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dit View History Bookmarks People Window Help Problem 3.77 (Go Multipart) Secure https://edugen.wileyplus.com/edugen/sharedwassignment/test/aprint.uni Print by: Karol Moises spring17-merged m221-01-02 wang: Spring Merged EGM221 01-02 Engineering Materials Wang Homew \"Problem 3.77 (Go Multipart) Part 1 First, review section 3.2 and Figure 3.1. Then, review section 3.10 and Figure 3.25. If you are having difficulty visualizing this problem, you may wish to review the following: Review of 2D and 3D Geometry solving right triangle (a) Derive planar density expressions for Fcc (100) and (111) in terms of the atomic radius R. (b) Compute and compare linear density values for these same two planes for aluminum. Click here to access the Characteristics of Metals table Consider the (100) plane in Fcc: How many atoms are centered on the 100] in FCC? exact number, no tolerance Microsoft Excel Part 2 What is the correct PD expression for the (100) in Fcc?

Solution

In (100) plane one atom at each of the four cube corners and each of that is shared with four adjacent

unit cells, while the center atom lies entirely within the unit cell. (Figure in part-1)

The planar section represented in the above figure is a square, wherein thes side lengths are equal to the unit cell edge length, 2R^2 and the area of this square is

(2R 2)^2 = 8R^2

. Hence, the planar density for this (100) plane is just

PD100 = number of atoms centered on (100) plane / area of (100) plane

= 2 atoms / 8*R^2 = 1 / 4R^2 (Ans part -2)

(Figure in part- 3)

In FCC (111) plane a unit cell contain six atoms whose centers lie on this plane, 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 of the base length and the height, h. We consider half of the triangle, then

(2R)^2+ h^2= (4R)^2

which leads to h = 2R 3 .

Thus, the area is equal to

Area = 4R(h) / 2 = (4R)(2R^2 3) / 2 = 4R^2 3 (Ans part-3)

And, thus, the planar density is

PD111 = number of atoms centered on (111) plane / area of (111) plane

= 2 atoms / 4R^2 3 = 1 / 2R^2 3 (Ans part-4)

If we took r = 0.143 nm for aluminium

PD100 (Al) = 1 / 4*R^2 = 1 / 4 (0.143nm)^2 = 12.23 nm^-2 (Ans part -5)

LD100(Al) = 1/2R2 = 1/2*0.143*2 = 2.47 nm^-1

PD111(Al) = 1/2R² 3= 1/2 3 (0.143 nm)² = 14.11 nm^-2 (Ans part -6)

LD111(Al) = 1/2R6 = 1/2*0.143*6 = 1.427 nm^-1