e To observe the rings on the film we need φ π 2 so � 2 a p 2 q 2 r 2 1 2 sin π

# E to observe the rings on the film we need φ π 2 so

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(e) To observe the rings on the film we need φ < π 2 , so λ 2 a ( p 2 + q 2 + r 2 ) 1 / 2 < sin π 4 = 2 - 1 / 2 . For a = 2 . 8 × 10 - 10 and λ = 0 . 7 × 10 - 10 , we obtain the condition p 2 + q 2 + r 2 < 32 . (12.13) The permutations of p , q , and r lead to the same difraction rings, so they must be excluded. In addition, the sum p + q + r must be an even number. The inspection shows that the number of relevant combinations is 16. (f) The primitive vectors of the tetragonal structure are t 1 = a 2 [ e 1 + e 2 - ( 1 + ε ) e 3 ] , t 2 = a 2 [ e 2 +( 1 + ε ) e 3 - e 1 ] , and t 3 = a 2 [ e 1 +( 1 + ε ) e 3 - e 2 ] . (12.14) 87 The volume of the primitive cell is V t = t 1 · t 2 × t 3 = ( 1 + ε ) a 3 2 , so the fundamental vectors of the reciprocal lattice are ˜ t 1 = 2 π a ( e 1 + e 2 ) (12.15) ˜ t 2 = 2 π a e 2 + e 3 1 + ε (12.16) ˜ t 3 = 2 π a e 1 + e 3 1 + ε . (12.17) So a general vector of the reciprocal lattice is G t = k ˜ t 1 + l ˜ t 2 + m ˜ t 3 = 2 π a ( k + m ) e 1 +( k + l ) e 2 +( l + m ) e 3 1 + ε 2 π a p e 1 + q e 2 + r e 3 1 + ε . Since G = 2 π a p p 2 + q 2 + l 2 ( 1 + ε ) - 2 , the maximum interference condition reads: 2 π a s p 2 + q 2 + r 2 ( 1 + ε ) 2 = 4 π λ sin φ 2 . In the case of the cubic lattice any permutation of the integers p , q , and r would lead to the formation of the same cone of maximum interference, while in the case of tetragonal lat- tice, the interchange of p and q leaves the condition of maximum interference unchanged, while the interchange of r with any of p or q leads to a different angle of maximum inter- ference. Beside this, the degeneracy is lifted for any other combination of p , q , and r and p 0 , q 0 , and r 0 , so that p 2 + q 2 + l 2 = ( p 0 ) 2 +( q 0 ) 2 +( l 0 ) 2 , while l 6 = l 0 . Problem 12.2: (a) The von Koch curve is constructed by dividing each side of equilateral triangle by 3 and replacing the central part by another equilateral triangle, see Fig. 12.4. Find fractal dimen- sion of the von Koch curve. (b) Consider a square lattice N × N of atoms with isotropic scattering form-factor. 88 CHAPTER 12. TESTS Figure 12.4: 1. Assume that a plane wave has a front parallel to the crystal plane, so the incident angle α = 0. Calculate the scattering intensity I as a function of the scattering angle, θ , for ka = π (Bragg conditions) and for N = 5, 10, and 20. Plot the dependence in polar coordinates. Show that the maximal scattering intensity is proportional to the total number of atoms squared. 2. Repeat the calculation for ka = 1 . 5 π and α = 0. Discuss the difference between the scattering curves. 3. Find the incident angle satisfying the Bragg conditions at ka = 1 . 5 π and repeat cal- culations of the item 1. How the Bragg condition manifest itself? Solution 12.2: (a) We have the recursive formula for the perimeter P ( ‘/ 3 ) = 4 3 P ( ) . Assuming P ( ) - ( D - 1 ) we get D = log4 log3 1 . 26 . This value is close to experimental one (1.3) for Norwegian coast. (b) For an incident wave with α = 0, the difference of the optical path corresponding to the adjacent atom to the right is - a sin θ . 89 Figure 12.5: Sketch of the angles.  #### You've reached the end of your free preview.

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