第一章 固体材料的结构 Chapter 1. The Structure of Materials

作业1：原版教材第105页第17题

17. Identify the plane in an FCC metal formed by the intercepts 1,-1,-2. What is the normal to this plane?

Solution:

1) the intercepts of this plane are 1,-1,-2; 2) The reciprocals of this plane are 1,-1,-reciprocals of the intercepts. 3) clear fractions, 4) site planes in parentheses (221) 5) The directions and planes which have same miller indices are perpendicular to each other.

作业2：原版教材第105页第18题

18. Calculate the angle between [100] and [111] in Al.

Solution:

The crystal structure of Al is Fcc. We can calculate the angle between [100] and [111] as

??z o y 1,taking the 2x cos??

uu'?vv'?ww'u?v?w?u'?v'?w'222222?1?1?0?1?0?11?0?0?1?1?1z 222222?13

??54.73?

o y x

作业3：原版教材第105页第19题

19. Construct a coordinate system at the center of a cubic unit cell with the axes parallel to the 100 directions. Determine the tetrahedral angle, the angle between directions from the origin to two ends of any face diagonal. Solution:

111??? OA:,,???111?

222??111??? OB:,,???111?

222??s? co?1?1?1?1?1?112?12?12?12?12?12?2??

33?3?2

作业4：原版教材第105页第21题

21. Give the indices of the points, directions, and planes in the cubic cells shown in Figure HP3.1.

Solution:

The first step is to select an origin and orient the coordinate axes. We have elected to use the bottom back left corner of the cube as our origin. z As shown in the figure; 1; 21Point B has coordinate 0,1,; 2Point A has coordinate 0,0,Plane F: The intercepts of F are 1,1, Take reciprocals: 1,1,2; Clear fraction: 1,1,2; Cite them into parentheses: ?112? Plane E: The intercepts of E: ?,?,1 Taking reciprocals: 0,0,1

Plane E 2 A B D 1; 2y Plane F x 1 C ???Direction C: 2号坐标 1,1,0 1号坐标：1,0,1 ?011?

??(1) Determine the coordinates of two points

(2) Subtract the coordinates of the second point from those of the first point; (3) Cleat fractions from the difference to give indices in lowest integer vales, 0,1,1

????(4) write the indices in square brackets without commas: ?011?

??????201??? (5) negative integer values are indicated by placing a bar over the integer. D: ?1） The coordinates of points G and H are 0,0,2) plane L:

(1) determine the intercepts of plane L:?11; 0,,0 respectively; 32111,, 223(2) take the reciprocals of the intercepts: -2,2,3 (3) clear fractions: -2,2,3 (4) cite planes in parentheses, placing bars over negative ???

indices: ?223?

??

???3) plane K: ?1,1,???110? ??Plane K J 114) directions: I 1点：?,1, 2点：0,1,0 22 2-1:

G I 11??,0,???101? 22???Plane L H J 2点：0,0,1 1点：-1,1,0 ??????? 1-2: -1,1,-1 ?111???111?

????作业5：原版教材第106页第22题

22. Give the indices of the points, directions, and planes in the cubic cells shown in Figure HP3.2.

Solution:

11???? Plane E: 1,?,? 1,-2,-2 ??122?

22??1??? Plane F: ?,?,1?0,?2,1??021?

2??A Plane E C B plane F D

1,0 211 Point B: ,?,?1

22 Point A: 0,? Direction B D:

D点坐标：0,0,-1

11??? D-B: ?,?,0??1,1,0??110?

22?? Direction C：（1） 1，-1，0 （2） 0,0,?1 21??? （1）-（2） 1,?1,?2,?2,1??221?

2??

Point H: 1,1,1

I 11Point G: ,,0

22Direction I: (1) 0,0,H Plane K 1 (2) 1,1,1 2Plane L G J 1????? (1)- (2) ?1,?1,???221? 2?? Direction J: (1) 1,1,0 (2) 0,1,0 (1)- (2) 1,0,0??100? Plane L: ?,11,?0,2,2??011? 22????Plane K: 1,1,1??111?

??

作业6：

Use a calculation to verify that the atomic packing factor for the FCC structure is 0.74. Solution:

In an FCC, there are four lattice points per cell: if there is one atom per lattice point, there are also four atoms per cell. The volume of one atom is 4πr3/3 and the volume of the unit cell is a3: Packing factor = 4×4πr3/3 a3

Since for FCC unit cell, a=4r/2, packing factor =o.74

作业7：

用金属键原理解释金属的以下几个特征：良好的导电导热性、正的电阻温度系数、不透明和良好的延展性。 解答：

1）导电导热性：

金属中的自由电子在外加电场作用下，能够沿着电场方向作定向运动，具有导电性。金属中正离子振动和自由电子的运动均可传递热量（能），表现出良好的导热性。 2）正的电阻温度系数：

温度升高时，金属正离子的振动加剧，与自由电子的碰撞几率加大，对自由电子的定向运动阻力增大，故具有电的电阻温度系数。 3) 不透明：

金属中自由电子能够吸收并随后辐射出投射到金属表面的光能。故金属不透明，且具有光泽。

4） 延展性：

金属键没有方向性和饱和性，对原子也没有选择性。当受外力原子发生相对移动时，金属正离子仍包围在电子气之中。金属键不会受到破坏。 作业8：

算出fcc、bcc晶体中四面体间隙和八面体间隙 大小，用原子半径rA表示，并注明原子间隙位置。 解答：

bcc：八面体间隙：最小rB?置（

a?rA?0.154rA，位2411rA,（面中心及各边中心） ，，1），a?223 四面体间隙：（

11，，1） 24?a??a?rB???????rA?0.25rA，

?2??4? fcc：八面体间隙：体心处（

22111，，），222rB?4arA ?rA?0.414rA，a?22 四面体间隙：体对角线

1313处：（，，），4444rB?3a?rA?0.25rA 4

作业9.

写出溶解在γ-Fe中碳原子所处的位置，若此类位置全部被碳原子占据，那么试问在这种情况下，γ-Fe能溶解多少重量百分数的碳？而实际上在γ-Fe中最大溶解度是多少？两者在数值上有差异的原因是什么？ 解答：

γ-Fe是fcc，八面体间隙尺寸大，故C的存在位置为八面体间隙。每个晶胞内有四个Fe原子，有4个八面体间隙，若全部被占满， 则原子百分数为：

4?C??100%?50%at 4?Fe??4?C? 重量百分数为：

4?12?100%?17.6%at

4?12?4?56而碳在γ-Fe中实际最大固溶度为2.11%wt。 因为碳原子半径0.077>八面体间隙（0.054nm）

故碳的溶入会引起γ-Fe晶格畸变，妨碍了碳原子进一步溶入。 rC=0.77? r八=0.535 ?

作业10：

若已知910℃时α-Fe和γ-Fe的晶格常数分别为， 试问γ-Fe在910℃转变为α-Fe时的体积是膨胀还是收缩，其体积变化率是多少？ 解答：

910℃： α-Fe的aα=0.2892nm

γ-Fe的aγ=.3633nm α-Fe γ-Fe 从晶胞体积来看：α-Fe：Vα=a?=2个原子 γ-Fe：Vγ=a?=4个原子 在γ-Fe中4个原子所占体积=a?=0.36333=0.0835477 α-Fe中4个原子所占体积=2 Vα=2?0.28923=0.0483754

3330.36333?2?0.28923?100%?0.885% 体积变化率=30.3633

作业11：

试在一个立方晶系晶胞中确定O（0，0，0）, A（1/2, 0, 1/2）, B（1/2, 1/2, 0）, C（0, 1/2, 1/2）四个点的位置，并写出由它们构成的正四面体各表面的晶面指数，以及各棱边的晶向指数。 解答：

各表面： OAB：(111) (原点1) （111） ABC：（111） (原点2) （111）

??????

OCA：（111） (原点3) （111） OBC：（111） (原点4) （111）

??????3??? 各棱边：OA：[101] AB:?011?

????? OB:[110] BC:?101?

????? OC:[011] BC:?110 ???1 ??C A o B 2 4

作业12：

已知碳原子半径为0.077nm, 在727℃时α-Fe原子半径为0.125nm, 在1148℃时，γ-Fe原子半径为0.129nm, 通过计算证明碳在γ-Fe中溶解度大于α-Fe中的溶解度。 解答：

fcc: 八面体 RB?0.414RA?0.414?0.129?0.053106nm?0.146a

Bcc： RB?0.154RA?0.154?0.125?0.01925nm?0.067a

Rc?0.077nm

作业13：

在立方晶胞中画出下列晶面与晶向，并指出相同指数的晶面与晶向的关系。（001）和 [001],（110）和 [110],（111）和 [111],（112）和 [112],（221）和 [221]. 解答： [112] [001] (001) (111) (112) (110) [110] [111]

作业14:

试计算体心立方晶格{100}， {110}， {111}等晶面的原子密度，以及[100], [110], [111]等晶向的原子密度，并指出其原子最密晶面和最密晶向。

晶胞类型 晶面 晶向

[100] [110] [111] （100） （110） （111）

11体心立方

112 221 ?2 ?3 aa22a3a6a?2aa?3 2a13 ?2a?2a?22

2a a 2a 2a 3a a a

21面心立方 2a 421 112a2 ?3?2?a 3 aa 232a3a62a?2 a13 2a?2a?

22

第二章 晶体结构的缺陷

Chapter 2 The Crystal Structure’s Defects

作业1：原版教材第180页第2题

2. Consider the FCC metal Cu, which has a lattice parameter a of 0.362nm. a. Calculate the length of the lowest-energy Burgers vector in this material. b. Express this length in terms of the radius of a Cu atom.

c. On which family of planes will slip occur in this material? Why? Solution:

0.362nm a. |b|2min=

2a?220.362nm?0.25597nm b. |b|2min=

2a?2r c. {111}. Because it needs the lowest energy to make dislocations slip on this family of planes.

作业2：原版教材第180页第3题

3. The lowest-energy Burgers vector in FCC Ag has length 0.288nm. Find the length of the unit cell edge in Ag. Solution: |b|2min=0.288nm=2a a?2?0.288nm

作业3：原版教材第181页第5题

5. A molybdenum (Mo) crystal has a Burgers vector of length 0.272nm. If the lattice parameter a in Mo is 0.314nm, determine whether Mo has the BCC or the FCC crystal structure. Solution:

If Mo has the FCC crystal structure, then: 0.272?22a a?2?0.272?0.3847

If Mo has the BCC structure, then: 0.272?So we know that Mo has the BCC structure.

作业4：原版教材第181页第12题

32a a??0.272?0.314 2312. Consider the 111 plane of an FCC crystal and a dislocation whose Burgers vector is parallel to

???110?. The dislocation itself is parallel to the intersection of 111 and (111).

??a. Give the Burgers vector of the dislocation.

b. Indicate the character of the dislocation. Solution:

??101?, the b). The intersection of ?111? and ?111? is ?101?, the angle between dislocation is parallel to this direction ? a). The Burgers vector of the dislocation is b?a110 2?111? t and b is 60°, so the character of dislocation is mixed dislocation.

也可计算： u-v+w=0 ?u=-w v=0, 最小整数为1，101or101。 u+v+w=0

作业5：原版教材第181页第13题

13. Show that the dislocation reaction given below is both vectorially proper and energetically favorable for a BCC metal: ?a2??111???a2?111?a?100? Solution: Vector：

[110] ??????a?111??a111?a?200??a?100? 222??The above dislocation reaction is vectorially proper.

a2222a2222a23??a2 ?6?a2 a?100Energy: 21?1?1?21?1?1?4222????b1?232a,b242?322a,b1?b242?322a?a2?b3 2So, the above dislocation reaction is energetically favorable.

The energy before the reaction is higher than that after reaction, and vectors before the

reaction equal to that after the reaction, so this dislocation reaction is both vectorially proper and energetically favorable.

作业6：原版教材第181页第18题

18. Suppose a metal single crystal is loaded (stressed) in the [101] direction.

a. If the critical resolved shear stress for this material is 0.34MPa, what magnitude of

10] slip system? applied stress is necessary for dislocations to begin to move in the (111) [1 b. If the slip system listed in part a is valid, does this metal have the FCC HCP, or BCC

crystal structure? Solution:

a. we know that：?k?0.34MPa loaded in the direction：[101] slip system：[111][110] ???k cos?cos?[101]?[111]12?02?12?1?12?122cos???1?12? 2?36?1?0?01? 22?2cos???101???110?12?02?12?1?12?02???0.34?0.34?6??0.833MPa 21?62So the magnitude of applied stress is 0.833MPa

b. If the slip system listed in part a is valid, then this metal has the FCC crystal structure.

作业7：原版教材第181页第19题

19. Suppose you have an FCC metal single crystal which is known to have a critical resolved shear stress of 55.2MPa.Find the largest normal stress that could be applied to a bar of this material in the [112] direction before dislocations begin to move in the 101 direction on the (111) plane. Solution:

??

We know that ?k?55.2MPa loaded in the direction：[112] slip system：(111) 101 ?????k

co?s?cos?then cos???112???111?12?12?22?12?12?12?1?1?24?

6?332 cos???112???101?12?12?22?12?02?12?1?21 ??6?223 ?max?55.255.2?36 ??202.8MPa42?1?32???????23?So the largest normal stress is 202.8MPa.

作业8：原版教材第181页第22题

22. The critical resolved shear stress of a BCC metal is 7 MPa. A single crystal of this alloy is stressed along the [001] direction. a. What slip systems will be activated? b. What normal stress will cause plastic deformation? Solution:

slip systems in BCC: {110}<111> {112}<111> {123}<111> 10) (101) (011) {110}： (110) (101) (011) (1s 若点乘[001]， 只有第三个位置不为0的cos?值不为0 co? cos??11??0.707

2?12{112}：(112) (112) (112) (112)

(121) (121) (121) (121)

1) (211) (211) (211) (21 点剩[001] 最大为第三位为2的

cos??2?0.816

6?1

{123}：(123) (132) (213) (231) (312) (321) (123) (132) (213) (231) (312) (321)

(123) (132) (213) (231) (312) (321)

(123) (132) (213) (231) (312) (321) 点剩[001]，最大为第三位为3的：

cos??33??0.8017

1?4?9?11411??0.577 3?13所有的λ相同：cos??故被首先激活的将为：红色所示，所需的??

作业9：原版教材第182页第24题 ?cos??cos??7?14.867

0.816?0.57724. Calculate the magnitude of the applied stress in the [123] direction necessary to promote slip in the [111] direction on the (110) plane in a BCC crystal with ?CR?5.5MPa. [It may or may not help you to know that a certain FCC single crystal with ?CR?0.55MPa slips in the [110] direction on the (111) plane under an applied stress of 9.63?10MPa applied in the [123] direction. ] Solution:

Loaded in the direction: [123] slip system: [111], 110,

?2???CR?5.5MPa

cos???123???110?12?22?32?12?12?02?1?21 ??13?226 cos???123???111??13?36 39

???CR?5.5?13?3?13?25.5?13?6 ???cos??cos?6?16

作业10：

分析下列位错反应能否进行： 几何条件：?b前??b后

22能量条件：?前??后

解答：

a???a???a???（1）?101???121???111?

2??6??3??a???a???a???a???a???a???几何：?101???121???303???121???222???111? 满足几何条件

2??6??6??6??6??3??a2a???22能量：b1??101?， 其长度=1?0???1?22??a2a???22 b2??121?， 其长度=1?2?166????12?2a 2??12?6a 63a 3a222a??? b3??111? ， 其长度=1?1?133????12?2b12?b2?22628222a?a?a?a 43612331b32?a2?a2

9322??b前??b后 故可以进行

aa???（2）a?100???101???101?

22??aa???a ?101???101???200??a?100?

22??2满足几何条件：b1?a

a22 b2?1?0?12??12?2a 2

b3?2a 223a2a2b?a b?b???a2

2221222能量相等

aaa???（3） ?112???111???111?

326??aaaa??? b1?b2??224???333???557???111? 故不能反应

6666??aa????（4）a?100???111???111?

22?? b1 b2 b3 b2?b3?212a?200??a?100? 满足几何条件 2223a2322 b?a b? b3?a 44 b2?b3?22622a?b1 不能反应 4

作业11：

在一个简单立方二维晶体中，画出一个正刃型位错和一个负刃型位错，并

（1） 用柏氏回路求出正刃型位错的柏氏矢量:

（2）若将正、负刃型位错反向时，其b是否也反向？ 是

（3）具体写出该柏氏矢量的大小和方向

正刃 b??a?100? ?100?

??????负刃 b??a?100? 长度为a， 方向?100?

????（4）求出两柏氏矢量之和：

b??b??0

作业12：

Assume that the triangle in the drawing lies on the (111) plane of a face-centered cubic crystal and that its edges are equal in magnitude to the Burgers vectors of the three total dislocations that can glide in this plane. Then if δ lies at the centroid of this triangle, lines Aδ, Cδ, Bδ accordingly correspond to the three possible partial dislocations of this plane. a. Identify each line with its proper Burgers vectors expressed in the vector notation. b. Demonstrate by vector addition that

C Bδ+ δC=BC

Solution: a：AB

以A为起始点，B点为坐标：B（1a,1a,0） ?δ o B

???AB?a?110?

??aaaA A?：δ点坐标：若以o点为原点，则δ：,, 3332aaa 若以A点为原点，则δ：?,, 333a???A???211?

3??a???a2aB???121? 以B为原点：δ：,?a,

3??333a????112a?C??112? 以δ为原点：C：?,?,

3?333???????BC?a?011? AC?a?101?

????a???a????a??????b: B???C??121???112???033??a?011??BC

3??3??3????

第三章 固体材料中的扩散

Chapter3 The Diffusion in Solid Materials

作业1：原版教材第143页第22题

22. Which type of diffusion do you think will be easier (have a lower activation energy)? a. C in HCP Ti b. N in BCC Ti c. Ti in BCC Ti

Explain your choice. Solution:

A and b interstitial solid solutions, but c is a substitutional solid solution. So the mechanism of diffusion of a and b is interstitial diffusion, and the mechanism of diffusion of c is the vacancy exchange. We have known that the activation energy for vacancy-assisted diffusion Qv are higher than those for interstitial diffusion Qi. So c is the most difficult one comparing a and b, HCP Ti is a close-packed structure, much closer than BCC, so b is the answer. The diffusion of N in BCC Ti will be easier (have a lower activation energy).

作业2：原版教材第143页第19题

19. Consider the possibility of solid solutions with Au acting as the solvent.

a. Which elements (N, Ag, or Cs) is most likely to form an interstitial solid solution with Au?

b. Which elements (N, Ag, or Cs) is most likely to form a substitutional solid solution with Au? Solution:

a. N is most likely to form an interstitial solid solution with Au; b. Ag is most likely to for a substitutional solid solution with Au.

作业3：原版教材第143页第23题

23.At one instant in time there is 0.19 atomic % Cu at the surface of Al and 0.18 atomic %

Cu at a depth of 1.2mm below the surface. The diffusion coefficient of Cu in Al is 44?10?14m2/s at the temperature of interest. The lattice parameter of FCC Al is 4.049?.

What is the flux of Cu atoms from the surface to the interior? Solution:

J?8.5?10atoms/cm?s

已知：Cu 的原子量（atomic mass）63.55 (Density of solid(g/cm3)) 8.93

Al的原子量（atomic mass）26.98 (Density of solid(g/cm3)) 2.70 换算成重量百分数：Cu%wt?92Cu%at?Cu原子量

Cu%at?Cu原子量?Al%at?Al原子量故c1???Cu%wt?density.of.FCC.Al?23203??2.7?6.02?10?1.1417?10atoms/cm

Mol.wt.Cu??203 c2?1.08176?10atoms/cm c2?c1?1.08176?1.1417??1020atoms/cm3?142J??D??4?10m/s?故 x1.2mm?1.998?1010atoms/cm2?s??

另一种算法： 每个Al晶胞有4个原子，晶胞体积为a3，故Al的原子密度为:

44223 ??6.026?10个/cm33?8a4.049?10cm??已知Cu的原子百分数为0.18%和0.19%，即0.0018，0.0019 故c1?0.0019?6.026?10个/cm c2?0.0018?6.026?10个/cm

223223c2?c1?0.0001?6.026?1022个/cm3?1442J??D??4?10?10cm/s? x0.12cm?2.0087?1010原子/cm2?s????

作业4：原版教材第143页第27题

27. Consider the diffusion of C into Fe. At approximately what temperature would a specimen of Fe have to be carburized for 2 hours to produce the same diffusion result as at 900℃ for 15 hours ? Solution:

?x?c?cs??cs?c0?erf????

?2Dt?The same diffusion result means that other variables are the same and D1t1=D2t2 900℃ D1?15?D2?2 T?

D9002? DT15We know that D?D0exp?QQ lnD?lnD0? RTRT?52查表可知： D0900℃ =0.20?10m/s Q900℃=84?10J/mol D0>912℃=2.0?10m/s Q>912℃=140?10J/mol R=8.314J/mol-K DT=D900?3?52315 lnDT?lnD900?ln15?ln2 2140?10315?84?103?? ln2.0?5? ??ln0.2?5???8.314T2?8.314?1173? T=18918k=1618.8℃

作业5：原版教材第143页第30题

30. A Ti rod (HCP) is to be placed in a furnace in order to increase its carbon content. If the initial carbon content of the rod is 0.2 weight % and the carbon content in the furnace is the equivalent of 1.0 weight %, find the temperature required to yield a carbon content of 0.5 weight % at a depth of 0.4mm below the surface of the rod in 48 hours. Solution:

c?cs??cs?c0?erf??We know that cs?1.0%wt

c0?0.2%wt

x?0.4mm

c?0.5%wt t?48hour s?x???

?2Dt?

0.4?10?3mThen 0.5?1.0??1.0?0.2?erf

2D?48?3600s erfZ?0.5?0.625 0.80.4?10?3?132 Z?0.625? D?5.9259?10m/s

60?248D

作业6：

将含碳0.2%的碳钢零件置于1.2%碳势的渗碳气氛中加热至930℃，经10小时保温后随炉冷却至室温，试分析在930℃和室温零件从表层到心部成分和组织的变化规律，并画出示意图。 解答：

过共析 共析 亚共析

作业7：

一根足够长的共析钢棒在800℃强脱碳气氛中从一端脱碳一段时间后，试画出沿长度方向碳浓度分布曲线及组织示意图。若将其缓慢冷却至室温，画出室温下组织示意图。 解答：

1.

作业8：

设有一个厚为0.1cm的Si图片，初始时有1千万个Si原子中含有一个磷原子。经加工处理后，在表面的每1千万个Si原子中含有400个磷原子。

Si的晶体结构为金刚石结构，每个晶胞中含有8个原子，晶胞常数a?0.543nm，试计算该硅圆片的线浓度梯度和体积梯度。 解答： 400个磷原子C? S计算P%： 107个硅原子1?100%?0.00001% 71cm 1?10400 Cs??100%?0.004% 71?10?C0.00001%?0.004%线浓度梯度???0.0399%/cm

?X0.1内部：Ci?体积浓度梯度：V晶胞?5.4307?10cm?1.6?10Ci?1个P 710个Si??8?3?22cm3 107?V晶胞?2?10?16cm3 10个Si原子的体积：87

1个/cm3?5?1015个/cm3 ?162?104003183 Cs? 个/cm?2?10个/cm?62?10此体积中的P浓度：Ci??C5?1015?2?1018???1.995?1019原子/cm3?cm 体积浓度梯度：?X0.1

第四章 结晶过程的基本原理 Chapter 4 Fundamentals of Solidification

作业1：

The solid-liquid interfacial （界面） energy of pure silver（纯银） is 0.126J/m2. The latent heat of melting（熔化潜热） is 104.6J/g. The melting point （熔点） of pure silver is 961℃. The density of solid/liquid silver at the melting point is 10.5g/cm3. (15points)

(1) What is the value of the critical radius（临界半径值） at 700℃?

(2) What is the value of ΔG* at this radius? Solution:

2?Tmr*=-=Lm?T2?0.126J/m2?(961?273)104.6J/g?10.5g/cm3?(961700)2=1.0848×10-7cm

?G??16??3Tm23Lm?T211??S??0.126J/m2?4?r2?6.21?10?19J 33作业2：

This diagram（图） is for a hypothetical embryo of silver growing against an arbitrary mold wall. With the aid of this diagram,

(a) Compute the angle of contact, θ, of the embryo with the mold wall.

(b) Determine the magnitude of the factor that may be used to convert the homogeneous free energy needed to obtain a nucleus into that of the corresponding heterogeneous free energy.

Solution:

(a) rlm?rsm?rlx?cos??0.214?0.100?0.123?cos? cos??0.114?0.9268 ??22℃ 0.123

(b) ?Ghet??Ghom,?2?3cos??cos3???? ??4???2?3cos??cos3??2?3?0.9268?0.92683????0.0039 ??44?? 作业3:

92

已知纯铜的熔化潜热为1.88×10J/M,熔点为1089℃，点阵常数为3.4167?, 发生均匀形核过冷度为230K，的铜原子数。 解答：

??12?。求铜的临界晶核半径及临界晶核中所含＝1.44*10J/m?SL2?Tm2?1.44?10?1??1089?273?r???0.9?10?9m 9Lm?T1.88?10?230?4*3?r33?4?313个 晶胞体积V胞?a，每个晶胞含4个Cu原子，晶核所含铜原子数n?V胞

作业4：

已知锌的熔点为419℃，其结晶潜热为7×10J/m,液固界面能?＝0.06J/m，锌的原子量为65.4，密度为7.18g/cm,试计算锌在350℃结晶后的临界晶核半径，并说明晶核内有多是个锌原子？ 解答：

3832r???2??Tm2?2?0.06??419?273??9???1.72?10m ?GVLm??T7?108??419?350?4V???r?3?2.13?10?26m3

3晶核重量W??V?7.18?2.13?10晶核中含有原子数 n???26?106?1.53?10?19

W?6.023?1023?1.4568?103个 M作业5：

金的熔点为1064℃，溶化潜热为12.8KJ/mol, 若液态金在1000℃时发生结晶，其临界晶核半径?解答： r???*?43.3*10?10m，试计算金属的液固界面能。

2??Tm2?? ?GVLm??T

Lm??T?r? ??2Tm?GS???S?4?r?2r??Lm??TL??TL?T ?4?r?3m?2?r?3m2Tm2TmTm?2?3.14?43.3?10?10???312.8?64?3.1238?10?25m3?KJ/mol

1064?273

作业6：

解释金属结晶时，为什么会产生过冷？ 解答：

金属结晶时，产生过冷是因为需要?G?Gs?Gl?0

作业7：

在均匀形核时，若设晶核形状为边长a的立方体，试求其临界晶核半径及形核功。 解答：

32? ?G??GV?V???S??GV?a???6a a??4? ?GV?4??G???GV?????GV?

作业8：

??4????6????GV??3?32?3??????G2 ?V2若在液态金属中以非均匀方式形成半径为r非的球冠形晶核，试证明?G非与临界晶核体积之间有：?G非??V?GV关系。 证明：

?G?V?GV??GS

???12??GS??L??r22?3cos??cos3?

???2?3cos??cos3???V??r? ??3??3??r非??2?L??GV

??2?L?V?????GV?

????33?2?3cos??cos3??8??L?????3??3?GV???2?3cos??cos3?????? 3????2?L??GS??L?????GV?

?3?????GV V?2?231???G非??GVV???GVV????GVV?

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