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 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:

s cm atoms J -⨯=29/105.8

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 %%%

故

3

20231/101417.11002.67.2.....%cm atoms Cu wt Mol Al FCC of density wt Cu c ⨯=⨯⨯⨯⎪⎭⎫ ⎝⎛⨯= 3202/1008176.1cm atoms c ⨯=

故

()

()s

cm atoms mm

cm atoms s m x c c D J -⨯=⨯-⨯

⨯-=--=-2103202

1412/10998.12.1/101417.108176.1/104

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