工程材料科学与设计原书第2版课后习题答案4—8章

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Solutions to Chapter 4

1. FIND: What material has a property that is hugely affected by a small impurity level?

SOLUTION: Electrical conductivity spans a wide range. Incorporation of a few parts per million impurities can change electrical conductivity orders of magnitude. Small cracks in brittle materials decrease their tensile strength by orders of magnitude. Small additions of impurity can change the color of gems.

COMMENTS: These are but a few examples.

2.

COMPUTE: The temperature at which the vacancy concentration is one half that of 25o C.

GIVEN: C 2 = C C 25v C 35v

o

o

EQUATION:????

??RT Q - = C fv v exp

where C v = vacancy concentration

Q fv = activation energy for vacancy information R = gas constant 8.314 J/mole-K

T = absolute temperature

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In the present problem C)25(C = C C);

35(C = C o v 2

v o v 1v

and T 1 = 35 + 273 = 308K

T 2 = 25 + 273 = 298K

also C v (35o C) = 2C v (25o C)

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Thus,

Solving for Q fv we get Q fv = 52893.5 J/mole.

Using this value of Q fv , the C v (25o C) can be calculated

The problem requires us to calculate the temperature at which the vacancy concentration is ? C v (25o C). ? C v (25o C) = 2.675 x 10-10

Thus

for solving T, we get: T = 288.63K or 15.63o C.

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3.

COMPUTE: C)80( C 3 = (T) C o v v

GIVEN: C) 80( C 41 = C) 25( C o

v o v

EQUATION:??

?

??298.R Q - C) 25( C Sv o v exp

Dividing (1) by (2) we get:

Solving for Q, we get: Q = 22033.56 J/mole = exp(-7.511)

= 5.46 x 10-4

The problem requires computing a temperature at which C v = 3C v (80o C).

3C v (80o C) = 3 x 5.46 x 10-4 = 1.63 x 10-3

??

?

??T x 8.3122033.56- = 10 x 1.633-ex p

solving for T, we get: T = 413.05K or 140.05o C

4.

5. FIND: Are Al and Zn completely soluble in solid solution?

If Al-Zn system obeys all the Hume-Rothery rules. Then it is expected to show complete solubility.

(i) The atomic radii of Al and Zn are 0.143nm and 0.133 nm respectively.

The difference in their radii is 7.5% which is less than 15%.

(ii) The electronegativities of Al and Zn are 1.61 an 1.65 respectively which are also very similar.

(iii) The most common valence of Al is +3 and +2 for Zn.

(iv) Al has an FCC structure where Zn has a HCP structure.

It appears that Al-Zn system obeys 3 out of 4 Hume-Rothery rules. In this case they are not expected to be completely soluble.

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6. SHOW: The extent of solid solution formation in the following systems using

Hume-Rothery Rules.

(a) Al in Ni

Size: r(Ni) = 0.125nm; r(Al) = 0.143nm difference = 14.4%

Electronegativity: Al = 1.61; Ni = 1.91

Most Common Valence: Al3+; Ni2+

Crystal Structure: Al: FCC; Ni:FCC

The crystal structure of Al and Ni are the same and the most common valencies are also comparable. However, the size difference is close to 15% and the

difference is electronegativities is rather significant.

Based on this, it appears that Ni and Al would not form a solid solution over the entire compositional range.

(b) Ti in Ni

Size: r(Ti) = 0.147 nm, r(Ni) = 0.125nm difference = 17.6%

Electronegativity: Ti: 1.54; Ni: 1.91

Valence: Ti4+; Ni2+

Crystal Structure: Ti:HCP; Ni FCC

Ti in Ni would not exhibit extensive solid solubility

(c) Zn in Fe

Size r(Zn) = 0.133nm; r(Fe) - 0.124nm difference = 7.25%

Electronegativity: Zn = 1.65; Fe = 1.83

Most Common Valence: Zn2+; Fe2+

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Crystal Structure: An: HCP; Fe: BCC

Since electronegativities and crystal structures are very different, Zn - Fe will not exhibit extensive solid solubility.

(d) Si in Al

Size r(Si) = 0.117 nm; r(Al) = 0.143nm; difference = 22.2% Electronegativity: (Si) = 1.90; Al = 1.61

Valence: Si4+; A;3+

Crystal Structure: Si: Diamond Cubic; Al: FCC

Since the size difference is greater than 15%, and the crystal structures are different, Si-Al would not exhibit extensive solid solubility.

(e) Li in Al

Size r(li): 0.152, r(Al): 0.143; difference - 6.29%

Electronegativity: Li: 0.98; Al: 1.61

Most Common Valence: Li1+; Al3+

Crystal Structure: Li:BCC; Al: FCC

Since electronegativity and crystal structures are very different, Li-Al will not exhibit extensive solid solubility.

(f) Cu in Au

Size r(Cu) = 0.125nm; r(au) = 0.144nm; difference = 12.5% Electronegativity: Cu = 1.90; Au = 1.93

Most Common Valence: Cu+; Au+

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Crystal Structure: Cu:FCC; Au:FCC

Cu-Au will exhibit extensive solid solubility.

(g) Mn in Fe

Size r(Mn) = 0.112, r(Fe) = 0.124 difference = 10.71%

Electronegativity: Mn 1.55; Fe 1.83

Most Common Valence: Mn2+; Fe2+

Crystal Structure: Mn:BCC; Fe BCC

The difference in electronegativity is high but Mn-Fe does obey the other 3 Hume-Rothery rules. Therefore, it will form solid solutions but not over the entire compositional range.

(h) Cr in Fe

Size r(Cr) = 0.125nm, Fe = 0.144nm difference = 12.5%

Electronegativity: Cr = 1.66; Fe = 1.83

Most Common Valence: Cr3+; Fe2+

Crystal Structure: Cr:BCC; Fe:BCC

Cr in Fe will exhibit extensive solid solubility but not over the entire compositional range since it obeys only 3 of 4 Hume-Rothery rules.

(i) Ni in Fe

Size r(Ni) = 0.125nm, r(Fe) = 0.124nm difference = 0.8%

Electronegativity: Ni: 1.91; Fe 1.83

Most Common Valence: Ni3+; Fe3+

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Crystal Structure: Ni:FCC; Fe: BCC

Ni and Fe obeys 3 of the 4 Hume-Rothery rules therefore, extensive solid

solution will be exhibited but not over the entire compositional range.

7. (a) When one attempts to add a small amount of Ni to Cu, Ni is the solute and

Cu is the solvent.

(b) Based on the relative sizes of Ni and Cu, radius of Ni = 0.128nm, radius of

Cu = 0.125nm, these two are expected to form substitutional solid solutions.

(c) Ni and Cu will be completely soluble in each other because they obey all four

Hume-Rothery rules.

8. FIND: Predict how Cu dissolves in Al.

DATA: Cu Al

atomic radius (A) 1.28 1.43

electronegativity 1.90 1.61

valence 1+,2+ 3+

crystal structure FCC FCC

SOLUTION: All of Hume-Rothery's rules must be followed for a substitutional

solution. In this case, the valences do not match. Cu will not go into substitutional positions in Al to a large extent.

COMMENTS: This principle is often used to precipitation harden Al using Cu.

9. What type of solid solution is expected to form when C is added to Fe?

The radius of carbon atom is 0.077nm and that of an Fe atom is 0.124nm. The size difference between these two is ~61% which is much grater than ~15%.

Thus, these two are not expected to form substitutional solid solution.

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If we compare the size ratio of C to Fe atoms with the size of tetrahedral and

octahedral interstitial sites in BCC iron, we find that C does not easily fit into

either type of interstitial position. C, however, forms an interstitial solid solution with Fe but the solubility is limited.

10. FIND: Calculate the activation for vacancy formation in Fe.

GIVEN: The vacancy concentration at 727 C = 1000K is 0.00022.

SOLUTION: We use equation 4.2-2 to solve this problem:

C v = exp (-Q fv/RT)

Solving for Q fv:

Q fv = -RT ln C v = -(8.31 J/mole-K)(1000K) ln 0.00022 = 7.0 x 104 J/mole

11. SHOW: A Schottky and Frenkel defect in MgF2 structures

A 2-D representation of the MgF2 structure containing a Schottky defect and a

Frenkel defect is shown below.

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