第 1章作业 答案.doc

Chapter one home work
1. (P80 3-3) Calculate the atomic radius in cm for the following:
(a) BCC metal with a0=0.3294nm and one atom per lattice point; and
(b) FCC metal with a0=4.0862Å and one atom per lattice point.
Solution:
(a) In BCC structures, atoms touch along the body diagonal, which isa0 in length. There are two atomic radii from the center atom and one atomic radius from each of the corner atoms on the body diagonal, so:
=0.14263nm=1.4263cm
(b) In FCC structures, atoms touch along the face diagonal of the cube, which is in length. There are four atomic radii along this length—two radii from the face-centered atom and one radius from each corner, so , =1.44447 Å=1.44447cm

2. (P80 3-4) determine the crystal structure for the following:
(a) a metal with a0=4.9489Å, r=1.75Å, and one atom per lattice point; and
(b) a metal with a0=0.42906nm, r=0.1858nm, and one atom per lattice point.
Solution:
We know the relationships between atomic radii and lattice parameters are in BCC and in FCC.
(a) 1.75 so its crystal structure is FCC;
(b) ==0.186nm so its crystal structure is BCC.

3. (P80 3-5) the density of potassium, which has the BCC structure and one atom per lattice point, is 0.855g/cm3. the atomic weight of potassium is 39.09g/mol. Calculate
(a) the lattice parameter; and
(b) the atomic radius of potassium.

Solution
(a) For a BCC unit cell, there are two atoms in per unit cell,
atomic mass is 39.09g/mol,
density ρ=0.855g/cm3
Avogadro’s number NA=6.02atoms/mol

0.855g/cm3=
So a==0.53=5.3Å

(b)then r==0.229cm=2.29Å

4. (P81 3-20) determine the indices for the directions in the cubic unit cell shown in Figure 3-32.

The procedure for finding the Miller indices for directions is as follows:
1. Using a right-handed coordinate system, determine the coordinates of two points, which lie on the direction.
2. Subtract the coordinates of the “tail” point from the coordinates of the “head” point to obtain the number of lattice parameters traveled in the direction of each axis of the coordinate system.

3. Clear fractions and/or reduce the results obtained from the subtraction to lowest integers.
4. Enclose the number in square brackets [ ]. If a negative sign is produced, represent the negative sign with a bar over the number.

Solution
Direction A
1. Two points are 0,0,1 and 1,0,0
2. 0,0,1-1,0,0=-1,0,1
3. no fraction to clear or integers to reduce
4.
Direction B
1. Two points are 1,0,1 and ,1,0
2. 1,0,1-,1,0=,-1,1
3. 2(,-1,1)=1,-2,2
4.

Direction C
1. Two points are 1,0,0 and 0,,1
2. 1,0,0-0,,1=1, -,-1
3. 4(1, -,-1)=4, -3, -4
4.

Direction D
1. Two points are 0,1, and 1,0,0
2. 0,1, -1,0,0=-1,1,
3. 2(-1,1, )=-2,2,1
4.
5. (P82 3-22) Determine the indices for the planes in the cubic unit cell shown in Figure 3-34.

The procedure for finding the Miller indices for planes is as follows:
1. Identify the points at which the plane intercepts the x, y, and z coordinates in terms of the number of lattice parameters. If the plane passes through the origin, the origin of the coordinate system must be moved!
2. Take reciprocals of these intercepts.
3. Clear fractions but not reduce to lowest integers.
4. Enclose the resulting numbers in parentheses (). Again, negative numbers should be written with a bar over the number.

Solution

Plane A
1. x=-1, y=, z=
2.
3. Clear fractions: -3, 6, 4
4. ()
Plane B
1. x=1, y=-, z=∞
2.
3. Clear fractions: 3, -4, 0
4.
Plane C
1. x=2, y=, z=1
2.
3. Clear fractions: 3, 4, 6
4. (346)

6. (P82 3-23) Sketch the following planes and directions within a cubic unit cell:
(a) [101] (b) [00] (c) [12] (d) [301] (e) [01] (f) [23]
(g) (0) (h) (102) (i) (002) (j) (10) (k) (12) (l) (3)

7. 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

8. 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/, packing factor =o.74
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
：
重量百分数为：

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）