

atom

amu107.9

cellunit

atoms4 22

23

m

323383 cm106.827cm104.087 aV

cm104.087m 1

cm 001

A101

m 1 A4.087 8

10

a

3

323

22

g/cm 10.5cell/unitcm106.827

cellg/unit107.167

V

md

Mass of unit cell

Volume of unit cell

Density of unit cell


Self-Test 5.5A

The atomic radius of silver is 144 pm and its density is

10.5 g cm-3. Is the stucture face-centered cubic (fcc;

close packed) or body-centered cubic (bcc)?

Solution

We begin by assuming fcc;

d =4M

83/2NAr3=

4 x (107.9 g mol-1)

83/2 x (6.022 x 1023 mol-1) x (1.44 x 10-8 cm)3

= 10.6 g cm-3. Hence silver has fcc structure.


Nickel crystallizes in a face-centered cubic array of atoms in which the

edge of the unit cell is 351 pm. Calculate the density of the metal

8 corner atoms = (8 x 1/8) = 1 atom

6 face-centered atoms = (6 x ½) = 3 atoms

Number of atoms = 4

Structures Of Crystalline Solids

Nig10x3.898atoms10x6.022

mol1x

mol1

g58.69xNiatoms4Ni Mass 22

23

323

3

10cm10x4.32

pm10

cm1xpm351Volume

3

323

22

g/cm9.02cm10x4.32

g10x3.898

volume

massDensity

34


Structures Of Crystalline Solids

Nickel crystallizes in a face-centered cubic array of atoms

in which the edge of the unit cell is 351 pm. What is the

radius of an atom of nickel?

Atoms are assumed to be spherical and in contact

Length of diagonal of one face = 4 times radius

Using Pythagoras theorem

2aaadiagonalofLength 22

2arx4

pm1244

)2pm)((351

4

2ar

35


Calculate the number of cations, anions, and

formula units per unit cell in each of the

following solids:

(a) The cesium chloride unit cell

Solution


(b) The rutile (TiO2) unit cell

(c) What are the coordination numbers

of the ions in rutile?

Solution


Graphite forms extended two-dimensional layers. Draw the smallest

possible rectangular unit cell for a layer of graphite. (b) How many

carbon atoms are in your unit cell? (c) What is the coordination

number of carbon in a single layer of graphite?

(a) Solution


If the edge length of a fcc unit cell of RbI is 732.6 pm, how long would

the edge of a cubic single crystal of RbI be that contains 1.00 mol RbI?

Solution


Illustrative Problem

The planes of a crystal are 0.281 nm apart and angle 2θ =

11.80 then for first order diffraction, what is the wavelength

of X-ray used?

2 θ = 11.8o or θ = 5.9o and n = 1

1. = 2 x 0.281 Sin 5.9 = 0.058 pm

Solution :



A substance AxBy crystallizes in a face centered cubic (fcc) lattice

in which atoms ‘A’ occupy each corner of the cube and atoms B

occupy the centers of each face of the cube.

Identify the correct composition of the substance AxBy.

No. of ‘A’ atoms = 1

8 18

16 3

2

No. of ‘B’ atoms =

Hence formula is AB3

Solution:

41


Illustrative example

Three elements P,Q and R crystallize in a cubic lattice with P

atoms at the corners, Q atoms at the cube centre & R atoms

at the centre of the faces of the cube then what would be the

formula of the compound?.

Solution

Atoms P per unit cell =8x1/8 =1

Atoms Q per unit cell=1

Atoms R per unit cell =6x1/2 =3

Hence the formula of compound is PQR3

42



Sodium metal crystallites in bcc lattice with the cell edge 4.29 Å

.what is the radius of sodium atom ?

Solution :

3r a

4

34.29 A

4

1.86 A

43


Determine the density of BCC iron, which has a lattice

parameter of 0.2866 nm. SOLUTION

Atoms/cell = 2, a0 = 0.2866 nm = 2.866 10-8 cm

Atomic mass = 55.847 g/mol

Volume of unit cell = = (2.866 10-8 cm)3 = 23.54 10-24

cm3/cell

Avogadro’s number NA = 6.02 1023 atoms/mol

3

0a


3

24 23

(number of atoms/cell)(atomic mass of iron)Density

(volume of unit cell)(Avogadro's number)

(2)(55.847)7.882g/cm

(23.54 10 )(6.02 10 )

44



An element exists in bcc structure with a cell edge of 288 pm.

Density of the element is 7.2 g cm–3 what is the atomic mass of

the element?

bcc structure means Z = 2

Volume of the cell = (288)3 × 10–30

Solution:

3A

Z M

N aDensity =

Or 7.2 = 2 × M / 6.023 × 1023 × (288)3 × 10–30

M = 52 g mol–1


Illustrative Example

Density (ρ) = mass of contents of cell/volume of cell

Mass of atoms in the cell = 4 x 1 x 26.98/6.022 x 1023

= 17.922 x 10-23 g

Volume of unit cell = (4.041 x10-8)3

= 6.599 x 10-23 cm3

ρ = mass/volume = 17.922 x 10-23 g/6.599 x 10-23 cm3

= 2.715 g cm-3

Aluminum has the face-centered cubic structure with a unit cell

dimension of 4.041Å. What is density of aluminum?

Solution:



A metal of atomic mass = 75 form a cubic lattice of edge length

5Å and density 2 g cm-3. Calculate the radius of the atom.

Given Avogadro’s number, NA = 6 x 1023.

3

3A

Z MDensity( ) g / cm

a N

Solution:

23 242 6 10 125 10Z 2

75

It indicates that metal has bcc lattice. For bcc lattice, 3r a

4

Or r = 2.165 x 102 = 216.5 pm

We know that


A few typical unit cell calculations

Example #1: X-ray diffraction found that a metal has a simple cubic unit cell. If the volume of the cell is 4.38x10-23 cm3 then calculate the diameter of the metal atom in pm.

X (side of cube) = 2R = Diameter of the atom

X3 = Volume = 4.38x10-23 cm3

X = (4.38x10-23 cm3)1/3 = 3.52x10-8 cm

3.52x10-8 cm 1 m 1012 pm = 352 pm

100. cm 1 m

Note: 1 cm = 1010 pm



Example #2: X-ray diffraction reveals that Cr forms cubic unit cells with an edge

of 2.885x10-8 cm. The density of Cr is 7.20 g/cm3. Calculate the type of unit

cell (SC, BC or FC).

Cell Volume: (2.885x10-8 cm)3 / cell = 2.401x10-23 cm3 / cell

Mass Cell: 7.20 g Cr 2.401x10-23 cm3 = 1.73x10-22 g Cr

cm3 cell cell

Mass of 1 Cr Atom: 52.0 g Cr 1 mole = 8.63x10-23 g Cr

mole 6.02x1023 Cr atom Cr atom

Atoms per Cell & type of unit cell:

1.73x10-22 g Cr Cr atom = 2.00 Cr atoms Must be BC

cell 8.63x10-23 g Cr cell



Example #3: X-ray diffraction reveals that Cu forms a face centered cubic cell with an edge of 361 pm. Calculate the radius of a Cu atom.

Are four Cu radii (R) per short diagonal (C) for the FC; Note: C = 4R

C2 = (4R)2 = X2+X2 = 2X2 (4R)2 = 2X2 4R = √2X2

R = √2X2 = √ 2(361 pm)2 = 128 pm

4 4

50


Determination Of Atomic Radius

At room temperature iron crystallizes with a bcc unit cell.

X-ray diffraction shows that the length of an edge is 287 pm.

What is the radius of the Fe atom?

Edge Length (e)

3

4re

4

3 er

pmpm

r 1244

2873

51


Avogadro’s Number

Fe(s) is bcc Two atoms / unit cell

55.85 g Mole Fe

Sample Problem: Calculate Avogadro’s number of

iron if its unit cell length is 287 pm and it has a

density of 7.86 g/cm3.


Avogadro’s Number

The density of Fe(s) is 7.86 g/cm3.

V= e3 = (287pm)3 = 2.36x10-23cm3

Fe(s) is bcc Two atoms / unit cell

length of an edge is 287 pm.

55.85 g Mole Fe 7.86 g

cm3

(10 -12)3 cm3

pm3

(287 pm)3

unit cell

53


Avogadro’s Number

55.85 g Mole Fe 7.86 g

cm3

(10 -12)3 cm3

pm3

(287 pm)3

unit cell 2 atoms

unit cell

= 6.022 X 10 23 atoms/mole

54


Determine the density of BCC iron, which has a lattice parameter of

0.2866 nm.

SOLUTION

Atoms/cell = 2, a0 = 0.2866 nm = 2.866 10-8 cm

Atomic mass = 55.847 g/mol

Volume of unit cell = = (2.866 10-8 cm)3 = 23.54 10-24 cm3/cell

Avogadro’s number NA = 6.02 1023 atoms/mol

3

0a

3

2324/882.7

)1002.6)(1054.23(

)847.55)(2(

number) sadro'cell)(Avogunit of (volume

iron) of mass )(atomicatoms/cell of(number Density

cmg

55


One form of silicon has density of 2.33 gcm-3 and crystallizes in a cubic

Lattice with a unit cell edge of 543 pm. (a) What is the mass of each unit

cell? (b) How many silicon atoms does one unit cell contain?

Solution


Self-Test 5.6A

Predict (a) the likely structure and (b) the coordinationtype of ammonium chloride. Assume that theammonium ion can be approximated as a sphere with aradius of 151 pm.

Solution

Radius ratio, =Radius of smaller ion

Radius of larger ion

=151 pm

181 pm= 0.834

This indicates (a) a cesium chloride structure with (b) (8,8)-coordination.


Calculate the percent volume change as zirconia transforms from a

tetragonal to monoclinic structure. The lattice constants for the monoclinic

unit cells are: a = 5.156, b = 5.191, and c = 5.304 Å, respectively. The angle

β for the monoclinic unit cell is 98.9. The lattice constants for the

tetragonal unit cell are a = 5.094 and c = 5.304 Å, respectively Does the

zirconia expand or contract during this transformation? What is the

implication of this transformation on the mechanical properties of zirconia

ceramics?

Calculating Volume Changes in Polymorphs of

Zirconia

58


SOLUTION

The volume of a tetragonal unit cell is given by V = a2c = (5.094)2 (5.304) = 134.33 Å3.

The volume of a monoclinic unit cell is given by V = abc sin β = (5.156) (5.191) (5.304)

sin(98.9) = 140.25 Å3.

Thus, there is an expansion of the unit cell as ZrO2 transforms from a tetragonal to

monoclinic form.

The percent change in volume

= (final volume - initial volume)/(initial volume) x 100

= (140.25 - 134.33 Å3)/140.25 Å3 * 100 = 4.21%.

Most ceramics are very brittle and cannot withstand more than a 0.1%

change in volume. The conclusion here is that ZrO2 ceramics cannot be used in their

monoclinic form since, when zirconia does transform to the tetragonal form, it will

most likely fracture. Therefore, ZrO2 is often stabilized in a cubic form using different

additives such as CaO, MgO, and Y2O3.

59


Designing a Sensor to Measure Volume Change

To study how iron behaves at elevated temperatures, we would like to

design an instrument that can detect (within a 1% accuracy) the change

in volume of a 1-cm3 iron cube when the iron is heated through its

polymorphic transformation temperature. At 911oC, iron is BCC, with a

lattice parameter of 0.2863 nm. At 913oC, iron is FCC, with a lattice

parameter of 0.3591 nm. Determine the accuracy required of the

measuring instrument.

SOLUTION

The volume of a unit cell of BCC iron before transforming is:

VBCC = = (0.2863 nm)3 = 0.023467 nm3 3

0a

60


SOLUTION (Continued)

The volume of the unit cell in FCC iron is:

VFCC = = (0.3591 nm)3 = 0.046307 nm3

But this is the volume occupied by four iron atoms, as there

are four atoms per FCC unit cell. Therefore, we must compare two

BCC cells (with a volume of 2(0.023467) = 0.046934 nm3) with each

FCC cell. The percent volume change during transformation is:

3

0a

%34.11000.046934

0.046934) - (0.046307 change Volume

The 1-cm3 cube of iron contracts to 1 - 0.0134 = 0.9866 cm3

after transforming; therefore, to assure 1% accuracy, the instrument

must detect a change of:

ΔV = (0.01)(0.0134) = 0.000134 cm3

61


Determine the Miller indices of directions A, B, and C in Figure a1.

Determining Miller Indices of Directions

a1. Crystallographic directions and

coordinates.

62


SOLUTION

Direction A

1. Two points are 1, 0, 0, and 0, 0, 0

2. 1, 0, 0, - 0, 0, 0 = 1, 0, 0

3. No fractions to clear or integers to reduce

4. [100]

Direction B

1. Two points are 1, 1, 1 and 0, 0, 0

2. 1, 1, 1, - 0, 0, 0 = 1, 1, 1

3. No fractions to clear or integers to reduce

4. [111]

Direction C

1. Two points are 0, 0, 1 and 1/2, 1, 0

2. 0, 0, 1 - 1/2, 1, 0 = -1/2, -1, 1

3. 2(-1/2, -1, 1) = -1, -2, 2

2]21[ .4

63


Determine the Miller indices of planes A, B, and C in Figure a2.

Determining Miller Indices of Planes

(c) 2003 Brooks/Cole Publishing / Thomson Learning™

Figure a2 Crystallographic

planes and intercepts (for

Example 3.8)

64


SOLUTION

Plane A

1. x = 1, y = 1, z = 1

2.1/x = 1, 1/y = 1,1 /z = 1

3. No fractions to clear

4. (111)

Plane B

1. The plane never intercepts the z axis, so x = 1, y = 2, and z =

2.1/x = 1, 1/y =1/2, 1/z = 0

3. Clear fractions:

1/x = 2, 1/y = 1, 1/z = 0

4. (210)

Plane C

1. We must move the origin, since the plane passes through 0, 0, 0. Let’s move the origin one lattice parameter in the y-direction. Then, x = , y = -1, and z =

2.1/x = 0, 1/y = 1, 1/z = 0

3. No fractions to clear.

)010( .4

65


66


Calculate the planar density and planar packing fraction for the

(010) and (020) planes in simple cubic polonium, which has a lattice

parameter of 0.334 nm.

Calculating the Planar Density and Packing

Fraction

(c) 2003 Brooks/Cole Publishing /

Thomson Learning™

Figure a3: The planer

densities of the (010) and

(020) planes in SC unit cells

are not identical.

67


Packing fraction

• Packing fraction is the fraction of total volume of a

cube occupied by constituent particles.

Packing fraction(PF) =

Volume occupied by effective number of particles

Volume of the unit cell

68


SOLUTION

The total atoms on each face is one. The planar density is:

2142

2

atoms/cm 1096.8atoms/nm 96.8

)334.0(

faceper atom 1

face of area

faceper atom (010)density Planar

The planar packing fraction is given by:

79.0)2(

)(

)( atom) 1(

face of area

faceper atoms of area (010)fraction Packing

2

2

20

2

r

r

r

a

However, no atoms are centered on the (020) planes. Therefore, the

planar density and the planar packing fraction are both zero. The

(010) and (020) planes are not equivalent!

69


Drawing Direction and Plane

Draw (a) the direction and (b) the plane in a cubic unit

cell. 1]2[1 10]2[

Figure A4: Construction of

a (a) direction and (b) plane

within a unit cell (for

Example 3.10)

70


SOLUTION

a. Because we know that we will need to move in the negative y-

direction, let’s locate the origin at 0, +1, 0. The ‘‘tail’’ of the direction

will be located at this new origin. A second point on the direction can be

determined by moving +1 in the x-direction, 2 in the y-direction, and +1

in the z direction [Figure 3.24(a)].

b. To draw in the plane, first take reciprocals of the indices to

obtain the intercepts, that is:

x = 1/-2 = -1/2 y = 1/1 = 1 z = 1/0 =

Since the x-intercept is in a negative direction, and we wish to draw the

plane within the unit cell, let’s move the origin +1 in the x-direction to

1, 0, 0. Then we can locate the x-intercept at 1/2 and the y-intercept at

+1. The plane will be parallel to the z-axis [Figure A4(b)].

10]2[

71


Determine the Miller-Bravais indices for planes A and B and

directions C and D in Figure A5.

Determining the Miller-Bravais Indices for Planes

and Directions

(c) 2003 Brooks/Cole Publishing / Thomson Learning™

Figure A5: Miller-Bravais indices are

obtained for crystallographic planes

in HCP unit cells by using a four-axis

coordinate system. The planes labeled

A and B and the direction labeled C

and D are those discussed.

72


SOLUTION

Plane A

1. a1 = a2 = a3 = , c = 1

2. 1/a1 = 1/a2 = 1/a3 = 0, 1/c = 1


4. (0001)

Plane B

1. a1 = 1, a2 = 1, a3 = -1/2, c = 1

2. 1/a1 = 1, 1/a2 = 1, 1/a3 = -2, 1/c = 1


4.

Direction C

1. Two points are 0, 0, 1 and 1, 0, 0.

2. 0, 0, 1, -1, 0, 0 = -1, 0, 1

3. No fractions to clear or integers to reduce.

4.

)1211(

113]2[or ]011[

73



Direction D

1. Two points are 0, 1, 0 and 1, 0, 0.

2. 0, 1, 0, -1, 0, 0 = -1, 1, 0

3. No fractions to clear or integers to reduce.

4. 100]1[or ]101[

74


Calculating Octahedral Sites

Calculate the number of octahedral sites that uniquely belong to one

FCC unit cell.

SOLUTION

The octahedral sites include the 12 edges of the unit cell, with the

coordinates

2

11,,0

2

11,,1

2

10,,1

2

1,0,0

,12

1,0 ,1

2

1,1 ,0

2

1,1 ,0

2

1,0

1,1,2

1 0,1,

2

1 1,0,

2

1 0,0,

2

1

plus the center position, 1/2, 1/2, 1/2.

75



Each of the sites on the edge of the unit cell is shared between four

unit cells, so only 1/4 of each site belongs uniquely to each unit cell.

Therefore, the number of sites belonging uniquely to each cell is:

(12 edges) (1/4 per cell) + 1 center location

= 4 octahedral sites

76


We wish to produce a radiation-absorbing wall composed of 10,000

lead balls, each 3 cm in diameter, in a face-centered cubic

arrangement. We decide that improved absorption will occur if we fill

interstitial sites between the 3-cm balls with smaller balls. Design the

size of the smaller lead balls and determine how many are needed.

Design of a Radiation-Absorbing Wall

Figure A5: Calculation of an

octahedral interstitial site .

77


SOLUTION

First, we can calculate the diameter of the octahedral

sites located between the 3-cm diameter balls. Figure A5 shows the

arrangement of the balls on a plane containing an octahedral site.

Length AB = 2R + 2r = 4R/

r = R – R = ( - 1)R

r/R = 0.414

22

2

78

Since r /R = 0.414, the radius of the small lead balls is

r = 0.414 * R = (0.414)(3 cm/2) = 0.621 cm.

Earlier we find that there are four octahedral sites in the FCC arrangement,

which also has four lattice points. Therefore, we need the same number of

small lead balls as large lead balls, or 10,000 small balls.


For potassium chloride (KCl), (a) verify that the compound has the

cesium chloride structure and (b) calculate the packing factor for the

compound.

SOLUTION

a. One may know, rK+ = 0.133 nm and rCl- = 0.181 nm, so:

rK+/ rCl- = 0.133/0.181 = 0.735

Since 0.732 < 0.735 < 1.000, the coordination number for each type of

ion is eight and the CsCl structure is likely.

Radius Ratio for KCl 79


SOLUTION KCl packing fraction

b. The ions touch along the body diagonal of the unit cell, so:

Diagonal (d) = 2rK+ + 2rCl- = 2(0.133) + 2(0.181) = 0.628 nm

a0 = 0.363 nm (d2=2a02)

725.0)363.0(

)181.0(3

4 )133.0(

3

4

ion) Cl 1(3

4 ion)K 1(

3

4

factor Packing

3

33

3

0

33

a

rrClK

80


Show that MgO has the sodium chloride crystal structure and calculate

the density of MgO.

SOLUTION

One may know that, rMg+2 = 0.066 nm and rO-2 = 0.132 nm, so:

rMg+2 /rO-2 = 0.066/0.132 = 0.50

Since 0.414 < 0.50 < 0.732, the coordination number for each

ion is six, and the sodium chloride structure is possible.

Illustrating a Crystal Structure and Calculating

Density

81


SOLUTION

The atomic masses are 24.312 and 16 g/mol for magnesium and oxygen,

respectively. The ions touch along the edge of the cube, so:

Diagonal = 2 rMg+2 + 2rO-2 = 2(0.066) + 2(0.132)

a0 = 0.396 nm = 3.96 10-8 cm

3

2338

-22

/31.4)1002.6(cm) 1096.3(

)16)(O4()312.24)(Mg4(cmg

82


The lattice constant of gallium arsenide (GaAs) is 5.65 Å. Show

that the theoretical density of GaAs is 5.33 g/cm3.

SOLUTION

For the ‘‘zinc blende’’ GaAs unit cell, there are four Ga and

four As atoms per unit cell.

From the periodic table :

Each mole (6.023 1023 atoms) of Ga has a mass of 69.7 g.

Therefore, the mass of four Ga atoms will be (4 * 69.7/6.023

1023) g.

Calculating the Theoretical Density of GaAs 83



Each mole (6.023 1023 atoms) of As has a mass of 74.9 g.

Therefore, the mass of four As atoms will be (4 * 74.9/6.023

1023) g. These atoms occupy a volume of (5.65 10-8)3 cm3.

38

23

cm) 1065.5(

10023.6/)9.747.69(4

volume

mass density

Therefore, the theoretical density of GaAs will be 5.33 g/cm3.

84


(c) 20

03

Bro

oks/C

ole P

ub

lishin

g / T

hom

son L

earnin

g

Figure A6 (a) Tetrahedron and (b) the diamond cubic (DC) unit cell. This

open structure is produced because of the requirements of covalent bonding.

85


Determine the packing factor for diamond cubic silicon.

Determining the Packing Factor for Diamond

Cubic Silicon

Figure A7 Determining the

relationship between lattice

parameter and atomic radius

in a diamond cubic cell.

86


SOLUTION

We find that atoms touch along the body diagonal of the

cell (Figure 3.39). Although atoms are not present at all

locations along the body diagonal, there are voids that

have the same diameter as atoms. Consequently:

34.0

)3/8(

)3

4)(8(

)3

4)(atoms/cell (8

factor Packing

83

3

3

3

0

3

0

r

r

a

r

ra

Compared to close packed structures this is a relatively open structure.

87


Calculating the Radius, Density, and Atomic Mass

of Silicon

The lattice constant of Si is 5.43 Å . What will be the radius of a

silicon atom? Calculate the theoretical density of silicon. The

atomic mass of Si is 28.1 gm/mol.

SOLUTION

For the diamond cubic structure,

Therefore, substituting a = 5.43 Å, the

radius of silicon atom = 1.176 Å .

There are eight Si atoms per unit cell.

ra 83 0

3

38

23

/33.2cm) 1043.5(

10023.6/)1.28(8

volume

mass density cmg

88


Theoretical Density,

where n = number of atoms/unit cell

A = atomic weight

VC = Volume of unit cell = a3 for cubic

NA = Avogadro’s number

= 6.023 x 1023 atoms/mol

Density = =

VC NA

n A =

Cell Unit of Volume Total

Cell Unit in Atoms of Mass


Ex: Cr (BCC)

A = 52.00 g/mol

R = 0.125 nm

n = 2

theoretical

a = 4R/ 3 = 0.2887 nm

actual

a

R

=

a 3

52.00 2

atoms

unit cell mol

g

unit cell

volume atoms

mol

6.023 x 1023

Theoretical Density,

= 7.18 g/cm3

= 7.19 g/cm3


Crystallographic Planes

Miller Indices: Reciprocals of the (three) axial intercepts for a plane, cleared of fractions & common multiples. All parallel planes have same Miller indices.

Algorithm 1. Read off intercepts of plane with axes in terms of a, b, c 2. Take reciprocals of intercepts 3. Reduce to smallest integer values 4. Enclose in parentheses, no commas i.e., (hkl)


PRACTICE

92

z

x

y a b

c

example a b c z

x

y a b

c

4. Miller Indices

1. Intercepts

2. Reciprocals

3. Reduction

example a b c

4. Miller Indices

1. Intercepts

2. Reciprocals

3. Reduction


Crystallographic Planes

z

x

y a b

c

4. Miller Indices (110)

example a b c z

x

y a b

c

4. Miller Indices (100)

1. Intercepts 1 1

2. Reciprocals 1/1 1/1 1/ 1 1 0

3. Reduction 1 1 0

1. Intercepts 1/2

2. Reciprocals 1/½ 1/ 1/ 2 0 0

3. Reduction 1 0 0

example a b c


The density of copper is 8.93 gcm-3 and its atomic radius is 128 pm.

Is the metal (a) close-packed or (b) body-centered cubic?

(a) Fcc (ccp) and hcp cannot be distinguished by density only.

For 4 atoms in a fcc cell,


(b) For 2 atoms in a bcc shell,

close-packed cubic (fcc)


96

Self-Test 5.7B

Estimate the density of cesium iodide from its crystalstructure.

Solution

r(Cs+) = 170 pm; r(I-) = 220 pm

Length of diagonal, b = 170 + 2(220) + 170 pm = 780 pm

Length of side a = 780/3 = 450 pm12

CsI has a cesium chloride (bcc) type structure.

Hence unit cell volume is = 9.11 x 107 pm3 or 9.11 x 10-23 cm3

(1 pm3 = 10-30 cm3)

Each bbc unit cell has one Cs+ ion and one I- ion,

Density = mass/volume =

(132.91 + 126.90) g mol-1

(6.022 x 1023 mol-1)9.11 x 10-23 cm3

= 4.74 g cm-3

C


97

Classify each of the following solids as ionic, network, metallic, or

molecular: (a) quartz, SiO2; (b) limestone, CaCO3; (c) dry ice, CO2;

(d) sucrose, C12H22O11; (e) polyethylene, a polymer with molecules

consisting of chains of thousands of repeating –CH2CH2- units.

Solution


What percentage of space is occupied by close-packed cylinders of

length l and radius r ?

190s

Solution


Question: When silver crystallizes, it forms face-centered cubic cells.

The unit cell edge length is 409 pm. Calculate the density of silver.

d = m

V V = a3 = (409 pm)3 = 6.83 x 10-23 cm3

4 atoms/unit cell in a face-centered cubic cell

m = 4 Ag atoms 107.9 g

mole Ag x

1 mole Ag

6.022 x 1023 atoms x = 7.17 x 10-22 g

d = m

V

7.17 x 10-22 g

6.83 x 10-23 cm3 = = 10.5 g/cm3

99


100

Bonding in Solids

A group IVA element with a density of 11.35 g/cm3

crystallizes in a face-centered cubic lattice whose unit cell

edge length is 4.95 Å. Calculate the element’s atomic weight.

What is the atomic radius of this element?

Face centered cubic unit cells have 4 atoms, ions, or molecules per unit cell.

Problem solution pathway:

1. Determine the volume of a single unit cell.

2. Use the density to determine the mass of a single unit cell.

3. Determine the mass of one atom in a unit cell.

4. Determine the mass of 1 mole of these atoms


101

Bonding in Solids

1. Determine the volume of a single unit cell.

322-38-

3

8-0

8-0

cm 101.21 cm 104.95

V so cubic are cellsunit cubic centered Face

cm 104.95 A 4.95 thuscm 10 A 1

2. Use the density to determine the mass of a unit cell.

cellunit one

g 101.38

cm

g 35.11 cm 101.21 21-

3

322-


102

Bonding in Solids

3. Determine the mass of one atom in the unit cell.

atomg

1044.3 4

101.38

fashion. in this determined becan atom one of mass the

cellunit per atoms 4 has cubic centered face Because

22

cellunit atoms

cellunit g21-

4. Determine the mass of one mole of these atoms.

g/mole 207mole

atoms10022.6atom

g1044.3 2322


Strategy Determine the contribution of each ion in the unit cell based on its position. Referring to

the figure, the unit cell has four Zn2+ ions completely contained within the unit cell, and S2- ions at

each of the eight corners and at each of the six faces. Interior ions (those completely contained within

the unit cell) contribute one, those at the corners each contribute one-eighth, and those on the faces

contribute one-half.

How many of each ion are contained within a unit cell of ZnS?

Solution The ZnS unit cell contains four Zn2+ ions (interior) and four S2- ions [8 × (corners)

and 6 × (faces)]

1

8 1

2

Think About It Make sure that the ratio of cations to anions that you determine for a unit

cell matches the ratio expressed in the compound’s empirical formula.


(a) Calculate the amount of heat deposited on the skin of a person burned by 1.00 g

of liquid water at 100.0°C and

(b) (b) the amount of heat deposited by 1.00 g of steam at 100.0°C.

(c) (c) Calculate the amount of energy necessary to warm 100.0 g of water from

0.0°C to body temperature and

(d) (d) the amount of heat required to melt 100.0 g of ice 0.0°C and then warm it to

body temperature. (Assume that body temperature is 37.0°C.)

Strategy For the purpose of following the sign conventions, we can designate the water as the system and the

body as the surroundings. (a) Heat is transferred from hot water to the skin in a single step: a temperature

change. (b) The transfer of heat from steam to the skin takes place in two steps: a phase change and a

temperature change. (c) Cold water is warmed to body temperature in a single step: a temperature change.

(d) The melting of ice and the subsequent warming of the resulting liquid water takes place in two steps: a

phase change and a temperature change. In each case, the heat transferred during a temperature change

depends on the mass of the water, the specific heat of the water, and the change in temperature. For the phase

changes, the heat transferred depends on the amount of water (in moles) and the molar heat of vaporization

(ΔHvap) or molar heat of fusion (ΔHfus). In each case, the total energy transferred or required is the sum of

the energy changes for the individual steps.

The specific heat is 4.184 J/g∙°C for water and 1.99 J/g∙°C for steam. ΔHvap is 40.79 kJ/mol and ΔHfus is 6.01

kJ/mol. Note: The ΔHvap of water is the amount of heat required to vaporize a mole of water. However, we

want to know how much heat is deposited when water vapor condenses, so we use −40.79 kJ/mol.


Worked Example 12.7 (cont.)

Solution (a) ΔT = 37.0°C – 100.0°C = –63.0°C

q = msΔT = 1.00 g × ×–63.0°C

Thus, 1.00 g of water at 100.0°C deposits 0.264 kJ of heat on the skin. (The

negative sign indicates that heat is given off by the system and absorbed by the

surroundings.)

(b)

q1 = nΔHvap = 0.0555 mol ×

q2 = msΔT = 1.00 g × ×–63.0°C

The overall energy deposited on the skin by 1.00 g of steam is the sum of q1 and

q2:

–2.26 kJ + (–0.264 kJ) = –2.53 kJ

4.184 J

g∙°C = –2.64×102 J = –0.264 kJ

1.00 g

18.02 g/mol = 0.0555 mol water

−40.79 kJ

mol

= –2.26 kJ

4.184 J

g∙°C = –2.64×102 J = –0.264 kJ


Solution (c) ΔT = 37.0°C – 0.0°C = 37.0°C

q = msΔT = 1.00 g × ×37.0°C

The energy required to warm 100.0 g of water from 0.0°C to 37.0°C is 15.5 kJ.

(d)

q1 = nΔHfus = 5.55 mol ×

q2 = msΔT = 100.0 g × ×37.0°C

The energy required to melt 100.0 g of ice at 0.0°C and warm it to 37.0°C is the

sum of q1 and q2:

33.4 kJ + 15.5 kJ = 48.9 kJ

4.184 J

g∙°C = 1.55×104 J = 15.5 kJ

100.0 g

18.02 g/mol = 5.55 mol water

6.01 kJ

mol = 33.4 kJ

4.184 J

g∙°C = 1.55×104 J = 15.5 kJ

Think About It In problems that include phase changes, the q values

corresponding to the phase-change steps will be the largest contributions to

the total. If you find that this is not the case in your solution, check to see if

you have made the common error of neglecting to convert the q values

corresponding to temperature changes from J to kJ.


Phase Diagrams

A phase diagram summarizes the conditions at which a substance

exists as a solid, liquid, or gas.

The triple point is the

only combination of

pressure and temperature

where three phases of a

substance exist in

equilibrium.

triple point


The phase diagram of

water:

Phase Diagrams


Strategy Each point on the phase diagram corresponds to a pressure-

temperature combination. The normal boiling and melting points are the

temperatures at which the substance undergoes phase changes. These points fall

on the phase boundary lines. The triple point is where the three phase

boundaries meet.

Using the following phase diagram, (a) determine the normal boiling point

and the normal melting point of the substance, (b) determine the physical

state of the substance at 2 atm and 110°C, and (c) determine the pressure

and temperature that correspond to the triple point of the substance.


Solution By drawing lines corresponding to a given pressure and/or temperature,

we can determine the temperature at which a phase change occurs, or the physical

state of the substance under specified conditions.

(a)

The normal boiling and melting points are

~140°C and ~205°C, respectively.


Solution

(b)

At 2 atm and 110°C the substance

is a solid.


Solution

(c)

The triple point occurs at ~0.8 atm and

~115°C.

Think About It The triple point of this substance occurs at a pressure below

atmospheric pressure. Therefore, it will melt rather than sublime when it is heated

under ordinary conditions.

Documents

Prof.P. Ravindran, - folk.uio.nofolk.uio.no/ravi/CMP2013/7. problem_rf.pdf · · 2013-10-17P.Ravindran, PHY075- Condensed Matter Physics, Spring 2013 : Problems & Solutions -I Geometry