Problems & Solutions -I 1 Prof.P. Ravindran, Department of Physics, Central University of Tamil Nadu, India http://folk.uio.no/ravi/cmp2013
Geometry of a Cube 2 Face Diagonal f 2 a 2 f a 2 a 2 Body Diagonal b 2 a 2 b a 3 f 2 3a 2
Fundamental Properties of Matter 3 Matter: - Has mass, occupies space Mass measure of inertia - from Newton s first law of motion. It is one of the fundamental physical properties. 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.
Plasma: 4 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.
Condensate: 5 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.
6 Fermions Half integral Bosons Spin Occupancy Examples Integral spin Only one per state Many allowed electrons, protons, neutrons, quarks, neutrinos photons, 4 He atoms, gluons
7 The energy required to change one gram of a solid at its melting point into a liquid is called the heat of fusion.
How many calories of energy are needed to change 10.0 g of ice at 0.00 o C to water 20.0 o C? The heat of fusion of ice at 0 o C is 80. cal/g. Determine the calories necessary to melt 10.0 g of ice. 10.0 g 80. cal 1 g = 800. cal 8
How many calories of energy are needed to change 10.0 g of ice at 0.00 o C to water 20.0 o C? Determine the calories necessary to heat 10.0 g of water from 0.00 o C to 20.0 o C. The specific heat of water is 1.00 cal/g o C. 1.00 cal 10.0 g o 200 cal o 20.0 C = 1 g C 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 0.00 o C to 20 o C. 800 cal + 200 cal = 1000 cal 9
10 The energy required to change one gram of a liquid at its boiling point into a vapor is called the heat of vaporization.
How many calories of energy are needed to change 20.0 g of water at 20.0 o C to steam at 100.0 o C? The specific heat of water is 1.00 cal/g o C. (mass) (sp.ht) (Δt) = energy Determine the calories necessary to heat 20.0 g from 20 o C to 100 o C. 1.00 cal gc 20.0 g o o o 100. C - 20.0 C 1600 cal 11
How many calories of energy are needed to change 20.0 g of water at 20.0 o C to steam at 100.0 o C? The heat of vaporization of water at 100 o C is 540 cal/g. Determine the calories necessary to change 20.0 g of water at 100. o C to steam at 100 o C. 20.0 g 540 cal 1 g 10800 cal The total heat absorbed by the system is the heat required to raise the water temperature from 20 o C to 100 o C plus the heat required to change the water to steam. 1600 cal + 10800 cal = 12400 cal 12
Heating (cooling curve) 13
How much heat (in kj) is needed to convert 80.0 g of ethanol at -130 o C to vapor at 96 o C? The specific heat of solid, liquid, and vapor ethanol are 0.97, 2.4, and 1.2 J/g. o C, respectively. DHfus= 5.02 kj/mol, DHvap= 39.3 kj/mol. The mp is -114 o C and bp is 78 o C. 14 Step 1: heat solid up to mp (-114 o C), remember q=smdt q1=0.97 J/g. o C x 80.0 g x (-114 o C -(-130 o C))x 1 kj/1000j=1.24 kj Step 2: melt solid to liquid q2= 80.0g x 1 mol C 2 H 5 OH/46.07 g x 5.02 kj/mol=8.72 KJ Step 3: heat liquid from mp (-114 o C) to bp (78 o C) q3=2.4 J/g. o C x 80.0 g x (78 o C -(-114 o C))x 1 kj/1000j=36.9 kj Step 4: convert liquid to vapor q4= 80.0g x 1 mol C 2 H 5 OH/46.07 g x 39.3 kj/mol=68.2 KJ Step 5: heat vapor from 78 o C to 96 o C q5=1.2 J/g. o C x 80.0 g x (96 o C -78 o C)x 1 kj/1000j=1.73 kj Total heat=q1 +q2 + q3+ q4 + q5 =117 kj
The Liquid State How many joules of energy must be absorbed by 5.00 x 10 2 g of H 2 O at 50.0 o C to convert it to steam at 120 o C? 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 o C and 36.4 J/mol o C, respectively. 15
The Liquid State 16? mol = 500 g H O 1 mol H 2 O 2 27. 8 18 g H O 753. J... J o 100 0 50 0 105 10 5 mol C o C? J = 27.8 mol 40. 7 10 mol 3 mol H O 1st let's calculate the heat required to warm water from 50 to 100 C? J = 27.8 mol 2 Next, let s calculate the energy required to boil the water. J 1131. 10 Finally, let s calculate the heat required to heat steam from 100 to 120 o C. o? J = 27.8 mol 36. 4 J 120.0-100.0 C 0. 20 10 5 J o mol C 5 2 J o
The Liquid State The total amount of energy for this process is the sum of the 3 pieces we have calculated 5 5 105 10 1131 10 5. J. J 0. 20 10 J 12. 56 10 5 J or 1.26 3 10 kj 17
The Liquid State If 45.0 g of steam at 140 o C is slowly bubbled into 450 g of water at 50.0 o C in an insulated container, can all the steam be condensed? 18
The Liquid State 1mol 1mol 45.0 g steam 2.50 mol 450 g water 25.0 mol 18 g steam 18 g (1) Calculate the amount of heat required to condense the steam. 2.50 mol 36.4 J o 140.0-100.0 C 2.50 mol 40.7 kj o (2) Calculate the amount of 25.0 mol Amount of Amount of mol C 75.3 Thus all of J mol o C heat to condense all of heat available in the liquid water. (100.0-50.0 C) 94.1 kj the steam is105 kj. heat that the liquid water can absorb is 94.1 kj. the steam cannot be condensed. o mol 105. kj 19
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.
Element Aluminum Argon Barium Beryllium Boron Bromine Cadmium Calcium Carbon Cesium Chlorine Chromium Cobalt Copper Flourine Gallium Germanium Gold Helium Hydrogen Characteristics of Selected Elements at 20C 21 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 Density (g/cm 3 ) 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 ------ ------ 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 ------ ------ Adapted from Table, "Characteristics of Selected Elements", inside front cover, Callister 6e. 15
Solids - Unit Cell & Density 22 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 / N A d = ------------ a 3 Mass of an atom / unit cell Volume of a unit cell N A is the Avogadro s number If the cubic unit cell has n atoms, then density n M d = (N A a 3 ).
Solids - Unit Cell & Radius of atom 23 Simple cubic: Cubic edge a = 2 time radius of atoms, a = 2 r Body centre cubic: face diagonal df 2 = 2 a 2 body diagonal, db 2 = df 2 + a 2 = 3 a 2 = (4r) 2 Face centre cubic: face diagonal df 2 = 2 a 2 = (4r) 2 a = 22 r
DENSITY 24 We can calculate the density in a unit cell. Density mass volume Mass is the mass of the number of atoms in the unit cell. Mass of one atom =atomic mass/6.022x10 23 N0 = 6.022 x 10 23 atoms per mole Avogadro s Number!
Volume of a copper unit cell 25 Cu crystalizes as a fcc r= 128pm = 1.28x10-10 m = 1.28x10-8 cm Volume of unit cell is given by: V cell r 8 3 V cell 81.2810 8 cm) 3 ( V cell 23 4.75 10 cm 3
Copper Density Calculation 26 63.54 g Cu mole Cu = 8.89 g/cm 3 mole Cu 6.022 X 10 23 atoms 4 atoms Cu unit cell unit cell 4.75X10-23 cm 3 Laboratory measured density: 8.92 g/cm 3
Solids Density and Radii of Atoms 27 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 N A = 6.023e 23, atomic mass of Cu 63.5 and a = 22 r 4 * 63.5 g d = ------------------ = 8.92 g cm -3 6.02e 23 * a 3 Thus, a 3 = 4*63.5 / (6.02e 23 * 8.92) cm 3 = 4.73e -23 cm 3 And Thus, a = 3.617e -8 cm = 22 r r = 1.279e -8 cm or 0.127 nm volume
Geometry of a Cube 28 Face Diagonal f 2 a 2 f a 2 a 2 Body Diagonal b 2 a 2 b a 3 f 2 3a 2
Atomic Radius and Cell Dimensions 29 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)
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 A Cu = 63.5 g/mol 30
Solution: Unit cell volume = 16 R 3 2 R = Atomic Radius [16 2(1.2810 463.5g/mole ρ 8 3 23 cm) /unit cell](6.023 10 )atoms = 8.89 g/cm 3 Close to 8.94 g/cm 3 in the literature 31
When silver crystallizes, it forms face-centered cubic cells. The unit cell edge length is 4.087Å. Calculate the density of silver. m 4 atoms 107.9 amu o 1g 22 unit cell Volume of unit cell Mass of unit cell atom 6.02210 23 amu 7.16710 g a 4.087 V a 3 1m A 110 10 100 cm 1m A 4.08710 8 cm 8 3 23 3 4.087 10 cm 6.827 10 cm Density of unit cell d m V 7.167 10 6.827 10 22 23 g/unit cell 10.5 g/cm 3 cm /unit cell 3 32
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 8 3/2 N A r 3 = 4 x (107.9 g mol -1 ) 8 3/2 x (6.022 x 10 23 mol -1 ) x (1.44 x 10-8 cm) 3 = 10.6 g cm -3. Hence silver has fcc structure. 33
Structures Of Crystalline Solids 34 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 Mass Ni 4atoms Ni x 58.69g 1mol x 1mol 6.022 x10 23 atoms 3.898 x10 22 g Ni Volume 351pm x 1cm 10 pm 3 23 3 4.32 x10 cm 10 Density mass volume 3.898 x10 4.32 x10 23 g cm 22 3 9.02 g/cm 3
Structures Of Crystalline Solids 35 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 Length of diagonal a 2 a 2 a 2 4 x r a 2 r a 2 4 (351pm)( 4 2) 124pm
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 36
(b) The rutile (TiO 2 ) unit cell (c) What are the coordination numbers of the ions in rutile? Solution 37
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 38
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 39
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? Solution : 2 θ = 11.8 o or θ = 5.9 o and n = 1 1. = 2 x 0.281 Sin 5.9 = 0.058 pm
Illustrative Problem A substance A x B y 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 A x B y. 41 Solution: 1 No. of A atoms = 8 1 8 No. of B atoms = 1 6 3 2 Hence formula is AB 3
Illustrative example 42 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 PQR 3
Illustrative Problem 43 Sodium metal crystallites in bcc lattice with the cell edge 4.29 Å.what is the radius of sodium atom? Solution : r 3 4 a 3 4.29 A 4 1.86 A
Illustrative example 44 Determine the density of BCC iron, which has a lattice parameter of 0.2866 nm. SOLUTION Atoms/cell = 2, a 0 = 0.2866 nm = 2.866 10-8 cm Atomic mass = 55.847 g/mol 3 Volume of unit cell = a = (2.866 10-8 cm)3 = 23.54 10-24 0 cm3/cell Avogadro s number N A = 6.02 10 23 atoms/mol (number of atoms/cell)(atomic mass of iron) Density (volume of unit cell)(avogadro's number) (2)(55.847) 24 23 (23.54 10 )(6.02 10 ) 7.882g / cm 3
Illustrative example 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? Solution: bcc structure means Z = 2 Volume of the cell = (288) 3 10 30 Density = N Z M A a 3 Or 7.2 = 2 M / 6.023 10 23 (288) 3 10 30 M = 52 g mol 1
Illustrative Example Aluminum has the face-centered cubic structure with a unit cell dimension of 4.041Å. What is density of aluminum? Solution: 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 10 23 = 17.922 x 10-23 g Volume of unit cell = (4.041 x10-8 ) 3 = 6.599 x 10-23 cm 3 ρ = mass/volume = 17.922 x 10-23 g/6.599 x 10-23 cm 3 = 2.715 g cm -3
Illustrative example 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, N A = 6 x 10 23. Solution: We know that Density( ) a Z M 3 N A g / cm 3 23 24 2 6 10 125 10 Z 2 75 It indicates that metal has bcc lattice. For bcc lattice, 3 Or r = 2.165 x 10 2 = 216.5 pm r 4 a
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 cm 3 then calculate the diameter of the metal atom in pm. X (side of cube) = 2R = Diameter of the atom X 3 = Volume = 4.38x10-23 cm 3 X = (4.38x10-23 cm 3 ) 1/3 = 3.52x10-8 cm 3.52x10-8 cm 1 m 10 12 pm = 352 pm 100. cm 1 m Note: 1 cm = 10 10 pm
A few typical unit cell calculations 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/cm 3. Calculate the type of unit cell (SC, BC or FC). Cell Volume: (2.885x10-8 cm) 3 / cell = 2.401x10-23 cm 3 / cell Mass Cell: 7.20 g Cr 2.401x10-23 cm 3 = 1.73x10-22 g Cr cm 3 cell cell Mass of 1 Cr Atom: 52.0 g Cr 1 mole = 8.63x10-23 g Cr mole 6.02x10 23 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
A few typical unit cell calculations 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 C 2 = (4R) 2 = X 2 +X 2 = 2X 2 (4R) 2 = 2X 2 4R = 2X 2 R = 2X 2 = 2(361 pm) 2 4 4 = 128 pm 50
Determination Of Atomic Radius 51 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) r e 3 e 4 4r 3 3 287 pm r 124 4 pm
Avogadro s Number 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/cm 3. 55.85 g Mole Fe Fe(s) is bcc Two atoms / unit cell
Avogadro s Number 53 The density of Fe(s) is 7.86 g/cm 3. length of an edge is 287 pm. V= e3 = (287pm) 3 = 2.36x10-23 cm 3 55.85 g cm 3 pm 3 unit cell Mole Fe 7.86 g (10-12 ) 3 cm 3 (287 pm) 3 Fe(s) is bcc Two atoms / unit cell
Avogadro s Number 54 55.85 g cm 3 pm 3 unit cell Mole Fe 7.86 g (10-12 ) 3 cm 3 (287 pm) 3 2 atoms unit cell = 6.022 X 10 23 atoms/mole 54
55 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 3 a 0 Volume of unit cell = = (2.866 10-8 cm) 3 = 23.54 10-24 cm 3 /cell Avogadro s number N A = 6.02 10 23 atoms/mol Density (23.54 (number of atoms/cell)(atomic mass of iron) (volume of unit cell)(avogadro' s number) (2)(55.847) 10 7.882g / 24 23 )(6.02 10 ) cm 3
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 56
Self-Test 5.6A Predict (a) the likely structure and (b) the coordination type of ammonium chloride. Assume that the ammonium ion can be approximated as a sphere with a radius 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.
Calculating Volume Changes in Polymorphs of Zirconia 58 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?
SOLUTION 59 The volume of a tetragonal unit cell is given by V = a 2 c = (5.094) 2 (5.304) = 134.33 Å 3. 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 Å 3. Thus, there is an expansion of the unit cell as ZrO 2 transforms from a tetragonal to monoclinic form. The percent change in volume = (final volume - initial volume)/(initial volume) x 100 = (140.25-134.33 Å 3 )/140.25 Å 3 * 100 = 4.21%. Most ceramics are very brittle and cannot withstand more than a 0.1% change in volume. The conclusion here is that ZrO 2 ceramics cannot be used in their monoclinic form since, when zirconia does transform to the tetragonal form, it will most likely fracture. Therefore, ZrO 2 is often stabilized in a cubic form using different additives such as CaO, MgO, and Y 2 O 3.
Designing a Sensor to Measure Volume Change 60 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-cm 3 iron cube when the iron is heated through its polymorphic transformation temperature. At 911 o C, iron is BCC, with a lattice parameter of 0.2863 nm. At 913 o C, 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: 3 a 0 VBCC = = (0.2863 nm) 3 = 0.023467 nm 3
SOLUTION (Continued) 61 The volume of the unit cell in FCC iron is: VFCC = 3 = (0.3591 nm)3 = 0.046307 nm3 a 0 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: Volume change (0.046307-0.046934) 0.046934 100 1.34% The 1-cm 3 cube of iron contracts to 1-0.0134 = 0.9866 cm 3 after transforming; therefore, to assure 1% accuracy, the instrument must detect a change of: ΔV = (0.01)(0.0134) = 0.000134 cm 3
Determining Miller Indices of Directions 62 Determine the Miller indices of directions A, B, and C in Figure a1. a1. Crystallographic directions and coordinates.
63 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 4. [1 22]
Determining Miller Indices of Planes 64 Determine the Miller indices of planes A, B, and C in Figure a2. Figure a2 Crystallographic planes and intercepts (for Example 3.8) (c) 2003 Brooks/Cole Publishing / Thomson Learning
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. 4.(010) 65
66
Calculating the Planar Density and Packing Fraction 67 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. Figure a3: The planer densities of the (010) and (020) planes in SC unit cells are not identical. (c) 2003 Brooks/Cole Publishing / Thomson Learning
Packing fraction 68 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
SOLUTION 69 The total atoms on each face is one. The planar density is: Planar density (010) atom per face area of face 1atom per face (0.334) 2 8.96 atoms/nm 2 8.96 10 14 atoms/cm 2 The planar packing fraction is given by: Packing fraction (010) area of atoms per face area of face (1atom) ( r ( a 0 ) 2 2 ) r 2 (2r) 2 0.79 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!
Drawing Direction and Plane 70 [121] [210] Draw (a) the direction and (b) the plane in a cubic unit cell. Figure A4: Construction of a (a) direction and (b) plane within a unit cell (for Example 3.10)
71 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 [210] 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)].
Determining the Miller-Bravais Indices for Planes and Directions 72 Determine the Miller-Bravais indices for planes A and B and directions C and D in Figure A5. 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. (c) 2003 Brooks/Cole Publishing / Thomson Learning
SOLUTION 73 Plane A 1. a1 = a2 = a3 =, c = 1 2. 1/a1 = 1/a2 = 1/a3 = 0, 1/c = 1 3. No fractions to clear 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 3. No fractions to clear 4. (1121) 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. [101]or [2113]
74 SOLUTION (Continued) 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. [110]or [1100]
Calculating Octahedral Sites 75 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 1 1 1 1,0,0,1,0,0,1,1,1 2 2 2 2 1 1 1 1 0,,0 1,,0 1,,1 0,,1 2 2 2 2 1 1 1 1 0,0, 1,0, 1,1, 0,1, 2 2 2 2 plus the center position, 1/2, 1/2, 1/2.
76 SOLUTION (Continued) 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
Design of a Radiation-Absorbing Wall 77 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. Figure A5: Calculation of an octahedral interstitial site.
78 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/ 2 r = 2 R R = ( - 1)R r/r = 0.414 2 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.
Radius Ratio for KCl 79 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, r K + = 0.133 nm and r Cl - = 0.181 nm, so: r K +/ r Cl - = 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.
80 SOLUTION KCl packing fraction b. The ions touch along the body diagonal of the unit cell, so: Diagonal (d) = 2r K + + 2r Cl - = 2(0.133) + 2(0.181) = 0.628 nm a 0 = 0.363 nm (d 2 =2a 02 ) Packing 4 3 factor 4 3 r (1K ion) 3 4 (0.133) (0.181) 3 3 (0.363) 3 K 3 a 3 0 4 3 r 0.725 3 Cl (1Cl ion)
Illustrating a Crystal Structure and Calculating Density 81 Show that MgO has the sodium chloride crystal structure and calculate the density of MgO. SOLUTION One may know that, r Mg +2 = 0.066 nm and r O -2 = 0.132 nm, so: r Mg +2 /r O -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.
82 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) a 0 = 0.396 nm = 3.96 10-8 cm (4Mg 2 (3.96 )(24.312) 10 8 cm) 3 (4O (6.02-2 )(16) 10 23 ) 4.31g / cm 3
Calculating the Theoretical Density of GaAs 83 The lattice constant of gallium arsenide (GaAs) is 5.65 Å. Show that the theoretical density of GaAs is 5.33 g/cm 3. 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 10 23 atoms) of Ga has a mass of 69.7 g. Therefore, the mass of four Ga atoms will be (4 * 69.7/6.023 10 23 ) g.
84 SOLUTION (Continued) Each mole (6.023 10 23 atoms) of As has a mass of 74.9 g. Therefore, the mass of four As atoms will be (4 * 74.9/6.023 10 23 ) g. These atoms occupy a volume of (5.65 10-8 ) 3 cm 3. density mass volume 4(69.7 74.9) / 6.023 (5.65 10 8 cm) 3 10 23 Therefore, the theoretical density of GaAs will be 5.33 g/cm 3.
85 (c) 2003 Brooks/Cole Publishing / Thomson Learning Figure A6 (a) Tetrahedron and (b) the diamond cubic (DC) unit cell. This open structure is produced because of the requirements of covalent bonding.
Determining the Packing Factor for Diamond Cubic Silicon 86 Determine the packing factor for diamond cubic silicon. Figure A7 Determining the relationship between lattice parameter and atomic radius in a diamond cubic cell.
SOLUTION 87 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: Packing 3 a 0 factor 8r (8 atoms/cell)( 4 3 (8)( r ) 3 3 (8r / 3) 0.34 Compared to close packed structures this is a relatively open structure. a 3 0 4 3 r 3 )
Calculating the Radius, Density, and Atomic Mass of Silicon 88 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, radius of silicon atom = 1.176 Å. There are eight Si atoms per unit cell. 3a0 8r the density mass volume 8(28.1) (5.43 / 6.023 10 8 cm) 10 3 23 2.33g / cm 3
Theoretical Density, Density = = = Mass of Atoms in Unit Cell Total Volume of Unit Cell n A V C N A where n = number of atoms/unit cell A = atomic weight V C = Volume of unit cell = a 3 for cubic NA = Avogadro s number = 6.023 x 10 23 atoms/mol 89
Theoretical Density, R a Ex: Cr (BCC) A = 52.00 g/mol R = 0.125 nm n = 2 a = 4R/ 3 = 0.2887 nm atoms unit cell = volume unit cell a 3 2 52.00 6.023 x 10 23 g mol theoretical actual atoms mol = 7.18 g/cm3 = 7.19 g/cm3 90
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) 91
PRACTICE example a b c 1. Intercepts 2. Reciprocals c z 3. Reduction 4. Miller Indices example a b c 1. Intercepts 2. Reciprocals 3. Reduction a x a c z b b y y 4. Miller Indices x 92
Crystallographic Planes z example a b c 1. Intercepts 1 1 2. Reciprocals 1/1 1/1 1/ 3. Reduction 1 1 1 1 0 0 4. Miller Indices (110) example a b c 1. Intercepts 1/2 2. Reciprocals 1/½ 1/ 1/ 2 0 0 3. Reduction 1 0 0 4. Miller Indices (100) a x a x c c z b b y y 93
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, 94
(b) For 2 atoms in a bcc shell, close-packed cubic (fcc) 95
Self-Test 5.7B Estimate the density of cesium iodide from its crystal structure. Solution CsI has a cesium chloride (bcc) type structure. r(cs + ) = 170 pm; r(i - ) = 220 pm Length of diagonal, b = 170 + 2(220) + 170 pm = 780 pm 1 2 Length of side a = 780/3 = 450 pm Hence unit cell volume is = 9.11 x 10 7 pm 3 or 9.11 x 10-23 cm 3 (1 pm 3 = 10-30 cm 3 ) Each bbc C unit cell has one Cs + ion and one I - ion, Density = mass/volume = (132.91 + 126.90) g mol -1 (6.022 x 10 23 mol -1 ) 9.11 x 10-23 cm 3 = 4.74 g cm -3 96
Classify each of the following solids as ionic, network, metallic, or molecular: (a) quartz, SiO 2 ; (b) limestone, CaCO 3 ; (c) dry ice, CO 2 ; (d) sucrose, C 12 H 22 O 11 ; (e) polyethylene, a polymer with molecules consisting of chains of thousands of repeating CH 2 CH 2 - units. 97 Solution
190s What percentage of space is occupied by close-packed cylinders of length l and radius r? Solution 98
99 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 = a 3 = (409 pm) 3 = 6.83 x 10-23 cm 3 4 atoms/unit cell in a face-centered cubic cell m = 4 Ag atoms 107.9 g x mole Ag x 1 mole Ag 6.022 x 10 23 atoms = 7.17 x 10-22 g d = m V 7.17 x 10-22 g = = 10.5 g/cm3 6.83 x 10-23 cm 3
Bonding in Solids A group IVA element with a density of 11.35 g/cm 3 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 100
Bonding in Solids 1. Determine the volume of a single unit cell. 0 1A 10-8 cm thus Face centered cubic unit cells are cubic so V -8 3-22 3 4.9510 cm 1.2110 cm 0 4.95 A 4.9510-8 cm 3 2. Use the density to determine the mass of a unit cell. 1.2110-22 cm 3 11.35 g 3 cm 1.3810-21 g one unit cell 101
Bonding in Solids 3. Determine the mass of one atom in the unit cell. Because face centered cubic has 4 atoms per unit cell the mass of one atom can be determined in this fashion. 1.3810 4-21 atoms unit cell g unit cell 3.4410 22 g atom 4. Determine the mass of one mole of these atoms. 3.4410 22 g atom 23 6.02210 atoms 207 g/mole mole 102
How many of each ion are contained within a unit cell of ZnS? 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. Solution The ZnS unit cell contains four Zn2+ ions (interior) and four S2- ions [8 1 (corners) and 6 1 (faces)] 8 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. ΔH vap is 40.79 kj/mol and ΔH fus is 6.01 kj/mol. Note: The ΔH vap 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 4.184 J q = msδt = 1.00 g g C 63.0 C = 2.64 102 J = 0.264 kj 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) 1.00 g 18.02 g/mol = 0.0555 mol water q1 = nδh vap = 0.0555 mol 40.79 kj = 2.26 kj mol q2 = msδt = 1.00 g 4.184 J 63.0 C = 2.64 10 2 J = 0.264 kj g 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
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) 4.184 J g C = 1.55 10 4 J = 15.5 kj 100.0 g 18.02 g/mol = 5.55 mol water q1 = nδh fus = 5.55 mol 6.01 kj mol = 33.4 kj 4.184 J q2 = msδt = 100.0 g 37.0 C = 1.55 10 g C 4 J = 15.5 kj 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 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
Phase Diagrams The phase diagram of water:
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. Strategy Each point on the phase diagram corresponds to a pressuretemperature 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.
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.