Question

Calculate (a) LD for FCC Al and (b) for BCC Cr

Problem 4 on Linear density of atoms (LD) LD along a certain direction vector of a crystal is defined as LDZ= number of atoms centered along the x direction vector length of the x direction vector For example, along 100 direction of a cubic crystal Linear density is defined as LD 100 = 1 number of atoms centered on [100] direction vector length of [100] direction vector Similarly LD number of atoms centered on [111] direction vector = length of [111] direction vector = LD 111 = If R is given then it is a simple matter of plugging in that number. If it is not given and only "a" (lattice parameter) is available then you can calculate R from the lattice parameter. HW #3. Calculate (a) LD for FCC AI and (b) for BCC Cr

Answer 1

Solution 7 LD for FCC Al Let dosection (110) for Al R = 0.1431 nm LD = Number of atoms centered on direction Vector dosection 1ło) x= R +2RAR x=4R length of dosetion Vector Lou 2 saling - batera 20:10 = FR ER LP210 = 2 x 0 1431 km Chertholl 0440 = 3-5 must alfiore 2 = 3.494060098 nm-1 2x0.1431 nm

Answer b-

Dissection (011) H Solution ? Cr> B C c unit cell R= 0.1249 nm , a Linear density=_ Number of atoms Centered, on closection vector length of direction Vector L Doze = 1 atans - * Eat +* = 48) *( * = 29319 But in 1 Qurum . Lu trening the I Dony = 1 73_ 42 R 13 42 x 01249 nm v2 LP 11 = 2:45145 1. nm-2 Answer #LP 11 = 2.5 nm - 1 | Approx

I have answered for two different directions. For FCC Al (110) and for BCC Cr (011).

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