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MMAT 201-02 Materials Science and Engineering I Spring 2006 Tue-Thu 3:30 pm - 4:45 pm, BRON 124

Solutions to Homework 2 1. (a) In the figure below is shown a [110] direction within a BCC unit cell.

For this [110] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalence of 1 atom that is centered on the direction vector. The length of thisdirection vector is denoted by x in this figure, which is equal to

x= z 2 − y2
4R 3

where y is the unit cell edge length, which, from Equation (3.3) is equal to

.

Furthermore, z is the length of the unit cell diagonal, which is equal to 4R Thus, using the above equation, the length x may be calculate as follows:
2 ⎛ 4R ⎞ ⎟ = 32R = 4R 2 x = (4 R ) − ⎜ ⎜ 3⎟ 3 3 ⎝ ⎠ 2 2

Therefore, theexpression for the linear density of this direction is

LD110 =

number of atoms centered on [110] direction vector length of [110] direction vector 1atom 4R 2 3 3 4R 2

=

=

A BCC unit cell within which is drawn a [111] direction is shown below.

For although the [111] direction vector shown passes through the centers of three atoms, there is an equivalence of only two atoms associatedwith this unit cell—onehalf of each of the two atoms at the end of the vector, in addition to the center atom belongs entirely to the unit cell. Furthermore, the length of the vector shown is equal to 4R, since all of the atoms whose centers the vector passes through touch one another. Therefore, the linear density is equal to number of atoms centered on [111] direction vector length of [111]direction vector = 2 atoms 1 = 4R 2R

LD111 =

(b) From the table inside the front cover, the atomic radius for iron is 0.124 nm. Therefore, the linear density for the [110] direction is
3 3

LD110 ( Fe) =

4 R 2 (4)(0.124 nm) 2

=

= 2.47 nm −1 = 2.47 x10 −9 m −1

While for the [111] direction 1 1 = = 4.03 nm −1 = 4.03 x10 −9 m −1 2 R (2)(0.124 nm)

LD111 ( Fe) =

2. (a) In thefigure below is shown a (100) plane for an FCC unit cell.

For this (100) plane there is one atom at each of the four cube corners, each of which is shared with four adjacent unit cells, while the center atom lies entirely within the unit cell. Thus, there is the equivalence of 2 atoms associated with this FCC (100) plane. The planar section represented in the above figure is a square, wherein theside lengths are equal to the unit cell edge length, 2R 2 [Equation (3.1)]; and, thus,
2 2R the area of this square is just ( 2 ) = 8R2. Hence, the planar density for this

(100) plane is just number of atoms centered on (100) plane area of (100) plane
= 2 atoms 8R 2 = 1 4R 2

PD100 =

That portion of an FCC (111) plane contained within a unit cell is shown below.

There are six atomswhose centers lie on this plane, which are labeled A through F. One-sixth of each of atoms A, D, and F are associated with this plane (yielding an equivalence of one-half atom), with one-half of each of atoms B, C, and E (or an equivalence of one and one-half atoms) for a total equivalence of two atoms. Now, the area of the triangle shown in the above figure is equal to one-half of the product ofthe base length and the height, h. If we consider half of the triangle, then

(2 R) 2 + h 2 = (4 R) 2 which leads to h = 2R 3 . Thus, the area is equal to
4 R ( h) ( 4 R ) 2 R 3 = = 4R 2 3 2 2

Area=

(

)

And, thus, the planar density is

PD111 =

number of atoms centered on (111) plane area of (111) plane = 2 atoms 4R
2

3 2R 2 3

=

1

(b) From the table inside the frontcover, the atomic radius for aluminum is 0.143 nm. Therefore, the planar density for the (100) plane is

PD100 ( Al ) =

1 4R

= 2

1 4(0.143 nm)
2

=12.23 nm − 2 =1.223 x10 −17 m −2

While for the (111) plane PD111 ( Al ) = 1 2R
2

3 2 3 (0.143 nm)

=

1
2

=14.12 nm − 2 =1.412 x10 −17 m − 2

3.

(a) From the data given in the problem, and realizing that 36.12 = 2...