Ch18 Solution Manual Material Science and Engineering 8th Edition PDF

Preview

Summary

Download Ch18 Solution Manual Material Science and Engineering 8th Edition PDF

Description

CHAPTER 18

ELECTRICAL PROPERTIES

PROBLEM SOLUTIONS

Ohm’s Law Electrical Conductivity 18.1 (a) Compute the electrical conductivity of a 5.1-mm (0.2-in.) diameter cylindrical silicon specimen 51 mm (2 in.) long in which a current of 0.1 A passes in an axial direction. A voltage of 12.5 V is measured across two probes that are separated by 38 mm (1.5 in.). (b) Compute the resistance over the entire 51 mm (2 in.) of the specimen. Solution This problem calls for us to compute the electrical conductivity and resistance of a silicon specimen. (a) We use Equations 18.3 and 18.4 for the conductivity, as

s =

1 Il = = r VA

Il æ d ö2 Vp ç ÷ è2 ø

And, incorporating values for the several parameters provided in the problem statement, leads to

s =

(0.1 A)(38 ´ 10 -3 m) æ 5.1 ´ 10 -3 m ö2 (12.5 V)(p )ç ÷ 2 ø è

= 14.9 (W - m) -1

(b) The resistance, R, may be computed using Equations 18.2 and 18.4, as

R=

=

l rl = = A sA

l æ d ö2 sp ç ÷ è2 ø

51 ´ 10 -3 m = 168 W æ 5.1 ´ 10 -3 m ö 2 -1 14.9 (W - m) (p) ç ÷ 2 ø è

[

]

18.2 A copper wire 100 m long must experience a voltage drop of less than 1.5 V when a current of 2.5 A passes through it. Using the data in Table 18.1, compute the minimum diameter of the wire. Solution For this problem, given that a copper wire 100 m long must experience a voltage drop of ess l than 1.5 V when a current of 2.5 A passes through it, we are to compute the minimum diameter of the wire. Combining Equations 18.3 and 18.4 and solving for the cross-sectional area Aleads to

Ilr Il = V Vs

A=

æ d ö2 -1 From Table 18.1, for copper s= 6.0 ´ 10 7 (W-m) . Furthermore, inasmuch as A = p ç ÷ for a cylindrical wire, then è 2ø æ d ö2 Il pç ÷ = è2 ø Vs or

4 Il pVs

d=

When values for the several parameters given in the problem statement are incorporated into this expression, we get

d =

(4)(2.5 A)(100 m)

[

(p)(1.5 V) 6.0 ´ 10 7 (W - m)-1

-3

= 1.88 ´ 10 m = 1.88 mm

]

18.3 An aluminum wire 4 mm in diameter is to offer a resistance of no more than 2.5 W. Using the data in Table 18.1, compute the maximum wire length. Solution This problem asks that we compute, for an aluminum wire 4 mm in diameter, the maximum length such that the resistance will not exceed 2.5 W. From Table 18.1 for aluminum, s = 3.8 ´ 107 (W-m)-1. If d is the diameter then, combining Equations 18.2 and 18.4 leads to

l=

æ d ö2 RA = RsA = Rsp ç ÷ r è2 ø

æ4 ´ 10 -3 m ö2 = (2.5 W) 3.8 ´ 10 7 (W - m)-1 (p) ç ÷ = 1194 m 2 è ø

[

]

18.4 Demonstrate that the two Ohm’s law expressions, Equations 18.1 and 18.5, are equivalent. Solution Let us demonstrate, by appropriate substitution and algebraic manipulation, that Equation 18.5 may be made to take the form of Equation 18.1. Now, Equation 18.5 is just

J = sE

(In this equation we represent the electric field with an “E”.) But, by definition, J is just the current density, the I V current per unit cross-sectional area, or J = . Also, the electric field is defined by E = . And, substituting these A l expressions into Equation 18.5 leads to

I V =s l A

But, from Equations 18.2 and 18.4

s=

l RA

and

æ l ö æV ö I = ç ÷ç ÷ A è RAø è l ø

Solving for Vfrom this expression gives V= IR, which is just Equation 18.1.

18.5 (a) Using the data in Table 18.1, compute the resistance of a copper wire 3 mm (0.12 in.) in diameter and 2 m (78.7 in.) long. (b) What would be the current flow if the potential drop across the ends of the wire is 0.05 V? (c) What is the current density? (d) What is the magnitude of the electric field across the ends of the wire? Solution (a) In order to compute the resistance of this copper wire it is necessary to employ Equations 18.2 and 18.4. Solving for the resistance in terms of the conductivity,

R=

rl A

=

l = sA

l æ dö2 sp ç ÷ è 2ø

From Table 18.1, the conductivity of copper is 6.0 ´107 (W-m)-1, and

R=

l æ dö sp ç ÷ è 2ø

2m

=

2

æ 3 ´ 10 -3 m ö2 6.0 ´ 10 7 (W - m)-1 (p)ç ÷ 2 è ø

[

]

= 4.7 ´ 10 -3 W

(b) If V= 0.05 V then, from Equation 18.1

I =

0.05 V V =10.6 A = R 4. 7 ´ 10 - 3 W

(c) The current density is just

J =

I = A

I æd ö 2 pç ÷ è 2ø

=

10.6 A æ 3 ´ 10 -3 m ö 2 pç ÷ 2 è ø

= 1.5 ´ 10 6 A/m 2

(d) The electric field is just

E=

V 0.05 V = = 2.5 ´ 10 -2 V/m l 2m

Electronic and Ionic Conduction 18.6 What is the distinction between electronic and ionic conduction? Solution When a current arises from a flow of electrons, the conduction is termed electronic; for ionic conduction, the current results from the net motion of charged ions.

Energy Band Structure s in Solids 18.7 How does the electron structure of an isolated atom differ from that of a solid material? Solution For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid.

Conduction in Terms of Band and Atomic Bonding Models 18.8

In terms of electron energy band structure, discuss reasons for the difference in electrical

conductivity between metals, semiconductors, and insulators. Solution For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors. There are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons will occur, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities.

Electron Mobility 18.9 Briefly tell what is meant by the drift velocity and mobility of a free electron. Solution The drift velocity of a free electron is the average electron velocity in the direction of the force imposed by an electric field. The mobility is the proportionality constant between the drift velocity and the electric field. It is also a measure of the frequency of scattering events (and is inversely proportional to the frequency of scattering).

18.10 (a) Calculate the drift velocity of electrons in germanium at room temperature and when the magnitude of the electric field is 1000 V/m. (b) Under these circumstances, how long does it take an electron to traverse a 25-mm (1-in.) length of crystal? Solution (a) The drift velocity of electrons in Ge may be determined using Equation 18.7. Since the room temperature mobility of electrons is 0.38 m2/V-s (Table 18.3), and the electric field is 1000 V/m (as stipulated in the problem statement),

vd = m e E = ( 0.38 m 2 /V - s)(1000 V/m) = 380 m/s

(b) The time, t , required to traverse a given length, l (= 25 mm), is just

t =

l 25 ´ 10 -3 m = 6.6 ´ 10 -5 s = vd 380 m /s

18.11 At room temperature the electrical conductivity and the electron mobility for copper are 6.0 ´ 10 7 1 2 (W-m) - and 0.0030 m /V-s, respectively. (a) Compute the number of free electrons per cubic meter for copper at

room temperature. (b) What is the number of free electrons per copper atom? Assume a density of 8.9 g/cm3. Solution (a) The number of free electrons per cubic meter for copper at room temperature may be computed using Equation 18.8 as

n=

=

s | e | me

6.0 ´ 107 (W - m)-1 (1.602 ´ 10 -19 C)(0.003 m 2 /V - s) = 1.25 ´ 10 29 m-3

(b) In order to calculate the number of free electrons per copper atom, we must first determine the number of copper atoms per cubic meter, NCu. From Equation 4.2 (and using the atomic weight value for Cu found inside the front cover—viz. 63.55 g/mol)

NCu =

=

(6.022

N A r¢ ACu

´ 1023 atoms / mol)(8.9 g/cm 3)(10 6 cm3 / m 3 ) 63.55 g/mol = 8.43 ´ 10 28 m-3

(Note: in the above expression, density is represented by r ' in order to avoid confusion with resistivity which is designated by r.) And, finally, the number of free electrons per aluminum atom is just n/NCu

n NCul

=

1. 25 ´ 10 29 m-3 = 1.48 8.43 ´ 10 28 m-3

18.12 (a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. The electrical conductivity and density for Au are 4.3 ´ 107 (W-m) -1 and 19.32 g/cm3, respectively. (b) Now compute the electron mobility for Au. Solution (a) This portion of the problem asks that we calculate, for gold, the number of free electrons per cubic meter (n) given that there are 1.5 free electrons per gold atom, that the electrical conductivity is 4.3 ´10 7 (W-m)-1, and that ' ) is 19.32 g/cm3. (Note: in this discussion, the density of silver is represented by r ' in order to the density (rAu Au avoid confusion with resistivity which is designated by r .) Since n = 1.5NAu, and NAu is defined in Equation 4.2 (and using the atomic weight of Au found inside the front cover—viz 196.97 g/mol), then

é r' N ù n = 1.5 N Au = 1.5 ê Au A ú êë AAu úû é (19.32 g/cm 3)(6.022 ´ 10 23 atoms / mol)ù = 1.5 ê ú 196 .97 g/mol û ë = 8.86 ´ 10 22cm-3 = 8.86 ´ 10 28 m-3 (b) Now we are asked to compute the electron mobility, me. Using Equation 18.8

me =

=

s n | e|

4.3 ´ 10 7 (W - m)-1 = 3.03 ´ 10 -3 m2 /V - s (8.86 ´ 10 28 m-3)(1.602 ´ 10 -19 C)

Electrical Resistivity of Metals 18.13 From Figure 18.38, estimate the value of A in Equation 18.11 for zinc as an impurity in copper–zinc alloys. Solution We want to solve for the parameter Ain Equation 18.11 using the data in Figure 18.38. From Equation 18.11

A=

ri ci (1 - ci )

However, the data plotted in Figure 18.38 is the total resistivity, r total, and includes both impurity (r i) and thermal (r t) contributions (Equation 18.9). The value of r t is taken as the resistivity at ci = 0 in Figure 18.38, which has a value of 1.7 ´ 10-8 (W-m); this must be subtracted out. Below are tabulated values of Adetermined at ci = 0.10, 0.20, and 0.30, including other data that were used in the computations. (Note: the ci values were taken from the upper horizontal axis of Figure 18.38, since it is graduated in atom percent zinc.) ci

1 – ci

r total (W-m)

r i ( W-m)

A (W-m)

0.10

0.90

4.0 ´ 10 -8

2.3 ´ 10- 8

2.56 ´ 10 -7

0.20

0.80

5.4 ´ 10 -8

3.7 ´ 10- 8

2.31 ´ 10 -7

0.30

0.70

6.15 ´ 10 -8

4.45 ´ 10- 8

2.12 ´ 10 -7

So, there is a slight decrease of Awith increasing ci.

18.14 (a) Using the data in Figure 18.8, determine the values of ρ0 and a from Equation 18.10 for pure copper. Take the temperature T to be in degrees Celsius. (b) Determine the value of A in Equation 18.11 for nickel as an impurity in copper, using the data in Figure 18.8. (c) Using the results of parts (a) and (b), estimate the electrical resistivity of copper containing 1.75 at% Ni at 100°C. Solution (a) Perhaps the easiest way to determine the values of r0 and a in Equation 18.10 for pure copper in Figure 18.8, is to set up two simultaneous equations using two resistivity values (labeled r1t and r t2) taken at two corresponding temperatures (T1 and T2). Thus,

r t 1 = r 0 + aT1 r t 2 = r0 + aT2 And solving these equations simultaneously lead to the following expressions for a and r 0:

a =

rt 1 - r t 2 T1 - T2

ér - r ù t2 ú r 0 = r t1 - T1 ê t1 êë T1 - T2 úû ér - r ù t 2ú = r t - T2 ê t1 2 êë T1 - T2 úû

From Figure 18.8, let us take T1 = –150°C, T2 = –50°C, which gives r t1 = 0.6 ´ 10 -8 (W-m), and r t2 = 1.25 ´ 10 - 8 (W-m). Therefore

a =

=

rt 1 - r t 2 T1 - T2

[( 0.6 ´ 10 -8 ) - (1.25

]

´ 10 -8) (W - m )

-150°C - (- 50°C)

6.5 ´ 10-11 (W-m)/°C

and ér - r ù t2 ú r 0 = r t1 - T1 ê t1 êë T - T2 úû 1

= (0.6 ´ 10 -8) - (-150)

[(0.6 ´ 10 -8) - ( 1.25

]

´ 10 -8 ) (W - m)

-150°C - (-50°C)

= 1.58 ´ 10 -8 ( W-m)

(b) For this part of the problem, we want to calculate Afrom Equation 18.11

r i = Aci (1 - ci ) In Figure 18.8, curves are plotted for three civalues (0.0112, 0.0216, and 0.0332). Let us find Afor each of these ci's by taking a r total from each curve at some temperature (say 0°C) and then subtracting out r i for pure copper at this same temperature (which is 1.7 ´ 10- 8 W-m). Below is tabulated values of A determined from these three ci values, and other data that were used in the computations.

ci

1 – ci

r total (W-m)

r i ( W-m)

A(W-m)

0.0112

0.989

3.0 ´ 10 - 8

1.3 ´ 10 -8

1.17 ´ 10 -6

0.0216

0.978

4.2 ´ 10 - 8

2.5 ´ 10 -8

1.18 ´ 10 -6

0.0332

0.967

5.5 ´ 10 - 8

3.8 ´ 10 -8

1.18 ´ 10 -6

The average of these three Avalues is 1.18 ´ 10 -6 ( W-m). (c) We use the results of parts (a) and (b) to estimate the electrical resistivity of copper containing 1.75 at% Ni (ci = 0.0175) at 100°C. The total resistivity is just

r total = r t + ri Or incorporating the expressions for r t and r i from Equations 18.10 and 18.11, and the values ofr 0, a , and A determined above, leads to

r total = (r0 + aT ) + Aci (1 - ci )

=

{1.58

´ 10

-8

(W - m) + [6.5 ´ 10 -11 (W - m) / °C] (100 °C)}

+ {[1.18 ´ 10

-6

(W - m) ] (0.0175 ) (1 - 0.0175)

= 4.25 ´ 10 -8 ( W-m)

}

18.15 Determine the electrical conductivity of a Cu-Ni alloy that has a yield strength of 125 MPa (18,000 psi). You will find Figure 7.16 helpful. Solution We are asked to determine the electrical conductivity of a Cu-Ni alloy that has a yield strength of 125 MPa. From Figure 7.16b, the composition of an alloy having this tensile strength is about 20 wt% Ni. For this composition, the resistivity is about 27 ´ 10-8 W- m (Figure 18.9). And since the conductivity is the reciprocal of the resistivity, Equation 18.4, we have

s=

1 1 = 3.70 ´ 10 6 (W - m)-1 = r 27 ´ 10 -8 W - m

18.16 Tin bronze has a composition of 92 wt% Cu and 8 wt% Sn, and consists of two phases at room temperature: an a phase, which is copper containing a very small amount of tin in solid solution, and an ephase, which consists of approximately 37 wt% Sn. Compute the room temperature conductivity of this alloy given the following data: Phase α

Electrical Resistivity (Ω-m) 1.88 × 10 –8

e

5.32 × 10

–7

3

Density (g/cm ) 8.94 8.25

Solution This problem asks for us to compute the room-temperature conductivity of a two-phase Cu-Sn alloy which composition is 92 wt% Cu-8 wt% Sn. It is first necessary for us to determine the volume fractions of the aand e phases, after which the resistivity (and subsequently, the conductivity) may be calculated using Equation 18.12. Weight fractions of the two phases are first calculated using the phase diagram information provided in the problem. We may represent a portion of the phase diagram near room temperature as follows:

Applying the lever rule to this situation

Wa =

Ce - C 0 37 - 8 = = 0.784 Ce - Ca 37 - 0

C - Ca 8 - 0 = 0.216 We = 0 = Ce - Ca 37 - 0

We must now convert these mass fractions into volume fractions using the phase densities given in the problem statement. (Note: in the following expressions, density is represented by r ' in order to avoid confusion with resistivity which is designated by r.) Utilization of Equations 9.6a and 9.6b leads to

Wa Va =

r'a Wa We + r'a r'e

0.784 8.94 g/cm3 = 0.784 0.216 + 8.94 g/cm3 8.25 g/cm 3

= 0.770

Ve = Wa r'a

We r'e +

We r'e

0.216 8.25 g/cm 3 = 0.784 0.216 + 8.94 g/cm3 8.25 g/cm 3

= 0.230

Now, using Equation 18.12

r = raVa + reVe = (1.88 ´ 10 -8 W - m)(0.770) + (5.32 ´ 10 -7 W - m) (0.230)

= 1.368 ´ 10 -7 W -m

Finally, for the conductivity (Equation 18.4)

s=

1 1 = 7.31 ´ 10 6 (W - m) -1 = r 1.368 ´ 10 -7 W - m

18.17 A cylindrical metal wire 2 mm (0.08 in.) in diameter is required to carry a current of 10 A with a minimum of 0.03 V drop per foot (300 mm) of wire. Which of the metals and alloys listed in Table 18.1 are possible candidates? Solution We are asked to select which of several metals may be used for a 2 mm diameter wire to carry 10 A, and have a voltage drop less than 0.03 V per foot (300 mm). Using Equations 18.3 and 18.4, let us determine the minimum conductivity required, and then select from Table 18.1, those metals that have conductivities greaterthan this value. Combining Equations 18.3 and 18.4, the minimum conductivity is just

s =

=

1 Il = = r VA

Il æ d ö2 Vp ç ÷ è 2ø

(10 A)(300 ´ 10 -3 m) = 3.2 ´ 10 7 (W - m) -1 æ 2 ´ 10 -3 m ö2 (0.03 V) (p) ç ÷ 2 ø è

Thus, from Table 18.1, only aluminum, gold, copper, and silver are candidates.

Intrinsic Semiconduction 18.18 (a) Using the data presented in Figure 18.16, determine the number of free electrons per atom for intrinsic germanium and silicon at room temperature (298 K). The densities for Ge and Si are 5.32 and 2.33 g/cm3, respectively. (b) Now explain the difference in these free-electron-per-atom values. Solution (a) For this part of the problem, we first read, from Figure 18.16, the number of free electrons (i.e., the intrinsic carrier concentration) at room temperature (298 K). These values are n i(Ge) = 5 ´ 1019 m-3 and n i(Si) = 7 ´ 1016 m- 3. Now, the number of atoms per cubic meter for Ge and Si (NGe and NSi, respectively) may be determined ' and r ' ) and atomic weights (A and A ). (Note: here we using Equation 4.2 which involves the densities (rGe Si Ge Si

use r ' to represe...