HW-4-solution - Solution of HW #4 PDF

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Homework #4 1. It is often convenient to represent a fcc lattice as simple cubic, with a cubic primitive cell of side a and a four-point basis. (a) Show that the structural factor is then either 4 or 0 at all points of the simple cubic reciprocal lattice. (b) Show that when points with zero structural factor are removed, the remaining points of the reciprocal lattice makes up a bcc cubic lattice with conventional cell of side 4π/a. Why is this expected? Solution: (a) r r r a1 = axˆ, a2 = aˆy, a3 = aˆz. r 2π r 2π r 2π ˆz. b1 = xˆ, b2 = yˆ , b3 = a a a There are four lattice sites within each cubic cell. r r r r a a a R1 = 0, R2 = ( xˆ + yˆ ), R3 = ( yˆ + zˆ), R4 = ( zˆ + xˆ ) . 2 2 2 4 r r The structure factor is S = ∑ exp(iG ⋅ R i ) i =1

r r r r 2π with G = hb1 + kb2 + lb3 = ( hxˆ + kˆy + lˆz) . a r r G ⋅ R1 = 0 r r G ⋅ R2 = π (h + k ) r r G ⋅ R3 = π ( k + l ) r r G ⋅ R4 = π ( l + h) 4 r r = S ∑ exp( iG ⋅ R i) = 1 + exp[ i π( h + k )] + exp[ i π ( k + l)] + exp[ iπ ( l + h)] . i= 1

S = 4 when h, k , l are all even or all odd. S = 0 otherwise.

r r r r 2π ( hxˆ + kyˆ + lzˆ ) with all even or all (b) Keep only those sites of G = hb1 + kb2 + lb3 = a odd h, k , l , such as (000), (200), (111), (002), …, plot the reciprocal lattice.

z

(022)

(002)

(202) (111)

(222) (020)

(000)

(200)

y

(220) 2π a

x

This is a bcc lattice with a lattice constant of 4 π / a . This is expected because the reciprocal lattice of a fcc should be a bcc lattice. 2. Structure factor of diamond. The crystal structure of diamond is described in Chapter 1. The basis consists of eight atoms if the cell is taken as the conventional cube. (a) Find the structural factor S of this basis. (b). Find the zeros of S and show that the allowed reflections of the diamond structure satisfy v 1 +v 2 + v 3 = 4 n , where all indices are even and n is any integer, or else all indices are odd (Notice that h, k, l may be written for v1 , v2 , v3 and this is often done.) Solution: 8 r r (a) S = ∑exp( iG ⋅ Ri ) i=1 r r r r r r r r R1 , R2 , R3 , R4 are the four atoms in a fcc cubic cell. R5 , R6 , R7 , R8 are the four atoms r in another fcc cubic cell which is shifted by R0 = ( xˆ + yˆ + zˆ) a / 4 . Thus r r r r r r r r r r r r R5 = R1 + R0 , R6 = R2 + R0 , R7 = R3 + R0 , and R8 = R4 + R0 . 4 8 4 r r r r r r r S = ∑ exp(iG ⋅ Ri ) = ∑ exp( iG ⋅ Ri ) + ∑ exp[ iG ⋅ ( Ri + R0 )] i =1

i =1

i =1

r r 4 r r = [1 + exp( iG ⋅ R0)] ∑exp( iG ⋅ Ri ) i= 1

r r The ∑exp(iG ⋅ Ri ) is the fcc structure factor. 4

i =1

r r r 2π r r r For G = hb1 + kb2 + lb3 = (hxˆ + kyˆ + lzˆ ) , G ⋅ R0 = π (h + k + l ) / 4 . a S = {1 + exp[ π (h + k + l ) / 2)]}{1 + exp[i π (h + k )] + exp[i π (k + l )] + exp[i π (l + h )]} (b) The zero of S occurs when h, k , l are not all even or odd due to the fcc h + k + l = 4n + 2 (n=integer) that gives structure factor, or when 1 + exp[ π (h + k + l ) / 2) = 0 . When h + k + l = 4 n and all h, k , l are even, S = 8 . When h + k + l = 4n ± 1 and all h , k , l are odd, S = 4(1 ± i ) .

3. Diatomic line. Consider a line of atoms ABAB…..AB, with A-B bond length of a/2. The form factors are fA and fB for atoms A, B, respectively. The incident beam of xrays is perpendicular to the line of atoms. (a) Show that the interference condition is nλ = a cosθ , where θ is the angle between the diffraction beam and the line of atoms. (b) Show that the intensity of the diffracted beam is proportional to | f A − f B |2 for n odd, and to | f A + f B |2 for n even. (c) Explain what happens if f A = fB . Solution:

l

a

A

B

A

B

A

B

A

B

A

B

r 2 πn The reciprocal lattice vector is G = xˆ with n being integers. a r r a Within a cell, R A = 0, RB = xˆ . Then the structure factor is 2 r r ⎧ f − fB n = odd S = f A + f B exp(iG ⋅ R B ) = f A + f B exp(inπ ) = ⎨ A . ⎩ f A + f B n = even (a) The path difference is l = a cosθ so the interference condition is a cos θ = λ . ⎧| f − f B | 2 n = odd (b) I ∝| S | 2 = ⎨ A . 2 ⎩| f A + f B | n = even

(c) If f A = f B , I =0 when n is odd. In this case, the lattice is a monatomic lattice with lattice constant of a/2. The reciprocal lattice is 2π m /( a / 2) = ( 2π / a )2m . Thus only n = 2m = even gives the diffraction.

4. Powder specimens of three different monoatomic cubic crystals are analyzed with a Debye-Scherrer camera. It is known that one sample is fcc, one is bcc, and one has the diamond structure. The approximate positions of the first four diffraction rings in each case are: A B C 42.2o 28.8o 42.8o 49.2o 41.0o 73.2o o o 72.2 50.8 89.0o 87.3o 59.6o 115.0o (a) Identify the crystal structures of A, B, and C. (b) If the wavelength of the incident x-ray beam is 1.5Å, what is the length of the side of the conventional cubic cell in each case?

φ Sample

Film

Solution: The Bragg law gives 2d sin θ = λ . For this problem, θ = φ / 2 . r λ λ |G| λ = = sin( φ / 2) = h2 + k 2 + l2 . × d 2 2 2π 2a For fcc lattice, h, k , l need to be all even or all odd. The first four diffraction peaks then correspond to (111), (200), (220), and (311) which give sin( φ1 / 2) =

λ

3 , sin(φ2 / 2) =

λ

2 , sin(φ3 / 2) =

λ

2 2 , sin(φ4 / 2) =

λ

11 . 2a 2a 2a 2a sin(φ 2 / 2) 2 sin(φ 3 / 2) 2 2 sin( φ4 / 2) 11 = = 1.155, = = 1.633, = = 1.915. or sin(φ 1 / 2) sin( φ1 / 2) 3 sin(φ1 / 2) 3 3 For bcc lattice, h + k + l needs to be even. The first four diffraction peaks then correspond to (110), (200), (112), and (220) which give sin( φ1 / 2) = or

λ

2a

2 , sin(φ2 / 2) =

λ

2a

2 , sin(φ 3 / 2) =

sin(φ 2 / 2) 2 sin(φ3 / 2) = = 1.414, = sin(φ 1 / 2) 2 sin(φ1 / 2)

λ

2a

6, sin(φ4 / 2) =

λ

2a

8.

6 sin(φ 4 / 2) 8 = 1.732, = = 2. 2 sin(φ 1 / 2) 2

For diamond structure, h, k , l need to be all even or all odd plus 1 + exp[ π (h + k + l ) / 2) ≠ 0 . The first four diffraction peaks then correspond to (110), (220), (311), (400) which give sin( φ1 / 2) =

λ

3 , sin(φ2 / 2) =

λ

8 , sin(φ3 / 2) =

2a 2a sin(φ 2 / 2) 8 sin(φ 3 / 2) = = 1.633, = or sin(φ 1 / 2) 3 sin(φ1 / 2)

λ

2a

11 , sin(φ 4 / 2) =

λ

2a

4.

11 sin( φ4 / 2) 4 = 1.915, = = 2.309. 3 sin(φ1 / 2) 3

A

sin (φi / 2 )

42.20 49.20 72.20 87.30

0.3616 0.4163 0.5892 0.6903

B

sin (φi / 2 )

28.80 41.00 50.80 59.60

0.2487 0.3502 0.4289 0.4970

C

sin (φ i / 2 )

42.80 73.20 89.00 115.0 0

0.3616 0.4163 0.5892 0.6903

sin(φi / 2) sin(φ1 / 2) 1.15 1.63 1.91 sin(φi / 2) sin(φ1 / 2) 1.41 1.72 2.00 sin(φi / 2) sin(φ1 / 2) 1.63 1.92 2.31

Comparing to the calculated ratio, it is easy to see A is fcc, B is bcc, and C has diamond structure. (b) For A, sin( φ1 / 2) = For B, sin( φ1 / 2) = For C, sin( φ1 / 2) =

λ 2a

λ 2a

λ 2a

3. a =

0 3λ 1.5 3 = = 3.59 A . 2 sin(φ 1 / 2) 2 × 0.3616

2. a =

0 2λ 1. 5 2 = = 4.26 A 2 sin(φ1 / 2) 2 × 0.2487

3. a =

0 3λ 1.5 3 = = 3.56 A 2 sin(φ 1 / 2) 2 × 0.3649...