Problem4.) - Home | University of Colorado Boulder · Kittel 2.6 ) € n(r)= e−2r/a 0 πa ...
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Solutions for Homework 5 1. Kittel 2.4 a) | F 2 | = 1 − e [ − iM ( a⋅Δk )] 1 − e [ −i( a⋅Δk )] 1 − e [ iM ( a⋅Δk )] 1 − e [ i( a⋅Δk )] Expanding the parentheses and using: cos( x ) = (e ix + e − ix )/2 We get: | F 2 | = 2 − 2cos( Ma⋅Δk ) 2 − 2cos( a⋅Δk ) = sin 2 ( Ma⋅Δk ) sin 2 ( a⋅Δk ) b) For F 2 to be zero the numerator must be zero. sin[(M(2πh+ε)/2]=0 Using trig. Identities: sin[(M(2πh+ε)/2]=sin(Mπh)cos(Mε/2)+cos(Mπh)sin(Mε/2)= sin(Mε/2)=0 Therefore Mε/2=nπ, and for the first zero n=1 and ε=2π/M. 2. Kittel 2.5 a) The basis points are: (0,0,0), a/2(1,1,0). a/2(1,0,1), a/2(0,1,1), a/4(1,1,1), a/4(1,1,1),a/4(1,1,1),a/4(1,1,1). We write G=(n1b1+n2b2 +n3b3) The structure factor is then given by: S G = 1 + e iπ ( n 1 + n 2 ) + e iπ ( n 1 + n 3 ) + e iπ ( n 3 + n 2 ) + e iπ ( n 1 + n 2 + n 3 )/2 + e iπ ( − n 1 − n 2 + n 3 )/2 + e iπ ( n 1 − n 2 − n 3 )/2 + e iπ ( −n 1 + n 2 − n 3 )/2 ) b) Let n= n1+n2 +n3 . Then we can write SG as: S G = (1 + e iπn /2 )(e iπ ( n 1 + n 3 ) + e iπ ( n 3 + n 2 ) + e iπ ( n 1 + n 2 ) ) Then the zeros are n=4m+2 with m –integer coming from the first factor. The second factor gives zeroes when one of the n1,n2 ,n3 is odd AND the others are even or one of the n1,n2 ,n3 is even AND the others are odd. So for reflections with the nonzero structure factor 1)n must be odd or 2)n=4m with all even n1,n2 ,n3 3. Kittel 2.6
Transcript of Problem4.) - Home | University of Colorado Boulder · Kittel 2.6 ) € n(r)= e−2r/a 0 πa ...
Solutions for Homework 5
1. Kittel 2.4
a)
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|F 2 |= 1− e[− iM (a⋅Δk )]
1− e[−i(a⋅Δk )]⎛
⎝ ⎜
⎞
⎠ ⎟ 1− e[iM (a⋅Δk )]
1− e[i(a⋅Δk )]⎛
⎝ ⎜
⎞
⎠ ⎟
Expanding the parentheses and using:
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cos(x) = (eix + e− ix ) /2
We get:
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|F 2 |= 2 − 2cos(Ma⋅ Δk)2 − 2cos(a⋅ Δk)
=sin2(Ma⋅ Δk)sin2(a⋅ Δk)
b) For F2 to be zero the numerator must be zero. sin[(M(2πh+ε)/2]=0 Using trig. Identities: sin[(M(2πh+ε)/2]=sin(Mπh)cos(Mε/2)+cos(Mπh)sin(Mε/2)= sin(Mε/2)=0 Therefore Mε/2=nπ, and for the first zero n=1 and ε=2π/M.
2. Kittel 2.5
a) The basis points are: (0,0,0), a/2(1,1,0). a/2(1,0,1), a/2(0,1,1), a/4(1,1,1), a/4(-‐1,1,-‐1),a/4(-‐1,-‐1,1),a/4(1,-‐1,-‐1).
We write G=(n1b1+ n2b2 + n3b3) The structure factor is then given by:
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SG =1+ eiπ (n1 +n2 ) + eiπ (n1 +n3 ) + eiπ (n3 +n2 ) +
eiπ (n1 +n2 +n3 ) / 2 + eiπ (−n1 −n2 +n3 ) / 2 + eiπ (n1 −n2 −n3 ) / 2 + eiπ (−n1 +n2 −n3 ) / 2)
b) Let n= n1+ n2 + n3 . Then we can write SG as:
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SG = (1+ eiπn / 2)(eiπ (n1 +n3 ) + eiπ (n3 +n2 ) + eiπ (n1 +n2 )) Then the zeros are n=4m+2 with m –integer coming from the first factor. The second factor gives zeroes when one of the n1, n2 , n3 is odd AND the others are even or one of the n1, n2 , n3 is even AND the others are odd. So for reflections with the nonzero structure factor 1)n must be odd or 2)n=4m with all even n1, n2 , n3
3. Kittel 2.6
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n(r) =e−2r / a0
πa03
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The form factor is:
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f = 4π drn(r) sin(Gr)Gr0
∞
∫ ;
Looking this up in an integral table (say in Wikipedia)
gives:
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f =16
(4 +G2a02)2
Problem 4.
a) 1st: 2acos(30º)=
2nd: -‐ acos(30º)+ (2a-‐asin(30º))=
-‐ /2+ 1.5a
3rd: 2c
b) =-2 c( cos(30º)+ cos(30º))=-2 c( +0.5 )=- c( +)
=-2 c
=1.5
-3ac =-33/2a2c
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b1 =2π(a2 × a3)a1⋅ (a2 × a3)
=
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b2 =2π(a1 × a3)a1⋅ (a2 × a3)
= =
= =
c) SHKL=
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eigHKL ⋅riunitcell
∑ , where gHKL are reciprocal lattice vectors, ri are coordinates of ith
atom in the unit cell. For graphite we have 4 atoms in the unit cell. The coordinates are:
r1=(0,0,0), r2=(-1/3,2/3,0), r3=(-1/3,2/3,0.5), r4=(-2/3,4/3,0.5) in the coordinates defined by normalized unit vectors.
r1=(0,0,0), r2=(0,a,0), r3=(0,a,c), r4=(0,2a,c) in the coordinates defined by x,y,z. (Note that the top layer is simply shifted by a in the y-direction.
SHKL=1+exp(i((2π/3)H+(4π/3)K))+ exp(i((2π/3)H+(4π/3)K+πL))+ exp(i((4π/3)H+(8π/3)K+πL))=
=1+exp(i((2π/3)H-(2π/3)K))+ exp(i((2π/3)H-(2π/3)K+πL))+ exp(i((-2π/3)H+(2π/3)K+πL))=
= 1+exp(i((2π/3)(H-K))+(-1)L2cos ((2π/3)(H-K)).
or:
SHKL=1+exp(i((2π/3)H+(4π/3)K))+ exp(i((2π/3)H+(4π/3)K+πL))+ exp(i((4π/3)H+(8π/3)K+πL))=
=1+exp(i((2π/3)H+(4π/3)K))+(1+exp(i((2π/3)H+(4π/3)K)))exp exp(i((2π/3)H+(4π/3)K)+πL)=
(1+exp(i((2π/3)H+(4π/3)K)))(1+exp exp(i((2π/3)H+(4π/3)K)+πL))= (1+exp(i((2π/3)(H-K))(1+exp(i((2π/3)(H-K)+πL))=
(1+exp(i((2π/3)(H-K))(1+(-1)Lexp(i((2π/3)(H-K)).
d) S10L=1+exp(i((2π/3)H)+( -1)L2cos ((2π/3)H).
