第三章 固体材料中的扩散作业答案

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 are 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 Q v are higher than those for interstitional diffusion Q i. 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
s m /104214-⨯
at the temperature of interest.
The lattice parameter of FCC Al is 4.049Å. What is the flux of Cu atoms from the surface to the interior?
Solution:
已知：Cu 的原子量（atomic mass ）63.55 (Density of solid(g/cm 3)) 8.93
Al 的原子量（atomic mass ）26.98 (Density of solid(g/cm 3)) 2.70 换
算成重量百分数：
原子量
原子量原子量
Al at l Cu at Cu Cu at Cu wt Cu ⨯+⨯⨯=
%A %%%
当Cu 故321231
/1096.681002.6.....%cm atoms Cu wt Mol Al FCC of density wt Cu c
⨯=⨯⨯⎪⎭
⎫ ⎝⎛⨯= 3212/1002.19cm atoms c ⨯=
故
()
()s
cm atoms mm
cm atoms s m x c c D J -⨯=⨯-⨯
⨯-=--=-2103212
1412/10988.12.1/1002.1996.68/104
另一种算法：
每个Al 晶胞有4个原子，晶胞体积为a 3，故Al 的原子密度为:
()
322383/10026.610049.444cm cm
a 个⨯=⨯=- 已知Cu 的原子百分数为0.18%和0.19%，即0.0018，0.0019 故3221
/10026.60019.0cm c 个⨯⨯=
3222/10026.60018.0cm c 个⨯⨯=
()
()
s
cm cm
cm s cm x c c D J ∙⨯=⨯⨯-⨯
⨯⨯-=--=-2103222
41412/100087.212.0/10026.60001.0/10104原子个
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:
()⎪⎪⎭
⎫
⎝⎛--=Dt x erf c c c c s s 20 The same diffusion result means that other variables are the same and D 1t 1=D 2t 2 900℃
21521⨯=⨯D D T?
15
2
900=T D D We know that
RT
Q
D D -=exp
RT
Q D D -
=0ln ln
查表可知： D 0900℃ =s m /1020.025
-⨯
Q 900℃=mol J /10
843
⨯
D 0>912℃=s m /100.225
-⨯
Q>912℃=mol J /101403
⨯
R=8.314J/mol-K
D T =2
15900⨯
D
2ln 15ln ln ln 900-+=D D T
⎪⎪⎭⎫
⎝⎛⨯⨯--=⨯--1173
314.8108452.0ln 215314.81014050.2ln 3
3T T=1891.8k=1618.8℃
The same diffusion result means that other variables are the same and D 1t 1=D 2t 2 即：
0102exp(
)2
15
exp()Q
D RT Q
D RT -=- 121115
ln 2.0152
Q R T T ⎛⎫-== ⎪⎝⎭ 查表可知：
Q 1=mol J /10
843
⨯
R=8.314J/mol-K
计算可得：
T 2=1453.73K=1180.73℃
作业5：
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nm a 543.0=，试计算该硅圆片的线浓度梯度和体积梯度。

解答： 1） 计算P%： 内部：%00001.0%10010
11
7
=⨯⨯=
i C
%004.0%10010
14007=⨯⨯=s C
线浓度梯度cm X C /%0399.01
.0%004.0%00001.0-=-=∆∆
体积浓度梯度：()
3223
8cm 106.1cm 104307.5--⨯=⨯=晶胞
V
107
个Si
原子的体积：3167
cm 102V 8
10-⨯=⨯晶胞
此体积中的P 浓度：3
15316
cm /105cm /10
21C 个个⨯=⨯=
-i
3
1836
cm /102cm /102400C 个个⨯=⨯=-s 体积浓度梯度：cm cm /10995.11
.0102105C 31918
15∙⨯-=⨯-⨯=
∆∆原子X
个磷原子
7
400。