The total scattering factor is given as S A N 2 B N 2 Performing the summation

The total scattering factor is given as s a n 2 b n 2

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The total scattering factor is given as S = | A ( N ) | 2 · B ( N ) | 2 . Performing the summation of the geometric series we get | A ( N ) | 2 = sin [( π N / 2 ) sin θ ] sin [( π / 2 ) sin θ ] 2 , | B ( N ) | 2 = sin ( π N cos θ / 2 ) sin ( π cos θ / 2 ) 2 . Substituting these expression and assuming the Bragg condition ka = π we obtain the graphs shown in Fig. 12.6. Similar curves for ka = 3 π / 2 are shown in Fig. 12.7. Figure 3.5: Plots of S ( N , θ ) / N 4 for ka = π ; N = 5 (red curve), 10 (blue curve), and 20 (green curve) Problem 3.4: Consider a random 1-dimensional chain along a straight line, the probability of finding the nearest neighbor at distance x being P ( x ) = 1 2 πσ e - ( x - a ) 2 / 2 σ . (3.1) Assuming that σ ¿ a 2 and the atoms are point-like find the density distribution function, g ( x ) . Plot g ( x ) for σ = 0 . 1 and σ = 0 . 3 and discuss the results. Hints: (i) Recall the rules of calculating probability of a sum of random quantities; (ii) Have in mind that the Fourier transform of the Gaussian probability is P ( k ) = e - k 2 σ / 2 . transform
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36 CHAPTER 3. FRACTALS AND DISORDERED SYSTEMS Figure 3.6: Plots of S ( N , θ ) / N 4 for ka = 3 π / 2; N = 5 (red curve), 10 (blue curve), and 20 (green curve) Solution 3.4: Let us define the distance between the initial and n -th atom as x = na + U n , n j = 1 u j . The probability of the the sum of of displacement are given by the general prescription W ( U n ) = n j = 1 Z du j P ( u j ) δ ˆ U n - n j = 1 u j ! = j Z dk j 2 π P ( k j ) Z du j e ik j u j Z dk 2 π e ik ( U n - u j ) = Z dk 2 π e ikU n [ P ( k )] n = Z dk 2 π e ikU n e - k 2 σ n / 2 . Here P ( k ) is Fourier transform of P ( u ) . We have also used the hints and the propeety Z due i ( k 1 - k 2 ) = 2 πδ ( k 1 - k 2 ) . Now we have to find the distribution g ( x ) , which is just g ( x ) = Z du 0 P ( u 0 ) W ( x - u 0 ) = Z dk 2 π P ( k ) W ( k ) .
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37 From this we get, g n ( x ) = Z dk 2 π e - kx e - k 2 σ ( n + 1 ) / 2 = 1 p 2 πσ ( n + 1 ) e - ( x - na ) 2 / 2 σ ( n + 1 ) . Since for small dispersion σ the Gaussian spikes do not overlap, the final answer can be expressed as g ( x ) = n = 1 1 p 2 πσ ( n + 1 ) exp - ( x - na ) 2 2 σ ( n + 1 ) . The plots are shown in Fig. 3.7 Figure 3.7: Plot of the distribution function for σ = 0 . 1 (red curve) and σ = 0 . 03 (green curve).
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38 CHAPTER 3. FRACTALS AND DISORDERED SYSTEMS
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Chapter 4 Defects and Diffusion Quick access: 1 2 3 4 5 Problem 4.1: Austenite: Austenite is a type of steel with less than 1% carbon. Call this concentration C 0 . It is desirable to increase the carbon content near the surface to give the steel a harder surface. This can be achieved by keeping the Austenite in a mixture of CO and CO 2 where the reaction 2CO CO 2 + C aust will produce a given concentration C s on the surface. This concentration will then diffuse into the steel. (a) Show that the expression C ( x , t ) = C s - ( C s - C 0 ) erf x 2 Dt (4.1) satisfies the diffusion equation C t = D 2 C x 2 , (4.2) where erf ( y ) 2 π R y 0 e - t 2 dt . Sketch the function C ( x , t ) at different times.
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