2.6 (a) Derive linear density expressions for FCC [100) and[111] directions in terms of the atomic…
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Question “2.6 (a) Derive linear density expressions for FCC [100) and[111] directions in terms of the atomic…”
Answer
Part a
i)
FCC Linear density expression [100]
Diagram of FCC [100]
Linear density
= (number of atoms centered in direction 100) /(length of vector).
= (number of atoms centered in direction 100) /(unit cell edge length).
The diagram is shown below
There is 1 atom in each corner of two unit cells.
Number of atoms centered in direction 100 = 1
Unit cell edge length = 2R(2) , 0.5
Linear density = 1/[2R(2) ,0.5]
ii)
FCC Linear density expression [111]
Diagram of FCC [111]
Linear density
= (number of atoms centered in direction 111) / (length vector).
At each end, vectors pass through the center of a single atom.
Number of atoms centered in direction 111 = 1
length of vector z = (x2 + y2)0.5
Length of bottom diagonal x = 4R
Unit cell edge length, y = 2R(2) 0.55
z = [(4R)2 + 2R(2)0.52 ]0.5
z = 2R(6)0.5
Linear density = 1 [2R(6) , 0.5]
Part b
i)
Planar density expression for FCC [100]
Diagram of FCC [100]
Planar density
= (number of atoms centered in direction 100) /(area of the plane)
Each corner has one atom, and each corner atom can be shared with unit cells. The middle atom remains in the unit cell.
Number of atoms centered in direction 100 = 2
Length of the unit cell edge = 2R(2) and 0.5
A plane is a square
Area = [2R(2) 0.50] =
Planar density = 2 / 8R2 = 1/(4R2)
ii)
Planar density expression for FCC [111]
Diagram of FCC [111]
Planar density
= (number of atoms centered in direction 111) / (area)
6 atoms are found on the plane: A, B C, D D, E, F
1/6th atom equivalent to 1/2 atom in A, D, F
1/2 atom B, C, or E equals 3/2 atom
Number of atoms that are centered in direction 111 = 1/2 + 1/3
= 2 Atoms
Area of triangle = 1/2x base length x height
(2R)2 + h2 = (4R)2
h = 2R(3) 0.5
Area = 1/2x (4R), x h = 1/2x (4R), x 2R(3)
= 4R2(3)0.5
Planar density = 2 / [4R2(3)0.5]
= 1/ [2R2(3)0.5]
Conclusion
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