solution manual of principles of electronic materials and devices by kasop (2nd ed)

Solutions Manual to accompany

Principles of Electronic

Materials and Devices

Second Edition

S.O. Kasap University of Saskatchewan

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Solutions Manual to accompany

PRINCIPLES OF ELECTRONIC MATERIALS AND DEVICES, SECOND EDITION

S.O. KASAP

Published by McGraw-Hill Higher Education, an imprint of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas,

New York, NY 10020. Copyright The McGraw-Hill Companies, Inc., 2002, 1997. All rights reserved.

The contents, or parts thereof, may be reproduced in print form solely for classroom use with PRINCIPLES OF ELECTRONIC

MATERIALS AND DEVICES, Second Edition by S.O. Kasap, provided such reproductions bear copyright notice, but may not be

reproduced in any other form or for any other purpose without the prior written consent of The McGraw-Hill Companies, Inc.,

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Solutions to Principles of Electronic Materials and Devices: 2nd Edition (Summer 2001) Chapter 1

1.1

Second Edition ( 2001 McGraw-Hill)

Chapter 1

1.1 The covalent bondConsider the H2 molecule in a simple way as two touching H atoms as depicted in Figure 1Q1-1. Doesthis arrangement have a lower energy than two separated H atoms? Suppose that electrons totallycorrelate their motions so that they move to avoid each other as in the snapshot in Figure 1Q1-1. Theradius ro of the hydrogen atom is 0.0529 nm. The electrostatic potential energy PE of two charges Q1and Q2 separated by a distance r is given by Q1Q2/(4or).

Using the Virial Theorem stated in Example 1.1 (in textbook) consider the following:

a. Calculate the total electrostatic potential energy (PE) of all the charges when they are arranged asshown in Figure 1Q1-1. In evaluating the PE of the whole collection of charges you must considerall pairs of charges and, at the same time, avoid double counting of interactions between the same pairof charges. The total PE is the sum of the following: electron 1 interacting with the proton at adistance ro on the left, proton at ro on the right, and electron 2 at a distance 2ro + electron 2 interactingwith a proton at ro and another proton at 3ro + two protons, separated by 2ro, interacting with eachother. Is this configuration energetically favorable?

b. Given that in the isolated H-atom the PE is 2 (13.6 eV), calculate the change in PE with respect totwo isolated H-atoms. Using the Virial theorem, find the change in the total energy and hence thecovalent bond energy. How does this compare with the experimental value of 4.51 eV?

Solution

Nucleus

Hydrogen

ro

eNucleus

Hydrogen

roe

2

1

Figure 1Q1-1 A simplified view of the covalent bond in H2: a snap shot at one instant in time. Theelectrons correlate their motions and avoid each other as much as possible.

a Consider the PE of the whole arrangement of charges shown in the figure. In evaluating the PE ofall the charges, we must avoid double counting of interactions between the same pair of charges. The totalPE is the sum of the following:

Electron 1 interacting with the proton at a distance ro on the left, with the proton at ro on theright and with electron 2 at a distance 2ro

+ Electron 2 on the far left interacting with a proton at ro and another proton at 3ro+ Two protons, separated by 2ro, interacting with each other


1.2

PE er

e

r

e

r

e

r

e

r

e

r

o o o o o o

o o o o

o o

=

+

+

2 2 2

2 2

2

4 4 4 2

4 4 3

4 2

( )

substituting and calculating we findPE = 1.0176 10-17 J or -63.52 eV

The negative PE for this particular arrangement indicates that this arrangement of charges is indeedenergetically favorable compared with all the charges infinitely separated (PE is then zero).b The potential energy of an isolated H-atom is 2 13.6 eV or 27.2 eV. The difference between thePE of the H2 molecule and two isolated H-atoms is,

PE = (63.52 eV) 2(27.2) eV = - 9.12 eVWe can write the last expression above as changes in the total energy as

E = = = 12 12 9 12PE ( . )eV 4.56 eV This is the change in the total energy which is negative. The H2 molecule has lower energy than

two H-atoms by 4.56 eV which is the bonding energy. This is very close to the experimental value of 4.51eV. (Note: We used an ro value from quantum mechanics - so the calculation was not totally classical).

1.2 Ionic bonding and NaClThe interaction energy between Na+ and Cl- ions in the NaCl crystal can be written as

E rr r

( ) . .= + 4 03 10 6 97 1028 96

8

where the energy is given in joules per ion pair, and the interionic separation r is in meters. Calculate thebinding energy and the equilibrium ionic separation in the crystal; include the energy involved in electrontransfer from Cl- to Na+. Also estimate the elastic modulus Y of NaCl given that

Yr

d Edro r ro

=

16

2

2

SolutionThe PE curve for NaCl is given by

E rr r

( ) . .= + 4 03 10 6 97 1028 96

8

where E is the potential energy and r represents interionic separation.We can differentiate this and set it to zero to find the minimum PE, and consequently the minimum

(equilibrium) separation (ro).


1.3

dE r

dro

o

( )= 0

+ =55 761

104 03

110

096 9 28 2. .r ro o

isolating ro: ro = 2.81 10-10 m or 2.81

The apparent ionic binding energy (Eb) for the ions alone, in eV, is:

EE r

q r r qbo

o o

= =

+

( ) . .4 03 10 6 97 10 128 96

8

Eb =

( ) +

( )

4 03 10

2 81 10

6 97 10

2 81 10

11 602 10

28

10

96

10 8 19

.

.

.

. . m m J/eV

E b = 7.83 eV

Note however that this is the energy required to separate the Na+ and Cl- ions in the crystal andthen to take the ions to infinity, that is to break up the crystal into its ions. The actual bond energyinvolves taking the NaCl crystal into its constituent neutral Na and Cl atoms. We have to transfer theelectron from Cl- to Na+. The energy for this transfer, according to Figure 1Q2-1, is -1.5 eV (negativerepresents energy release). Thus the bond energy is 7.83 - 1.5 = 6.33 eV as in Figure 1Q2-1.

Cl Na+

ro = 0.28 nm

6

6

0

6.3

0.28 nm

Potential energy E(r), eV/(ion-pair)

Separation, r

Coh

esiv

e E

nerg

y

1.5 eV

r=

Cl Nar=

Na+Cl

Figure 1Q2-1 Sketch of the potential energy per ion pair in solid NaCl. Zero energy

corresponds to neutral Na and Cl atoms infinitely separated.

If r is defined as a variable representing interionic separation, then Youngs modulus is given by:

Yr

d

dr

dE r

dr=

16

( )

Y r rr

=

+

16

8 061

10501 84

11028 3 96 10

. .

Yr r

= 8 364 101

1 3433 10195

1128

4. .


1.4

Substituting the value for equilibrium separation (ro) into this equation (2.81 10-10 m),

Y = 7.54 1010 Pa = 75 GPa

This value is somewhat larger than about 40 GPa in Table 1.2 (in the textbook), but not too farout.

*1.3 van der Waals bondingBelow 24.5 K, Ne is a crystalline solid with an FCC structure. The interatomic interaction energy peratom can be written as

E rr r

( ) . .=

2 14 45 12 136 12

(eV/atom)

where and are constants that depend on the polarizability, the mean dipole moment, and the extent of

overlap of core electrons. For crystalline Ne, = 3.121 10-3 eV and = 0.274 nm.

a. Show that the equilibrium separation between the atoms in an inert gas crystal is given by ro =(1.090). What is the equilibrium interatomic separation in the Ne crystal?

b. Find the bonding energy per atom in solid Ne.

c. Calculate the density of solid Ne (atomic mass = 20.18 g/mol).

Solutiona Let E = potential energy and x = distance variable. The energy E is given by

E xx x

( ) . .=

2 14 45 12 136 12

The force F on each atom is given by

F xdE x

dxxx

xx

( )( )

. .= =

2 145 56 86 7

11

2

5

2

F xx x

( ) . .=

2 145 56 86 7

12

13

6

7

When the atoms are in equilibrium, this net force must be zero. Using ro to denote equilibriumseparation,

F ro( ) = 0

2 145 56 86 7 012

13

6

7

. .

r ro o

=

145 56 86 712

13

6

7. .

r ro o=


1.5

r

ro

o

13

7

12

6

145 5686 7

= ..

ro = 1.090

For the Ne crystal, = 2.74 10-10 m and = 0.003121 eV. Therefore,

ro = 1.090(2.74 10-10 m) = 2.99 10-10 m for Ne.

b Calculate energy per atom at equilibrium:

E rr ro o o

( ) . .=

2 14 45 12 13

6 12

E ro( ) . .

.

.

= ( ) ( )

2 0 003121 1 602 10

14 45

12 13

19

6

12eV J/eV

2.74 10 m 2.99 10 m

2.74 10 m 2.99 10 m

-10

-10

-10

-10

E (ro) = -4.30 10-21 J or -0.0269 eV

Therefore the bonding energy in solid Ne is 0.027 eV per atom.

c To calculate the density, remember that the unit cell is FCC, and density = (mass of atoms in theunit cell) / (volume of unit cell). There are 4 atoms per FCC unit cell, and the atomic mass of Ne is 20.18g/mol.

a

a

a2R

FCC Unit Cell(a) (b) (c)

a

Figure 1Q3-1

(a) The crystal structure of copper which is face-centered-cubic (FCC). The atoms are positionedat well-defined sites arranged periodically, and there is a long-range order in the crystal.

(b) An FCC unit cell with close-packed spheres.

(c) Reduced-sphere representation of the FCC unit cell.

Examples: Ag, Al, Au, Ca, Cu, -Fe (>912 C), Ni, Pd, Pt, Rh.

Since it is an FCC crystal structure, let a = lattice parameter (side of cubic cell) and R = radius ofatom. The shortest interatomic separation is ro = 2R (atoms in contact means nucleus to nucleus separationis 2R (see Figure 1Q3-1).

R = ro/2

and 2a2 = (4R)2


1.6

a Rro

= = = ( )2 2 2 2 2 2 2 99 10 10. m

a = 4.228 10-10 m

Therefore, the volume (V) of the unit cell is:

V = a3 = (4.228 10-10 m)3 = 7.558 10-29 m3

The mass (m) of 1 Ne atom in grams is the atomic mass (Mat) divided by NA, because NA numberof atoms have a mass of Mat.

m = Mat / NA

m =( )( )

=

20 18 0 001. . g/mol kg/g

6.022 10 mol3.351 10 kg23 -1

26

There are 4 atoms per unit cell in the FCC cell. The density () can then be found by:

= (4m) / V = [4 (3.351 10-26 kg)] / (7.558 10-29 m3)

= 1774 kg/m3

In g/cm3 this density is:

= ( ) ( ) =1774100

10003

kg/mcm/m

g/kg3

1.77 g/cm3

The density of solid Ne is 1.77 g cm-3.

1.4 Kinetic Molecular TheoryCalculate the effective (rms) speeds of the He and Ne atoms in the He-Ne gas laser tube at roomtemperature (300 K).

Solution

Gas atoms

Figure 1Q4-1 The gas molecules in the container are in random motion.


1.7

Current regulated HV power supply

Flat mirror (Reflectivity = 0.999) Concave mirror(Reflectivity = 0.985)

He-Ne gas mixtureLaser beam

Very thin tube

Figure 1Q4-2 The He-Ne gas laser.

To find the root mean square velocity (vrms) of He atoms at T = 300 K:

The atomic mass of He is (from Periodic Table) Mat = 4.0 g/mol. Remember that 1 mole has amass of Mat grams. Then one He atom has a mass (m) in kg given by:

mM

Nat

A

= =

( ) =

4 06 022 10

0 001 6 642 1023 127.

. . .

g/molmol

kg/g kg

From kinetic theory:

12

32

2m v kTrms( ) =

vkT

mrms= =

( )( )( )3 3

3001.381 10 J K K

6.642 10 kg

-23 -1

-27

vrms = 1368 m/s

The root mean square velocity (vrms) of Ne atoms at T = 300 K can be found using the samemethod as above, changing the atomic mass to that of Ne, Mat = 20.18 g/mol. After calculations, the massof one Ne atom is found to be 3.351 10-26 kg, and the root mean square velocity (vrms) of Ne is found tobe vrms = 609 m/s.

*1.5 Vacuum depositionConsider air as composed of nitrogen molecules N2.

a. What is the concentration n (number of molecules per unit volume) of N2 molecules at 1 atm and 27C?

b. Estimate the mean separation between the N2 molecules.

c. Assume each molecule has a finite size that can be represented by a sphere of radius r. Also assumethat l is the mean free path, defined as the mean distance a molecule travels before colliding withanother molecule, as illustrated in Figure 1Q5-1a. If we consider the motion of one N2 molecule,with all the others stationary, it is apparent that if the path of the traveling molecule crosses the cross-sectional area S =(2r)2, there will be a collision. Since l is the mean distance between collisions,there must be at least one stationary molecule within the volume Sl, as shown in Figure 1Q5-1a.Since n is the concentration, we must have n(Sl) = 1 or l =1/(4r2n). However, this must becorrected for the fact that all the molecules are in motion, which only introduces a numerical factor,so that


1.8

l =

12 41 2 2/ r n

Assuming a radius r of 0.1 nm, calculate the mean free path of N2 molecules between collisions at 27C and 1 atm.

d. Assume that an Au film is to be deposited onto the surface of an Si chip to form metallicinterconnections between various devices. The deposition process is generally carried out in avacuum chamber and involves the condensation of Au atoms from the vapor phase onto the chipsurface. In one procedure, a gold wire is wrapped around a tungsten filament, which is heated bypassing a large current through the filament (analogous to the heating of the filament in a light bulb)as depicted in Figure 1Q5-1b. The Au wire melts and wets the filament, but as the temperature of thefilament increases, the gold evaporates to form a vapor. Au atoms from this vapor then condenseonto the chip surface, to solidify and form the metallic connections. Suppose that the source(filament)-to-substrate (chip) distance L is 10 cm. Unless the mean free path of air molecules is muchlonger than L, collisions between the metal atoms and air molecules will prevent the deposition of theAu onto the chip surface. Taking the mean free path l to be 100L, what should be the pressure insidethe vacuum system? (Assume the same r for Au atoms).

v

S = (2r)2

Any molecule withcenter in S gets hit.

Molecule

Molecule

(a)

Vacuumpump

Vacuum Hotfilament

Evaporatedmetal atoms

Semiconductor

Metal film

(b)

Figure 1Q5-1

(a) A molecule moving with a velocity v travels a mean distance l between collisions. Since thecollision cross-sectional area is S, in the volume Sl there must be at least one molecule.Consequently, n(Sl) = 1.

(b) Vacuum deposition of metal electrodes by thermal evaporation.

SolutionAssume T = 300 K throughout. The radius of the nitrogen molecule (given approximately) is r =

0.1 10-9 m. Also, we know that pressure P = 1 atm = 1.013 105 Pa.

Let N = total number of molecules , V = volume and k = Boltzmanns constant. Then:

PV = NkT

(Note: this equation can be derived from the more familiar form of PV = RT, where is the totalnumber of moles, which is equal to N/NA, and R is the gas constant, which is equal to k NA.)

The concentration n (number of molecules per unit volume) is defined as:

n = N/V

Substituting this into the previous equation, the following equation is obtained:

P = nkT

a What is n at 1 atm and T = 27 C + 273 = 300 K?


1.9

n = P/(kT)

n =

( )( )1 013 10

300

5.

Pa

1.381 10 J K K-23 -1

n = 2.45 1025 molecules per m3

b What is the mean separation between the molecules?

Consider a crystal of the material which is a cube of unit volume, each side of unit length as shownin Figure 1Q5-2. To each atom we can attribute a portion of the whole volume which for simplicity is acube of side d. Thus, each atom is considered to occupy a volume of d3.

Length = 1

Length = 1

Length = 1

Each atom has this portionof the whole volume. Thisis a cube of side d. .d

d

d

Interatomic separation = d

Volume of crystal = 1

Figure 1Q5-2 Relationship between interatomic separation

and the number of atoms per unit volume.

The actual or true volume of the atom does not matter. All we need to know is how much volumean atom has around it given all the atoms are identical and that adding all the atomic volumes must give thewhole volume of the crystal.

Suppose that there are n atoms in this crystal. Then n is the atomic concentration, number of atomsper unit volume. Clearly, n atoms make up the crystal so that

n d 3 = Crystal volume = 1

Remember that this is only an approximation. The separation between any two atoms is d. Thus,

d

n

= =

( )1 1

2 45 101

325 33 . m

d = 3.44 10-9 m or 3.4 nm

c Assuming a radius, r, of 0.1 nm, what is the mean free path, l, between collisions?

l =

=

( ) ( ) 1

2 41

2 4 0 1 10 2 45 102 9 2 25 3 r n . . m m

l = 2.30 10-7 m or 230 nm

d We need the new mean free path, l = 100L, or 0.1 m 100 (L is the source-to-substrate distance)

l = 10 m


1.10

The new l corresponds to a new concentration n of nitrogen molecules.

=

l1

2 4 2r n

=

=

( ) ( )n r18

2 18

2

0 1 10 102 9 2 l . m m

= 5.627 1017 m-3

This new concentration of nitrogen molecules requires a new pressure, P:

P = nkT = (5.627 1017 m-3)(1.381 10-23 J K-1)(300 K) = 0.00233 Pa

In atmospheres this is:

=

= P

0 002331 013 105

. .

PaPa/atm

2.30 10 atm8

In units of torr this is:

= ( )( ) = P 2 30 10 7608. atm torr/atm 1.75 10 torr5There is an important assumption made, namely that the cross sectional area of the Au atom is

about the same as that of N2 so that the expression for the mean free path need not be modified to accountfor different sizes of Au atoms and N2 molecules. The calculation gives a magnitude that is quite close tothose used in practice , e.g. a pressure of 10-5 torr.

1.6 Heat capacitya. Calculate the heat capacity per mole and per gram of N2 gas, neglecting the vibrations of the

molecule. How does this compare with the experimental value of 0.743 J g-1 K-1?

b. Calculate the heat capacity per mole and per gram of CO2 gas, neglecting the vibrations of themolecule. How does this compare with the experimental value of 0.648 J K-1 g-1? Assume that CO2molecule is linear (O-C-O), so that it has two rotational degrees of freedom.

c. Based on the Dulong-Petit rule, calculate the heat capacity per mole and per gram of solid silver.How does this compare with the experimental value of 0.235 J K-1 g-1?

d. Based on the Dulong-Petit rule, calculate the heat capacity per mole and per gram of the siliconcrystal. How does this compare with the experimental value of 0.71 J K-1 g-1?

Solution

a N2 has 5 degrees of freedom: 3 translational and 2 rotational. Its molar mass is Mat = 2 14.01g/mol = 28.02 g/mol.

Let Cm = heat capacity per mole, Cs = specific heat capacity (heat capacity per gram), and R = gasconstant, then:

Cm = = ( ) = 5252

R 8.315 J K mol-1 -1 20.8 J K mol1 1

Cs = Cm/ Mat = (20.8 J K-1 mol-1)/(28.02 g/mol) = 0.742 J K-1 g-1

This is close to the experimental value.

b CO2 has the linear structure O=C=O. Rotations about the molecular axis have negligible rotationalenergy as the moment of inertia about this axis is negligible. There are therefore 2 rotational degrees of


1.11

freedom. In total there are 5 degrees of freedom: 3 translational and 2 rotational. Its molar mass is Mat = 12.01 + 2 16 = 44.01 g/mol.

Cm = = ( ) = 5252

8 315R . J K mol1 1 20.8 J K mol1 1


This is smaller than the experimental value 0.648 J K-1 g-1, because vibrational energy wasneglected in the 5 degrees of freedom assigned to the CO2 molecule.

c For solid silver, there are 6 degrees of freedom: 3 vibrational KE and 3 elastic PE terms. Its molarmass isMat = 107.87 g/mol.

Cm = = ( ) = 6262

8 315R . J K mol1 1 24.9 J K mol1 1


This is very close to the experimental value.

d For a solid, heat capacity per mole is 3R. The molar mass of Si is Mat = 28.09 g/mol.

Cm = = ( ) = 6262

8 315R . J K mol1 1 24.9 J K mol1 1


The experimental value is substantially less and is due to the failure of classical physics. One hasto consider the quantum nature of the atomic vibrations and also the distribution of vibrational energyamong the atoms. The student is referred to modern physics texts (under heat capacity in the Einsteinmodel and the Debye model of lattice vibrations).

1.7 Thermal expansion

a. If is the thermal expansion coefficient, show that the thermal expansion coefficient for an area is2. Consider an aluminum square sheet of area 1 cm2. If the thermal expansion coefficient of Al atroom temperature (25 C) is about 24 10-6 K-1, at what temperature is the percentage change in thearea +1%?

b. The density of silicon at 25 C is 2.329 g cm-3. Estimate the density of silicon at 1000 C given that

the thermal expansion coefficient for Si over this temperature range has a mean value of about 3.5 10-6 K-1.

c. The thermal expansion coefficient of Si depends on temperature as,

= ( )[ ]{ } + 3 725 10 1 0 00588 124 5 548 106 10. exp . .T Twhere T is in Kelvins. The change in the density due to a change T in temperature is given by

= = 0 03V T T

By integrating this equation , calculate the density of Si at 1000 C and compare with the result in b.

Solutiona Consider an rectangular area with sides xo and yo. Then at temperature T0,


1.12

A x y0 0 0=

and at temperature T,

A x T y T x y T= +( )[ ] +( )[ ] = +( )0 0 0 0 21 1 1 that is

A x y T To= + + ( )[ ]0 21 2 .We can now use that A x y0 0 0= and neglect the term T( )2 because it is very small in comparison withthe linear term T (


1.13

fixed, cylindrical iron core. The iron core ensures a strong magnetic field through the coil. The coil ishinged to rotate in the air gap and around the iron core. When a current I is passed through the coil, itexperiences a torque and tries to rotate, but the rotation is restricted by the spring. A needle attached tocoil indicates the amount of rotation. The deflection of the needle is proportional to the torque , byvirtue of Hooke's Law

= Kwhere, from electromagnetic theory, the torque is proportional to the product of the current I and themagnetic field B. Thus, measures the current I.

A given ammeter has a coil inductance of 0.1 mH and a coil resistance of 100 . The springconstant K is 1011 rad N-1 m-1 and the length of the needle is 10 cm.

a. What is the rms fluctuation in the position of the ammeter needle?

b. Assume the ammeter is placed in a closed circuit where the dc current is zero (I = 0), as shown inFigure 1Q8-1. Sketch the instantaneous current i(t) versus time. What is the minimum current thisammeter can measure (noise in the current)? How can this be improved?

SolutionWe can assume a room temperature of T = 300 K, and we are given the ammeters characteristics.

a The spring constant (K) in this case relates and (torque) by the equation: = K

Spring constant: K = 1011 N-1 m-1 (Remember that rads are dimensionless units, and donot need to be carried through calculations)

Length of pointer: d = 0.1 m

Inductance of coil: L = 0.0001 H

is the rms (root mean square) value of the angle fluctuations. It must be in radians. The averageenergy fluctuations in the spring must be of the order of (1/2)kT.

12

1 12

2

KkT =

= = ( )( ) ( ) TKk 300 10 1 381 1011 23 . K N m J K1 1 1 = 2.04 10-5 rad

As the needle rotates randomly left and right about its equilibrium position, its tip is displacedrandomly. These random fluctuations in the displacement have an rms value of (where d is the length ofthe needle):

xrms = d = (2.04 10-5 rad)(0.10 m/rad) = 2.04 10-6 m or 2 mCertainly this extent is visible under a microscope, so if you tried to measure the position of the

needle accurately by focusing a microscope on it to obtain the current accurately, youd be wasting time.


1.14

Ammeter

0.1 mH

R (Resistance of the coil)

I = 0

i(t)

N S

Spring

Permanentmagnet

Fixed iron core MovingCoil

NeedleScale

0

Figure 1Q8-1 Left: The ammeter. Right: The ammeter short circuited. It should be reading azero current at all times (short circuit). But is it?

b Suppose that the ammeter is short circuited as in Figure 1Q8-1 and the current through the ammeteris monitored by an invisible, non-intrusive, second noiseless ammeter. Let i be the instantaneous currentin the circuit. Its rms value is then irms. The average energy fluctuation due to inductance in the coil mustbe in the order of (1/2)kT:

12

12

2Li kTrms =

iTk

Lrms= =

( ) ( )( )

300 K 1.381 10 J K

0.0001 H

-23 -1

irms = 6.44 10-9 A or 6 nA

Noise in the current measurements is 6 nA. If you were to examine the current in the circuit youwould see the following:

irms = 6 nA

i(t)iaverage = 0

t

Figure 1Q8-2 Noise fluctuations of current in ammeter.

Assumption: When we considered the random fluctuations, we assumed that the mechanicalmotion containing the spring (energy storage element) was decoupled from the electrical circuit containingthe inductance (energy storage element). An exact analysis is quite difficult but this first order treatmentprovides rough values on the limits of the system. We should calculate how much current, denoted in, isneeded to provide a deflection that is equal to xrms. For this approach we have to know more about theelectromagnetic operation of the ammeter coil. For a more rigorous calculation see Electricity andMagnetism, Third Edition, B. I. Bleaney and B. Bleaney (Oxford University Press, Oxford, 1976), Ch.23.


1.15

1.9 Electrical noiseConsider an amplifier with a bandwidth B of 5 kHz, corresponding to a typical speech bandwidth.Assume the input resistance of the amplifier is 1 M. What is the rms noise voltage at the input? Whatwill happen if the bandwidth is doubled to 10 kHz? What is your conclusion?

Solution

Bandwidth: B = 5 103 Hz

Input resistance: Rin = 106

Assuming room temperature: T = 300 K

The noise voltage (or the rms voltage, vrms) across the input is given by:

v kTR Brms in= = ( )( )( ) ( )4 4 300 10 5 106 31.381 10 J K K Hz-23 -1 vrms = 9.10 10

-6 V or 9.1 V

If the bandwidth is doubled: B = 2 5 103 = 10 103 Hz

= = vrms 4kTR Bin 1.29 10 V 12.9 V5 or

The larger the bandwidth, the greater the noise voltage. 5-10 kHz is a typical speech bandwidthand the input signal into the amplifier must be much greater than 13 V for amplification without noiseadded from the amplifier itself.

Voltage, v(t)

Time

vrms

Figure 1Q9-1 Random motion of conduction electrons in a conductor results in electrical noise.

1.10 Thermal activationA certain chemical oxidation process (e.g., SiO2) has an activation energy of 2 eV atom

-1.

a. Consider the material exposed to pure oxygen gas at a pressure of 1 atm. at 27 C. Estimate howmany oxygen molecules per unit volume will have energies in excess of 2 eV? (Consider numericalintegration of Equation 1.17, listed in textbook)

b. If the temperature is 900 C, estimate the number of oxygen molecules with energies more than 2 eV.What happens to this concentration if the pressure is doubled?


1.16

Solution 1: Method of EstimationThe activation energy (EA) of one atom is 2 eV/atom, or:

EA = (2 eV/atom)(1.602 10-19 J/eV) = 3.204 10-19 J/atom

a We are given pressure P = 1 atm = 1.013 105 Pa and temperature T = 300 K. If we consider aportion of oxygen gas of volume V = 1 m3 (the unit volume), the number of molecules present in the gas(N) will be equal to the concentration of molecules in the gas (n), i.e.: N = n. And since we know =N/NA, where is the total number of moles and NA is Avogadros number, we can make the followingsubstitution into the equation PV = RT:

PVn

NRT

A

=

isolating n, nN PV

RTA

= =

( ) ( )( )( )( )

6 022 10 1 013 10 1

300

123 5. .

mol Pa m

8.315 m Pa K mol K m

1 3

3 -1 -1 3

n = 2.445 1025 m-3

Therefore there are 2.445 1025 oxygen molecules per unit volume.For an estimation of the concentration of molecules with energy above 2 eV, we can use the

following approximation (remember to convert EA into Joules). If nA is the concentration of moleculeswith E > EA, then:

n

n

E

kTA A

= exp

n nE

kTAA

= = ( )

( )( )( )

exp . exp.

2 445 10

3 204 10

30025

19

mJ

1.381 10 J K K3

-23 -1

nA = 6.34 10-9 m- 3

However, this answer is only in the right order of magnitude. For a better calculation we need touse a numerical integration of n(E) from EA to .

b At T = 900 C + 273 = 1173 K and P = 1 atm, the same method as above can be used to find theconcentration of molecules with energy greater than EA. After calculations, the following numbers will beobtained:

n = 6.254 1024 m-3

nA = 1.61 1016 m- 3

This corresponds to an increase by a factor of 1025 compared to a temperature of T = 300 K.Doubling the pressure doubles n and hence doubles nA. In the oxidation of Si wafers,

high pressures lead to more rapid oxidation rates and a shorter time for the oxidation process.

Solution 2: Method of Numerical IntegrationTo find the number of molecules with energies greater than EA = 2 eV more accurately, numerical

integration must be used. Suppose that N is the total number of molecules. Let

y = nE/N

where nE is the number of molecules per unit energy, so that nE dE is the number of molecules in theenergy range dE.


1.17

Define a new variable x for the sake of convenience:

xE

kT=

E = Tkx

where E is the energy of the molecules. Using partial differentiation:

dE = Tk dx

The energy distribution function is given by:

ykT

EE

kT=

2 112

3

21

2

exp

substitute: y kTx

Tkx=

( )2

13

2exp

simplify: yx x

Tk=

2exp( )

Integration of y with respect to E can be changed to that with respect to x as follows:

ydEx xdE

Tk=

2 exp( )

ydEx xdx

=

2 exp( ) (substitute dE = Tk dx)We need the lower limit xA for x corresponding to EA:

xE

kTAA

= =

( )( ) =3.204 10 J

1.381 10 J K K

-19

-23 -1 30077 36

.

This is the activation energy EA in terms of x.

We also need the upper limit, xb which should be , but we will take it to be a multiple of xA:

xb = 2 xA = 154.72

We do this because the numerical integration is difficult with very small numbers, e.g. exp(-xA) =2.522 10-34.

The fraction, F, of molecules with energies greater than EA (= xA) is:

F ydEx x

dxE x

x

A A

B

= =

=

2 2 519 10 33exp( ) .At T = 300 K, P = 1 atm = 1.013 105 Pa, and V = 1 m3, the number of molecules per unit

volume is n:

PVn

NRT

A

=


1.18

nN PV

RTA

= = 2.445 10 m25 3

The concentration of molecules with energy greater than EA (nA) can be found using the fraction F :

nA = nF = (2.445 1025 m-3)(2.519 10-33 ) = 6.16 10-8 m-3

If we estimate nA by multiplying n by the Boltzmann factor as previously:

n nE

kTAA

= = exp 6.34 m 310 9

The estimate is out by a factor of about 10.

b The concentration of molecules with energy greater than EA (nE) can be found at T = 900 C + 273= 1173 K using the same method as in part a. After calculations, the following values will be obtained:

F = 1.3131 10-8

n = 6.254 1024 m-3

nA = 8.21 1016 m- 3

If we compare this value to the one obtained previously through estimation (1.61 1016 m-3), wesee the estimate is out by a factor of about 5. As stated previously, doubling the pressure doubles n andhence doubles nA. In the oxidation of Si wafers, high pressures lead to more rapid oxidation rates and ashorter time for the oxidation process.

1.11 Diffusion in Si

The diffusion coefficient of boron (B) atoms in a single crystal of Si has been measured to be 1.5 10 -18

m2 s-1 at 1000 C and 1.1 10 -16 m2 s-1 at 1200 C.

a. What is the activation energy for the diffusion of B, in eV/atom?

b. What is the preexponential constant Do?

c. What is the rms distance (in micrometers) diffused in 1 hour by the B atom in the Si crystal at 1200C and 1000 C?

d. The diffusion coefficient of B in polycrystalline Si has an activation energy of 2.4-2.5 eV atom-1 andDo = (1.5-6) 10

-7 m2 s-1. What constitutes the diffusion difference between the single crystalsample and the polycrystalline sample?

SolutionGiven diffusion coefficients at two temperatures:

T1 = 1200 C + 273 = 1473 K D1 = 1.1 10-16 m2/s

T2 = 1000 C + 273 = 1273 K D2 = 1.5 10-18 m2/s

a The diffusion coefficients (D1 and D2) at certain temperatures (T1 and T2) are given by:

D DE q

kToA

11

=

exp D D

E q

kToA

22

=

exp

where EA is the activation energy in eV/atom. Since we know the following:

exp( )exp( )

exp( )

=

u

ww u


1.19

we can take the ratio of the diffusion coefficients to express them in terms of the activation energy (EA):

D

D

DE q

kT

DE q

kT

E q

kT

E q

kT

q T T E

T T k

oA

oA

A A A1

2

1

2

2 1

1 2

1 2

=

=

=

[ ]

exp

exp

exp exp

E

T T kD

D

q T TA=

( )1 2

1

2

1 2

ln

EA =( )( ) ( )

( ) ( )

1473 1273 1 381 101 1 101 5 10

1 602 10 1473 1273

2316

18

19

. ln. .

.

K K J Km /sm /s

J/eV K K

12

2

EA = 3.47 eV/atom

b To find Do, use one of the equations for the diffusion coefficients:

D DE q

kToA

11

=

exp

DD

E q

kT

o

A

=

=

( )

( ) ( )( )( )

1

1

16

19

23

1 1 10

3 47 1 602 10

1 381 10 1473exp

.

exp. .

.

m /s

eV J/eV

J K K

2

1

D 0 = 8.12 10-5 m2/ s

c Given: time (t) = (1 hr) (3600 s/hr) = 3600 s

At 1000 C, rms diffusion distance (L1000 C) in time t is given by:

L1000 C = ( ) = ( )( )2 2 1 5 10 36002 18D t . m /s s2 L1000 C = 1.04 10

-7 m or 0.104 m

At 1200 C:

L1200 C = = ( ) = ( )( )2 2 1 1 10 36001 16D t . m /s s2 L1200 C = 8.90 10

-7 m or 0.89 m (almost 10 times longer than at 1000 C)d Diffusion in polycrystalline Si would involve diffusion along grain boundaries, which is easierthan diffusion in the bulk. The activation energy is smaller because it is easier for an atom to break bondsand jump to a neighboring site; there are vacancies or voids and strained bonds in a grain boundary.

1.12 Diffusion in SiO2The diffusion coefficient of P atoms in SiO2 has an activation energy of 2.30 eV atom

-1 and Do = 5.73

10-9 m2 s-1. What is the rms distance diffused in one hour by P atoms in SiO2 at 1200 C?


1.20

SolutionGiven:

Temperature: T = 1200 C + 273 = 1473 K

Damping constant: Do = 5.73 10-9 m2/s

Activation energy: EA = 2.30 eV/atom

Time: t = (1 hr) (3600 s/hr) = 3600 s

Using the following equation, find the diffusion coefficient D:

D DE q

kToA

= exp

D = ( ) ( ) ( )( )( )

5 73 102 30 1 602 10

1 381 10 14739

19

23. exp. .

. m /s

eV J/eV

J K K2

1

D = 7.793 10-17 m2/s

The rms diffusion distance (L1200 C) is given as:

L1200 C = 2 2 7 793 10 360017Dt( ) = ( )( ). m /s s2

L1200 C = 7.49 10-7 m or 0.75 m

Comment: The P atoms at 1200 C seems to be able to diffuse through an oxide thickness of

about 0.75 m.

1.13 BCC and FCC Crystals

a. Molybdenum has the BCC crystal structure, has a density of 10.22 g cm-3 and an atomic mass of95.94 g mol-1. What is the atomic concentration, lattice parameter a, and atomic radius ofmolybdenum?

b. Gold has the FCC crystal structure, a density of 19.3 g cm-3 and an atomic mass of 196.97 g mol-1.What is the atomic concentration, lattice parameter a, and atomic radius of gold?

Solutiona. Since molybdenum has BCC crystal structure, there are 2 atoms in the unit cell. The density is

= = ( ) ( )Mass of atoms in unit cellVolume of unit cell

Number of atoms in unit cell Mass of one atom

Volume of unit cell

that is, =

2

3

M

N

a

at

A .

Solving for the lattice parameter a we receive


1.21

aM

Nat

A

= =

( )( ) ( )

2 2 95 94 10

10 22 10 6 022 103

3 1

3 3 23 13

.

. .

kg mol

kg m mol = 3.147 10-10 m

The Atomic concentration is 2 atoms in a cube of volume a3, i.e.

naat

= =

( )2 2

3 147 103 10 3. m

= 6.415 1022 cm-3 = 6.415 1028 m- 3

For a BCC cell, the lattice parameter a and the radius of the atom R are in the following relation(listed in Table 1.3 in the textbook):

Ra

=

34

i.e.

R =( )3 147 10 3

4

10. m = 1.363 10-10 m

b . Gold has the FCC crystal structure, hence, there are 4 atoms in the unit cell (as shown in Table 1.3in the textbook).

The lattice parameter a is

aM

Nat

A

= =

( )( ) ( )

4 4 196 97 10

19 3 10 6 022 103

3 1

3 3 23 13

.

. .

kg mol

kg m mol = 4.077 10-10 m

The atomic concentration is

naat

= =

( )4 4

4 077 103 10 3. m

= 5.901 1022 cm-3 = 5.901 1028 m- 3

For an FCC cell, the lattice parameter a and the radius of the atom R are in the following relation(shown in Table 1.3 in the textbook):

Ra

=

24

i.e.

Rm

=

( )4 077 10 24

10. = 1.442 10-10 m

*1.14 BCC and FCC crystalsa. Consider iron below 912 C, where its structure is BCC. Given the density of iron as 7.86 g cm-3

and its atomic mass as 55.85 g/mol, calculate the lattice parameter of the unit cell and the radius of theFe atom.

b. At 912 C, iron changes from the BCC (-Fe) to the FCC (-Fe) structure. The radius of the Featom correspondingly changes from 0.1258 nm to 0.1291 nm. Calculate the density of -Fe andexplain whether there is a volume expansion or contraction during this phase change.


1.22

SolutionGiven:

Density of iron at room temperature: = 7.86 103 kg/m3

Atomic mass of iron: Mat = 55.85 g/mole

a For the BCC structure, the density is given by:

=

( )2

10 3

3

M

Na

at

A

kg/g

Thus the lattice parameter a is:

aM

Nat

A

= ( )

1500

1

3

g kg/

a = ( )( )

( ) ( )

1500

55 85

6 022 10 7 86 1023 3

1

3

.

. . g kg

g mol

mol kg m/

/

/1 3

a = 2.87 10-10 m

The radius of the Fe atom, R, and the lattice parameter, a, are related.

a R3 4=

R a= = ( )14

314

3 2 87 10 10. m

R = 1.24 10-10 m

b Fe has a BCC structure just below 912 C (-Fe). An Fe atom in the -Fe state has a radius ofRBCC = 0.1258 10

-9 m. The density of -Fe is therefore:

BCC

at

A

BCC

M

N

R=

( )

=

( )( )( )

( )

210

43

255 85 10

6 022 10

4 0 1258 10

3

3

3

3

23

9 3

.

.

.

kg/g g/mol kg/g

mol

m

1

BCC = 7564 kg/m3 (less than at room temperature)

Fe has a FCC structure just above 912 C (-Fe). An Fe atom in the -Fe state has a radius ofRFCC = 0.1291 10

-9 m (Remember that for a FCC structure, a RFCC2 4= ). The density of -Fe istherefore:

FCC

at

A

FCC

M

N

R=

( )

=

( )( )( )

( )

410

42

455 85 10

6 022 10

4 0 1291 10

2

3

3

3

23

9 3

.

.

.

kg/g g/mol kg/g

mol

m

1


1.23

FCC = 7620 kg/m3

As the density increases, the volume must contract (the Fe retains the same mass).

1.15 Planar and surface concentrationsNiobium (Nb) has the BCC crystal with a lattice parameter a = 0.3294 nm. Find the planar

concentrations as the number of atoms per nm2 of the (100), (110) and (111) planes. Which plane hasthe most concentration of atoms per unit area? Sometimes the number of atoms per unit area nsurface on thesurface of a crystal is estimated by using the relation nsurface = nbulk

2/3 where nbulk is the concentration ofatoms in the bulk. Compare nsurface values with the planar concentrations calculated above and commenton the difference. [Note: The BCC (111) plane does not cut through the center atom and the (111) has1/6th of an atom at each corner]

SolutionPlanar concentration (or density) is the number of atoms per unit area on a given plane in the

crystal. It is the surface concentration of atoms on a given plane. To calculate the planar concentrationn(hkl) on a given (hkl) plane, we consider a bound area A. Only atoms whose centers line on A areinvolved in n(hkl). For each atom, we then evaluate what portion of the atomic cross section cut by theplane (hkl) is contained within A.

For the BCC crystalline structure the planes (100), (110) and (111) are drawn in Figure 1Q15-1.

a

(100)

a

a

(110)a2

(100), (110), (111) planes in the BCC crystal

(111)

a2

Figure 1Q15-1

Consider the (100) plane.

Number of atoms in the area a a , which is the cube face

= (4 corners) (1/4th atom at corner) = 1.

Planar concentration is

na( ) .

100 2 10 2

414 1

3 294 10=

=

( ) m = 9.216 1018 atoms m-2

The most populated plane for BCC structure is (110).

Number of atoms in the area a a 2 defined by two face-diagonals and two cube-sides

= (4 corners) (1/4th atom at corner) + 1 atom at face center = 2

Planar concentration is


1.24

na( ) .

110 2 10 2

414

1

22

3 294 10 2=

+=

( ) m = 1.303 1019 atoms m-2

The plane (111) for the BCC structure is the one with rarest population. The area of interest is anequilateral triangle defined by face diagonals of length a 2 (Figure 1Q15-1). The height of the triangle

is a32

so that the triangular area is 12

232

32

2

=a aa

. An atom at a corner only contributes a

fraction (60/360=1/6) to this area.

So the planar concentration is

na a( ) .

111 2 2 10 2

16

3

32

13

1

3 294 10 3=

( )= =

( ) m = 5.321 1018 atoms m-2

For the BCC structure there are two atoms in unit cell and the bulk atomic concentration is

nvo abulk

= = =

( )number of atoms in unit cell

volume of the cell m

2 2

3 294 103 10 3.

= 5.596 1028 atoms m-3

and the surface concentration is

n n msurface bulk= ( ) = ( )23 28 3 235 596 10. = 1.463 1019 atoms m-2

1 .16 Diamond and zinc blendeSi has the diamond and GaAs has the zinc blende crystal structure. Given the lattice parameters of Siand GaAs, a = 0.543 nm and a = 0.565 nm, respectively, and the atomic masses of Si, Ga, and As as28.08 g/mol, 69.73 g/mol, and 74.92 g/mol, respectively, calculate the density of Si and GaAs. What isthe atomic concentration (atoms per unit volume) in each crystal?

SolutionReferring to the diamond crystal structure in Figure 1Q16-1, we can identify the following types of

atoms:

8 corner atoms labeled C,

6 face center atoms (labeled FC) and

4 inside atoms labeled 1,2,3,4.

The effective number of atoms within the unit cell is:

(8 Corners) (1/8 C-atom) + (6 Faces) (1/2 FC-atom) + 4 atoms within the cell (1,2,3,4) = 8


1.25

a

C

a

a

1

2

4

3

C C

C

CC

FC

FC

FC

FC

FCFC

Figure 1Q16-1 The diamond crystal structure.

The lattice parameter (length of a cube side) of the unit cell is a. Thus the atomic concentration inthe Si crystal (nSi) is

nSi = = 8 8

0 543 103 9a ( . m)3= 5.0 1028 atoms per m-3

If Mat is the atomic mass in the Periodic Table then the mass of the atom (mat) in kg is

mat = (10-3 kg/g)Mat/NA (1)

where NA is Avogadros number. For Si, Mat = MSi = 28.09 g/mol, so then the density of Si is

= (number of atoms per unit volume) (mass per atom) = nSi mat

or =

8 103

3

a

M

NA

( )kg/g Si

i.e. =( )

( )( )

8

0 543 10

10 28 09

6 022 109 3

3

23.

( ) .

. m

kg/g g mol

mol

-1

1

calculating, = 2.33 103 kg m-3 or 2.33 g cm-3

In the case of GaAs, it is apparent that there are 4 Ga and 4 As atoms in the unit cell. Theconcentration of Ga (or As) atoms per unit volume (nGa) is

naGa

-3

mm= =

=

4 40 565 10

2 22 103 9 328

( . ).

Total atomic concentration (counting both Ga and As atoms) is twice nGa.

nTotal = 2nGa = 4.44 1028 m-3

There are 2.22 1028 Ga-As pairs per m3. We can calculate the mass of the Ga and As atomsfrom their relative atomic masses in the Periodic Table using Equation (1) with Mat = MGa = 69.72 g/molfor Ga and Mat = MAs = 74.92 g/mol for As. Thus,

= +

4 103

3

a

M M

NA

( )( )kg/g Ga As

or =

+

40 565 10

10 69 72 74 926 022 109 3

3

23( . )( )( . . )

. mkg/g g/mol g/mol

mol 1

i.e. = 5.33 103 kg m-3 or 5.33 g cm-3


1.26

1.17 Zinc blende, NaCl and CsCl

a. InAs is a III-V semiconductor that has the zinc blende structure with a lattice parameter of 0.606 nm.Given the atomic masses of In (114.82 g mol-1) and As (74.92 g mol-1) find the density.

b. CdO has the NaCl crystal structure with a lattice parameter of 0.4695 nm. Given the atomic massesof Cd (112.41 g mol-1) and O (16.00 g mol-1) find the density.

c. KCl has the same crystal structure as NaCl. The lattice parameter a of KCl is 0.629 nm. The atomicmasses of K and Cl are 39.10 g mol-1 and 35.45 g mol-1 respectively. Calculate the density of KCl.

Solutiona For zinc blende structure there are 8 atoms per unit cell (as shown in Table 1.3 in the textbook). Inthe case of InAs, it is apparent that there are 4 In and 4 As atoms in the unit cell. The density of InAs isthen

d

M

N

M

N

a

M M

N aInAs

at In

A

at As

A at In at As

A

=

+

=

+( )=

4 44

3 3

=+( ) ( )

( ) ( )

4 114 82 74 92 10

6 022 10 0 606 10

3 1

23 1 9 3

. .

. .

kg mol

mol m = 5.663 103 kg m-3 = 5.663 g cm- 3

b For NaCl crystal structure, there are 4 cations and 4 anions per unit cell. For the case of CdO wehave 4 Cd atoms and 4 O atoms per unit cell and the density of CdO is

d

M

N

M

N

a

M M

N aCdO

at Cd

A

at O

A at Cd at O

A

=

+

=

+( )=

4 44

3 3

=+( ) ( )

( ) ( )

4 112 41 16 00 10

6 022 10 0 4695 10

3 1

23 1 9 3

. .

. .

kg mol

mol m = 8.241 103 kg m-3 = 8.241 g cm- 3

c Analogously to b, for the density of KCl we receive

d

M

N

M

N

a

M M

N aKCl

at K

A

at Cl

A at K at Cl

A

=

+

=

+( )=

4 44

3 3

=+( ) ( )

( ) ( )

4 39 1 35 45 10

6 022 10 0 629 10

3 1

23 1 9 3

. .

. .

kg mol

mol m = 1.99 103 kg m-3 = 1.99 g cm- 3

1.18 Crystallographic directions and planesConsider the cubic crystal system.

a. Show that the line [hkl] is perpendicular to the (hkl) plane.

b. Show that the spacing between adjacent (hkl) planes is given by

da

h k l=

+ +2 2 2


1.27

SolutionThis problem assumes that students are familiar with three dimensional geometry and vector

products.

Figure 1Q18-1(a) shows a typical [hkl] line, labeled as ON, and a (hkl) plane in a cubic crystal.ux, uy and uz are the unit vectors along the x, y, z coordinates. This is a cubic lattice so we have Cartesiancoordinates and uxux = 1 and uxuy = 0 etc.

N

O

az1

ay1

ax1

ah

akal

A

B

C

ux

uy

uz (a)

D

O

az1

ay1

ax1A

B

C

(b)

Figure 1Q18-1 Crystallographic directions and planes

a Given a = lattice parameter, then from the definition of Miller indices (h = 1/x1, k = 1/y1 and l =1/z1) , the plane has intercepts: xo = ax1 =a/h; yo = ay1 = a/k; zo = az1 = a/l.

The vector ON = ahux + akuy + aluzIf ON is perpendicular to the (hkl) plane then the product of this vector with any vector in the (hkl)

plane will be zero. We only have to choose 2 non-parallel vectors (such as AB and BC) in the plane andshow that the dot product of these with ON is zero.

AB = OB OA = (a/k)uy (a/h)ux

ONAB = (ahux + akuy + aluz) ( (a/k)uy (a/h)ux) = a2 a2 = 0

Remember that: uxux = uyuy =1 and uxuy = uxuz = uyuz = 0

Similarly, ONBC = (ahux + akuy + aluz) ( (a/l)uz (a/k)uy) = 0

Therefore ON or [hkl] is normal to the (hkl) plane.

b Suppose that OD is the normal from the plane to the origin as shown in Figure 1Q18-1(b).Shifting a plane by multiples of lattice parameters does not change the miller indices. We can thereforeassume the adjacent plane passes through O. The separation between the adjacent planes is then simply thedistance OD in Figure 1Q18-1(b).

Let , and be the angles of OD with the x, y and z axes. Consider the direction cosines of theline OD: cos = d/(ax1) = dh/a; cos = d/(ay1) = dk/a; cos = d/(az1) = dl/aBut in 3 dimensions, (cos)2 + (cos)2 + (cos)2 = 1Thus, (d2h2/a2) + (d2k2/a2) + (d2l2/a2) = 1


1.28

Rearranging, d2 = a2 / [h2 + k2 + l2]

or, d = a / [h 2 + k 2 + l2]1/2

1.19 Si and SiO2a. Given the Si lattice parameter a = 0.543 nm, calculate the number of Si atoms per unit volume, in

nm-3.

b. Calculate the number of atoms per m2 and per nm2 on the (100), (110) and (111) planes in the Sicrystal as shown on Figure 1Q19-1. Which plane has the most number of atoms per unit area?

c. The density of SiO2 is 2.27 g cm-3. Given that its structure is amorphous, calculate the number of

molecules per unit volume, in nm-3. Compare your result with (a) and comment on what happenswhen the surface of an Si crystal oxidizes. The atomic masses of Si and O are 28.09 g/mol and 16g/mol, respectively.

a

a

(100) plane (110) plane (111) plane

a

Figure 1Q19-1 Diamond cubic crystal structure and planes. Determine what

portion of a black-colored atom belongs to the plane that is hatched.

Solutiona Si has the diamond crystal structure with 8 atoms in the unit cell, and we are given the latticeparameter a = 0.543 10-9 m and atomic mass Mat = 28.09 10

-3 kg/mol. The concentration of atoms perunit volume (n) in nm-3 is therefore:

na

= ( ) = ( ) ( )8 1

10

8

0 543 10

1

103 9 3 9 3 9 3 . nm/m m nm/m

n = 50.0 atoms/nm3

If desired, the density can be found as follows:

= =

( )

8 828 09 106 022 100 543 10

3

3

23

9 3

M

Na

at

A

. . .

kg/molmolm

1

= 2331 kg m-3 or 2.33 g cm-3

b The (100) plane has 4 shared atoms at the corners and 1 unshared atom at the center. The corneratom is shared by 4 (100) type planes. Number of atoms per square nm of (100) plane area (n) is shownin Fig. 1Q19-2:


1.29

(100)

a

a

A B

CD

E

a

Ge

a

a

A B

CD

E

Figure 1Q19-2 The (100) plane of the diamond crystal structure.

The number of atoms per nm2, n100, is therefore:

na100 2 9 2 9 2 9 2

414

11

10

414

1

0 543 10

1

10=

+( ) =

+( ) ( ) . nm/m m nm/m

n100 = 6.78 atoms/nm2 or 6.78 1018 atoms/m2

The (110) plane is shown below in Fig. 1Q19-3. There are 4 atoms at the corners and shared withneighboring planes (hence each contributing a quarter), 2 atoms on upper and lower sides shared withupper and lower planes (hence each atom contributing 1/2) and 2 atoms wholly within the plane.

(110)a

a2A B

CD

(110)

A

B

C

D

Figure 1Q19-3 The (110) plane of the diamond crystal structure.


na a

110 9 2

414

212

2

2

1

10=

+ +( )[ ] ( )

nm/m

n110 9 9 9 2

414

212

2

0 543 10 0 543 10 2

1

10=

+ +( ) ( )( )[ ] ( )

. . m m nm/m


This is the most crowded plane with the most number of atoms per unit area.

The (111) plane is shown below in Fig. 1Q19-4:


1.30

A

C

BD

a2a2

a22

a22

6060

30 30

a32

A

BC

D

Figure 1Q19-4 The (111) plane of the diamond crystal structure


n

a a111 9 2

360

3603

12

212

22

32

1

10=

+

( )

nm/m

n1119 9

9 2

360

3603

12

212

0 543 102

20 543 10

32

1

10=

+ ( )

( )

( )

. . m m nm/m


c Given:

Molar mass of SiO2: Mat = 28.09 10-3 kg/mol + 2 16 10-3 kg/mol = 60.09 10-3 kg/mol

Density of SiO2: = 2.27 103 kg m-3

Let n be the number of SiO2 molecules per unit volume, then:

= n MN

at

A

nN

MA

at

= =

( ) ( )( )

6 022 10 2 2760 09 10

23

3

. .

.

mol 10 kg m

kg/mol

1 3 3

n = 2.27 1028 molecules per m3

Or, converting to molecules per nm3:

n =

( ) =2 27 10

10

28

9 3

.

molecules m

nm m

/

/

3

22.7 molecules per nm3

Oxide has less dense packing so it has a more open structure. For every 1 micron of oxide formedon the crystal surface, only about 0.5 micron of Si crystal is consumed.


1.31

1.20 VacanciesEstimate the equilibrium vacancy concentration in the Si crystal at room temperature and at 1200 C,given that the energy of vacancy formation is 3.6 eV.

Solution

To find the vacancy concentration in Si at T = 27 C + 273 = 300 K:

Si has a diamond crystal structure (as shown in Table 1.3 in the textbook) and therefore theconcentration of atoms per unit volume (n) is given as:

na

= =

( ) = 8 8

0 543 104 997 103 9 3

28

. .

matoms/m3

where a = 0.543 nm is the lattice parameter of Si, given in Q1.19.

We are given that the energy of vacancy formation (EV) in Si is 2.4 eV, or:

EV = (2.4 eV/atom)(1.602 10-19 J/eV) = 3.845 10-19 J/atom

The concentration of vacancies (nV) is then given by:

n nE

kTVV

= exp

nV = ( ) ( )( )

4 997 103 845 10

30028

19

. exp.

m

J

1.381 10 J K K3

-23 1

nV = 2.47 10-12 vacancies / m3 (at room temperature)

At T = 1200 C + 273 = 1473 K:

=

n n EkTV Vexp

nV = ( ) ( )( )

4 997 103 845 10

147328

19

. exp.

m

J

1.381 10 J K K-3

-23 1

nV = 3.09 1020 vacancies / m3 (at 1200 C)

Note: Strictly we should use the concentration of Si (n) at 1473 K by taking into account the unitcell expansion and recalculating n = 8/a3 to get a better value for n. Given that nV = n exp(-EV/kT) has theentropic pre-exponential term missing (i.e. it is already an approximation) the calculation above is morethan sufficient.

1.21 Pb-Sn solderConsider the soldering of two copper components. When the solder melts, it wets both metal surfaces.If the surfaces are not clean or have an oxide layer, the molten solder cannot wet the surfaces and thesoldering fails. Assume that soldering takes place at 250 C, and consider the diffusion of Sn atoms intothe copper (the Sn atom is smaller than the Pb atom and hence diffuses more easily).


1.32

a. The diffusion coefficient of Sn in Cu at two temperatures is D = 1.69 10-9 cm2 hr-1 at 400 C and D

= 2.48 10-7 cm2 hr-1 at 650 C. Calculate the rms distance diffused by an Sn atom into the copper,assuming the cooling process takes 10 seconds.

b. What should be the composition of the solder if it is to begin freezing at 250 C?

c. What are the components (phases) in this alloy at 200 C? What are the compositions of the phasesand their relative weights in the alloy?

d. What is the microstructure of this alloy at 25 C? What are weight fractions of the and phasesassuming near equilibrium cooling?

Solutiona Given information:

Temperatures: T1 = 400 C + 273 = 673 K T2 = 650 C + 273 = 923 K

Diffusion coefficients: D1 = 1.69 10-9 cm2/hr = (1.69 10-9 cm2/hr)(0.01 m/cm)2 / (1 hr) (3600 sec/hr)

D1 = 4.694 10-17 m2/s

D2 = 2.48 10-7 cm2/hr = (2.48 10-7 cm2/hr)(0.01 m/cm)2 / (1 hr) (3600 sec/hr)

D2 = 6.889 10-15 m2/s

The diffusion coefficients at certain temperatures are given by:

D DE q

kToA

11

=

exp D D

E q

kToA

22

=

exp

where EA is the activation energy in eV/atom. We can take the ratio of the diffusion coefficients to expressthem in terms of the activation energy (EA):

D

D

DE q

kT

DE q

kT

E q

kT

E q

kT

q T T E

T T k

oA

oA

A A A1

2

1

2

2 1

1 2

1 2

=

=

=

[ ]

exp

exp

exp exp

E

T T kD

D

q T TA=

( )1 2

1

2

1 2

ln

EA =

( )( ) ( )

( ) ( )673 K 923 K 1.381 10 J K

4.694 10 m /s6.889 10 m /s

1.602 10 J/eV 673 K 923 K

-23 -1-17 2

-15 2

-19

ln

EA = 1.068 eV/atom

Now the diffusion coefficient Do can be found as follows:

D DE q

kToA

11

=

exp


1.33

DD

E q

kT

o

A

=

=

( ) ( )( )( )

1

1

exp exp

4.694 10 m /s

1.068 eV 1.602 10 J/eV

1.381 10 J K 673 K

-17 2

-19

-23 -1

Do = 4.638 10-9 m2/s

To check our value for Do, we can substitute it back into the equation for D2 and compare values:

D DE q

kToA

22

=

= ( )

( ) ( )( )( )

exp exp4.638 10 m /s

1.068 eV 1.602 10 J/eV

1.381 10 J K 923 K-9 2

-19

-23 -1

D2 = 6.870 10-15 m2/s

This agrees with the given value of 6.889 10-15 m2/s for D2.

Now we must calculate the diffusion coefficient D3 at T3 = 250 C + 273 = 523 K (temperature atwhich soldering is taking place).

D DE q

kToA

33

=

= ( )

( ) ( )( )( )

exp exp4.638 10 m /s

1.068 eV 1.602 10 J/eV

1.381 10 J K 523 K-9 2

-19

-23 -1

D3 = 2.391 10-19 m2/s

The rms distance diffused by the Sn atom in time t = 10 s (Lrms) is given by:

L D trms = = ( )( )2 2 103 2.391 10 m /s s-19 2 Lrms = 2.19 10

-9 m or 2 nm

b From Figure 1Q21-1, 250 C cuts the liquidus line approximately at 33 wt.% Sn composition(Co).

C o = 0.33 (Sn)

c For -phase and liquid phase (L), the compositions as wt.% of Sn from Figure 1Q21-1 are:

C = 0.18 CL = 0.56

The weight fraction of and L phases are:

W =

=

=

C C

C CL o

L

0 56 0 330 56 0 18

. .

. .0.605 or 60 wt.% -phase

WL =

=

=

C C

C Co

L

0 33 0 180 56 0 18

. .

. .0.395 or 39.5 wt.% liquid phase


1.34

LIQUIDA

B

0 20 40 60 80 1000

100

200

300

400

Pure Pb Pure SnComposition in wt. % Sn

Tem

pera

ture

(C

)

97.5%61.9%

L +

+

+ LCL

C

C CCo

Figure 1Q21-1 The equilibrium phase diagram of the Pb-Sn alloy.

d The microstructure is a primary -phase and a eutectic solid ( + ) phase. There are two phasespresent, + . See Figure 1Q21-2.

Primary Eutectic

Figure 1Q21-2 Microstructure of Pb-Sn at temperatures less than 183C.

Assuming equilibrium concentrations have been reached:

C = 0.02 C = 1

The weight fraction of in the whole alloy is then:

=

=

=WC C

C Co

1 0 331 0 02

.

.0.684 or 68.4 wt.% -phase

The weight fraction of in the whole alloy is: =

=

=WC C

C Co

0 33 0 021 0 02. .

.0.316 or 31.6 wt.% -phase

1.22 Pb-Sn solderConsider the 50% Pb-50% Sn solder.

a. Sketch the temperature-time profile and the microstructure of the alloy at various stages as it is cooledfrom the melt.

b. At what temperature does the solid melt?

c. What is the temperature range over which the alloy is a mixture of melt and solid? What is thestructure of the solid?


1.35

d. Consider the solder at room temperature following cooling from 182 C. Assume that the rate ofcooling from 182 C to room temperature is faster than the atomic diffusion rates needed to change thecompositions of the and phases in the solid. Assuming the alloy is 1 kg, calculate the masses ofthe following components in the solid:

1. The primary .

2. in the whole alloy.

3. in the eutectic solid.

4. in the alloy (Where is the -phase?).e. Calculate the specific heat of the solder given the atomic masses of Pb (207.2) and Sn (118.71).

Solutiona

L

LT

t

L

L +

L + +

+

183 C

~210 C

Eutectic

L (61.9% Sn) (19% Sn)

Eutectic ( + ) solidifying

Proeutecticsolidifying

Proeutectic (primary)

50% Pb-50% Sn

Figure 1Q22-1 Temperature - time profile and microstructure diagram of 50% Pb-50% Sn.

All compositions are in weight %.

b When 50% Pb-50% Sn is cooled from the molten state down to room temperature, it begins tosolidify at point A at about 210 C. Therefore at 210 C, the solid begins to melt.

c Between 210 C and 183 C, the liquid has the eutectic composition and undergoes the eutectictransformation to become the eutectic solid. Below 183 C, all the liquid has solidified, and the structure

is a combination of the solid -phase and the eutectic structure (which is composed of and layers).d At 182 C, the composition of the proeutectic or primary is given by the solubility limit of Sn in: 19.2% Sn.


1.36

The primary or proeutectic (pro-) exists just above and below 183 C (eutectic temperature),

i.e. it is stable just above and below 183 C. Thus the mass of pro- at 182 C is the same as at 184 C.Applying the lever rule (at 50% Sn):

Wpro =

=

=

C C

C CL o

L

61 9 5061 9 19 2

.. .

0.279

Assume that the whole alloy is 1 kg. The mass of the primary or proeutectic is thus 27.9% ofthe whole alloy; or 0.279 kg. The mass of the total eutectic solid ( + ) is thus 1 - 0.279 = 0.721or 72.1% or 0.721 kg.

If we apply the lever rule at 182 C at 50% Sn, we obtain the weight percentage of in the wholesolid:

W =

=

=

C C

C Co

97 5 5097 5 19 2

.. .

0.607

Thus the mass of in the whole solid is 0.607 kg. Of this, 0.28 kg is in the primary(proeutectic) phase. Thus 0.607 - 0.279 or 0.328 kg of is in the eutectic solid.

Since the total mass of in the solid is 0.607 kg, the remainder of the mass must be the -phase.Thus the mass of the -phase is 1 kg - 0.607 kg or 0.393 kg. The -phase is present solely in theeutectic solid, with no primary , because the solid is forming from a point where it consisted of the andliquid phases only.

e Suppose that nA and nB are atomic fractions of A and B in the whole alloy,

nA + nB = 1

Suppose that we have 1 mole of the alloy. Then it has nA moles of A and nB moles of B (atomicfractions also represent molar fractions in the alloy). Suppose that we consider 1 gram of the alloy. SincewA is the weight fraction of A, wA is also the mass of A in grams in the alloy. The number of moles of A inthe alloy is then wA/MA where MA is the atomic mass of A. Thus,

Number of moles of A = wA/MA.

Number of moles of B = wB/MB.

Number of moles of the whole alloy = wA/MA + wB/MB.

Molar fraction of A is the same as nA. Thus,

nw M

w

M

w

M

AA A

A

A

B

B

=

+

/and n

w Mw

M

w

M

BB B

A

A

B

B

=

+

/

We are given the molar masses of Pb and Sn:

MPb = 207.2 g/mol MSn = 118.71 g/mol

Let n = mole (atomic) fraction and W = weight fraction. Then WPb = weight fraction of Pb in thealloy. We know this to be 0.5 (50% Pb).

The mole fractions of Pb and Sn are nPb and nSn and are given by:

nM W

M M W MPbSn Pb

Pb Sn Pb Pb

=

( )


1.37

nPb118.71 g/mol

207.2 g/mol 118.71 g/mol 207.2 g/mol=

( )( )( )( )

0 5

0 5

.

.= 0.3642

nSn = 1 - nPb = 1 - 0.3642 = 0.6358

The mole fraction of Pb is 0.3642 and that of Sn is 1 - 0.3642 = 0.6358.

The heat capacity of a metal is 3R per mole, where R is the gas constant. We want the specificheat capacity Cs (heat capacity per gram of alloy). 1 mole of the alloy has a mass MPbSn:

MPbSn = nPb MPb + nSn MSnThus the specific heat capacity Cs (i.e. heat capacity per gram) is:

CR

M

R

n M n Ms=

3 3

PbSn Pb Pb Sn Sn

= +

Cs =( )

( ) ( ) ( ) ( )3 8.315 J K mol

0.3642 207.2 g/mol + 0.6358 118.71 g/mol

-1 -1

C s = 0.165 J K-1 g- 1

"If your experiment needs statistics, you ought to have done a better experiment."

Ernest Rutherford (1871-1937)


2.1

Second Edition ( 2001 McGraw-Hill)

Chapter 2

2.1 Electrical conductionNa is a monovalent metal (BCC) with a density of 0.9712 g cm-3. Its atomic mass is 22.99 g mol1.The drift mobility of electrons in Na is 53 cm2 V-1 s-1.

a. Consider the collection of conduction electrons in the solid. If each Na atom donates one electron tothe electron sea, estimate the mean separation between the electrons?

b. What is the approximate mean separation between an electron (e-) and a metal ion (Na+), assumingthat most of the time the electron prefers to be between two neighboring Na+ ions. What is theapproximate Coulombic interaction energy (in eV) between an electron and an Na+ ion?

c. How does this electron/metal-ion interaction energy compare with the average thermal energy perparticle, according to the kinetic molecular theory of matter? Do you expect the kinetic moleculartheory to be applicable to the conduction electrons in Na? If the mean electron/metal-ion interactionenergy is of the same order of magnitude as the mean KE of the electrons, what is the mean speed ofelectrons in Na? Why should the mean kinetic energy be comparable to the mean electron/metal-ioninteraction energy?

d. Calculate the electrical conductivity of Na and compare this with the experimental value of 2.1 107

-1 m-1 and comment on the difference.

Solutiona If D is the density, Mat is the atomic mass and NA is Avogadro's number, then the atomicconcentration nat is

nDN

MatA

at

= =

( . )( . )( .

971 2 6 022 1022 99 10

23 1

3

kg m molkg mol )

-3

1

i.e. nat = 2.544 1028 m-3

which is also the electron concentration, given that each Na atom contributes 1 conduction electron.

If d is the mean separation between the electrons then d and nat are related by (see Chapter 1Solutions, Q1.5; this is only an estimate)

d =

1 12 544 101 3 28nat

/ ( . m )3 1/3= 3.40 10-10 m or 0.34 nm

b Na is BCC with 2 atoms in the unit cell. So if a is the lattice constant (side of the cubic unit cell),the density is given by

D

M

N

a

at

A= =

(atoms in unit cell)(mass of 1 atom)

volume of unit cell

2

3

isolate for a: aM

DNat

A

=

=

2 2 22 99 100 9712 10 6 022 10

1 3 3

3 23 1

1 3/ /( .

( . )( . )kg mol )

kg m mol

1

-3

so that a = 4.284 10-10 m or 0.4284 nm


2.2

For the BCC structure, the radius of the metal ion R and the lattice parameter a are related by (4R)2

= 3a2, so that,

R = (1/4)[3a2] = 1.855 10-10 m or 0.1855 nmIf the electron is somewhere roughly between two metal ions, then the mean electron to metal ion

separation delectron-ion is roughly R. If delectron-ion R, the electrostatic potential energy PE between aconduction electron and one metal ion is then

PEe e

do=

+=

+

( )( ) ( . ( . ( . )( . )4

1 602 10 1 602 104 8 854 10 1 855 10

19 19

12 1 10 electron ion

C) C)F m m

(1)

PE = -1.24 10-18 J or -7.76 eV

c This electron-ion PE is much larger than the average thermal energy expected from the kinetictheory for a collection of free particles, that is Eaverage = KEaverage = 3(kT/2) 0.039 eV at 300 K. In thecase of Na, the electron-ion interaction is very strong so we cannot assume that the electrons are movingaround freely as if in the case of free gas particles in a cylinder. If we assume that the mean KE is roughlythe same order of magnitude as the mean PE,

KE m u PEaverage e average= = 1

22 181 24 10. J (2)

where u is the mean speed (strictly, u = root mean square velocity) and me is the electron mass.

Thus, uPE

maverage

e

=

2 2 1 24 109 109 10

1 2 18

31

1 2/ /( .

( . )J)

kg(3)

so that u = 1.65 106 m/s

There is a theorem in classical physics called the Virial theorem which states that if the interactionsbetween particles in a system obey the inverse square law (as in Coulombic interactions) then themagnitude of the mean KE is equal to the magnitude of the mean PE. The Virial Theorem states that:

KE PEaverage average= -12

Indeed, using this expression in Eqn. (2), we would find that u = 1.05 106 m/s. If theconduction electrons were moving around freely and obeying the kinetic theory, then we would expect(1/2)meu

2 = (3/2)kT and u = 1.1 105 m/s, a much lower mean speed. Further, kinetic theory predictsthat u increases as T1/2 whereas according to Eqns. (1) and (2), u is insensitive to the temperature. Theexperimental linear dependence between the resistivity and the absolute temperature T for most metals(non-magnetic) can only be explained by taking u = constant as implied by Eqns. (1) and (2).

d If is the drift mobility of the conduction electrons and n is their concentration, then the electricalconductivity of Na is = en. Assuming that each Na atom donates one conduction electron (n = nat), wehave

= = en ( . 1 602 10 2 10 5 1019 28 3 4 2C)( .544 m )( 3 m V s )-1 -1

i.e. = 2.16 107 -1 m-1

which is quite close to the experimental value.

Nota Bene: If one takes the Na+-Na+ separation 2R to be roughly the mean electron-electron separationthen this is 0.37 nm and close to d = 1/(n1/3) = 0.34 nm. In any event, all calculations are onlyapproximate to highlight the main point. The interaction PE is substantial compared with the mean thermalenergy and we cannot use (3/2)kT for the mean KE!


2.3

2.2 Electrical conduction

The resistivity of aluminum at 25 C has been measured to be 2.72 10-8 m. The thermal coefficientof resistivity of aluminum at 0 C is 4.29 10-3 K1. Aluminum has a valency of 3, a density of 2.70 gcm-3, and an atomic mass of 27.

a. Calculate the resistivity of aluminum at 40 C.

b. What is the thermal coefficient of resistivity at 40 C?

c. Estimate the mean free time between collisions for the conduction electrons in aluminum at 25 C,and hence estimate their drift mobility.

d. If the mean speed of the conduction electrons is 1.5 106 m s-1, calculate the mean free path andcompare this with the interatomic separation in Al (Al is FCC). What should be the thickness of anAl film that is deposited on an IC chip such that its resistivity is the same as that of bulk Al?

e. What is the percentage change in the power loss due to Joule heating of the aluminum wire when thetemperature drops from 25 C to 40 C?

Solution

a Apply the equation for temperature dependence of resistivity, (T) = o[1 + o(T-To)]. We havethe temperature coefficient of resistivity, o, at To where To is the reference temperature. The two given

reference temperatures are 0 C or 25 C, depending on choice. Taking To = 0 C + 273 = 273 K,

(-40 C + 273 = 233 K) = o[1 + o(233 K - 273 K)]

(25 C + 273 = 298 K) = o[1 + o(298 K - 273 K)]

Divide the above two equations to eliminate o,

(-40 C)/(25 C) = [1 + o(-40 K)] / [1 + o(25 K)]

Next, substitute the given values (25 C) = 2.72 10-8 m and o = 4.29 10-3 K-1 to obtain

(-40 C) = (2.72 10 m)[1+ (4.29 10 K )(-40 K)][1+ (4.29 10 K )(25 K)]

-8-3 -1

-3 -1

i.e. (-40 C) = 2.03 10-8 m

b In (T) = o[1 + o(T - To)] we have o at To where To is the reference temperature, for example,0 C or 25 C depending on choice. We will choose To to be first at 0 C = 273 K and then at -40 C =233 K so that

(-40 C) = (0 C)[1 + o(233 K - 273 K)]

and (0 C) = (-40 C)[1 + -40(273 K - 233 K)]Multiply and simplify the two equations above to obtain

[1 + o(233 K - 273 K)][1 + -40(273 K - 233 K)] = 1

or [1 - 40o][1 + 40-40] = 1

Rearranging,


2.4

-40 = (1 / [1 - 40o] - 1)(1 / 40)

-40 = o / [1 - 40o]

i.e. -40 = (4.29 10-3 K-1) / [1 - (40 K)(4.29 10-3 K-1)] = 5.18 10-3 K-1

Alternatively, consider, (25 C) = (-40 C)[1 + -40(298 K - 233 K)] so that

-40

= [(25 C) - (-40 C)] / [(-40 C)(65 K)]

-40

= [2.72 10-8 m - 2.03 10-8 m] / [(2.03 10-8 m)(65 K)]

-40

= 5.23 10-3 K-1

c We know that 1/ = = en where is the electrical conductivity, e is the electron charge, and is the electron drift mobility. We also know that = e/me, where is the mean free time between electroncollisions and me is the electron mass. Therefore,

1/ = e2n/me = me/e2n (1)

Here n is the number of conduction electrons per unit volume. But, from the density d and atomicmass Mat, atomic concentration of Al is

nN d

MA

atAl

-- = =

6. mol kg/m

. kg/mol= . m

022 10 2700

0 0276 022 10

23 1 328 3

( )( )( )

so that n = 3nAl = 1.807 1029 m-3

assuming that each Al atom contributes 3 "free" conduction electrons to the metal. Substitute into Eqn.(1):

= =

m

e ne2

(9.109 10 kg)(2.72 10 m)(1.602 10 C) (1.807 10 m )

-31

-8 -19 2 29 -3

= 7.22 10-15 s

(Note: If you do not convert to meters and instead use centimeters you will not get the correctanswer because seconds is an SI unit.)

The relation between the drift mobility d and the mean scattering time is given by Equation 2.5 (intextbook), so that

de

e

m

C s

kg= =

( ) ( )( )

1 602 10 7 22 10

9 109 10

19 15

31

. .

.

d = 1.27 10-3 m2 V-1s-1 = 12.7 cm2 V-1s-1

d The mean free path is l = u where u is the mean speed. With u 1.5 106 m s-1 we find themean free path:

l = u = (1.5 106 m s-1)(7.22 10-15 s) 1.08 10-8 m 10.8 nm

A thin film of Al must have a much greater thickness than l to show bulk behavior. Otherwise,scattering from the surfaces will increase the resistivity by virtue of Matthiessen's rule.


2.5

e Power P = I2R and is proportional to resistivity , assuming the rms current level stays relativelyconstant. Then we have

[P(-40 C) - P(25 C)] / P(25 C) = P(-40 C) / P(25 C) - 1

= (-40 C) / (25 C) - 1

= (2.03 10-8 m / 2.72 10-8 m) - 1

= -0.254, or -25.4%

(Negative sign means a reduction in the power loss).

2.3 TCR and Matthiessens ruleDetermine the temperature coefficient of resistivity of pure iron and of electrotechnical steel (Fe with 4%C), which are used in various electrical machinery, at two temperatures: 0 C and 500 C. Comment onthe similarities and differences in the resistivity versus temperature behavior shown in Figure 2Q3-1 for

the two materials.

Figure 2Q3-1 Resistivity versus temperaturefor pure iron and 4% C steel.

Solution

The temperature coefficient of resistivity o (TCR) is defined as follows:

o o T

o

o

d

dTo

= =1 Slope

where the slope is d/dT at T = To and o is the resistivity at T = To.

To find the slope, we draw a tangent to the curve at T = To (To = 0 C and then To = 500 C) and

obtain /T d/dT. One convenient way is to define T = 400 C and find on the tangent lineand then calculate /T. Iron at 0 C ,

Slopeo (0.23 10-6 m - 0 m) / (400 C) = 5.75 10-10 m C -1

0

0.5

1

1.5

400 0 400 800 1200

Pure Fe

Fe + 4%C

Temperature ( C)

Tangent

0.57

400 C0.23

0.11

0.96

0.53

0.85

0.68

1.05

400 C0.4

500 C

Resistivity ( m)


2.6

Since o 0.11 10-6 m, o = Slopeo/o 0.00523 C -1

Fe + 4% C at 0 C ,

Slopeo (0.57 10-6 m - 0.4 10-6 m) / (400 C) = 4.25 10-10 m C -1


Iron at 500 C ,

Slopeo (0.96 10-6 m - 0.4 10-6 m) / (400 C) = 1.40 10-9 m C -1


Fe + 4% C at 500 C ,

Slopeo (1.05 10-6 m - 0.68 10-6 m) / (400 C) = 9.25 10-10 m C -1


*2.4 TCR of isomorphous alloysa. Show that for an isomorphous alloy A%-B% (B% solute in A% solvent), the temperature coefficient

of resistivity AB is given by

AB

A A

AB

where AB is the resistivity of the alloy (AB) and A and A are the resistivity and TCR of pure A .What are the assumptions behind this equation?

b. Estimate the composition of the Cu-Ni alloy that will have a TCR of 4 104 K1, that is, a TCR thatis an order of magnitude less than that of Cu.

Solution

a By the Nordheim rule, the resistivity of the alloy is alloy = o + CX(1X). We can find the TCRof the alloy from its definition

alloyalloy

alloy

alloy

= = +1 1

1d

dT

d

dTCX Xo[ ( )]

To obtain the desired equation, we must assume that C is temperature independent (i.e.the increase in the resistivity depends on the lattice distortion induced by the impurity) so that d[CX(1 -X)]/dT = 0, enabling us to substitute for do/dT using the definition of the TCR: o =(do/dT)/o.Substituting into the above equation:

alloyalloy alloy

= =

1 1ddT

oo o

i.e. alloy alloy = o o or AB AB A A=Remember that all values for the alloy and pure substance must all be taken at the same

temperature, or the equation is invalid.

b Assume room temperature T = 293 K. Using values for copper from Table 2.1 in Equation 2.17(both in the textbook), Cu = 17.1 n m and Cu = 4.0 10-3 K-1, and from Table 2.3 (in the textbook)


2.7

the Nordheim coefficient of Ni dissolved in Cu is C = 1570 n m. We want to find the composition ofthe alloy such that CuNi = 4 10

-4 K-1. Then,

alloyCu Cu

alloy

1

1

K n mK

= =

( . )( . )( . )

0 0040 17 10 0004

= 171.0 n m

Using Nordheims rule:

alloy = Cu + CX(1 - X)

i.e. 171.0 n m = 17.1 n m + (1570 n m)X(1 - X)

X2 - X + 0.0879 = 0

solving the quadratic, we find X = 0.11.

Thus the composition is 89% Cu-11% Ni. However, thi

solution manual of principles of electronic materials and devices by kasop (2nd ed)

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