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    ch02 Solucionário Do Kittel

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    Eloise Rodrigues
    1. Miller indices (hkA) define crystal planes, with the interplanar spacing d given by 1/d=|G|/2π, where G is the reciprocal lattice vector. The volume of the primitive cell is a3 for simple cubic, and the volume of the first Brillouin zone is (2π)3/VC. 2. The primitive reciprocal lattice vectors bi are given by 2π(a2×a3)/|a1·(a2×a3)| etc. Six vectors form the first Brillouin zone. 3. The structure factor vanishes for certain reflections in diamond due to interference between the face-centered cubic lattice and basis structure factors

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    ch02 Solucionário Do Kittel

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    Eloise Rodrigues
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    1. Miller indices (hkA) define crystal planes, with the interplanar spacing d given by 1/d=|G|/2π, where G is the reciprocal lattice vector. The volume of the primitive cell is a3 for simple cubic, and the volume of the first Brillouin zone is (2π)3/VC. 2. The primitive reciprocal lattice vectors bi are given by 2π(a2×a3)/|a1·(a2×a3)| etc. Six vectors form the first Brillouin zone. 3. The structure factor vanishes for certain reflections in diamond due to interference between the face-centered cubic lattice and basis structure factors

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    ch02 solucionário do Kittel

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    1. Miller indices (hkA) define crystal planes, with the interplanar spacing d given by 1/d=|G|/2π, where G is the reciprocal lattice vector. The volume of the primitive cell is a3 for simple cubic, and the volume of the first Brillouin zone is (2π)3/VC. 2. The primitive reciprocal lattice vectors bi are given by 2π(a2×a3)/|a1·(a2×a3)| etc. Six vectors form the first Brillouin zone. 3. The structure factor vanishes for certain reflections in diamond due to interference between the face-centered cubic lattice and basis structure factors

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    51 views3 pages

    ch02 Solucionário Do Kittel

    Uploaded by

    Eloise Rodrigues
    1. Miller indices (hkA) define crystal planes, with the interplanar spacing d given by 1/d=|G|/2π, where G is the reciprocal lattice vector. The volume of the primitive cell is a3 for simple cubic, and the volume of the first Brillouin zone is (2π)3/VC. 2. The primitive reciprocal lattice vectors bi are given by 2π(a2×a3)/|a1·(a2×a3)| etc. Six vectors form the first Brillouin zone. 3. The structure factor vanishes for certain reflections in diamond due to interference between the face-centered cubic lattice and basis structure factors

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 3

    CHAPTER 2

    1. The crystal plane with Miller indices hkA is a plane defined by the points a1/h, a2/k, and a3 / A . (a)
    Two vectors that lie in the plane may be taken as a1/h – a2/k and a1 / h − a3 / A . But each of these vectors
    gives zero as its scalar product with G = ha1 + ka 2 + Aa3 , so that G must be perpendicular to the plane
    hkA . (b) If n̂ is the unit normal to the plane, the interplanar spacing is nˆ ⋅ a1/h . But nˆ = G / | G | ,
    whence d(hkA) = G ⋅ a1 / h|G| = 2π / | G| . (c) For a simple cubic lattice G = (2π / a)(hxˆ + kyˆ + Azˆ ) ,
    whence

    1 G 2 h 2 + k 2 + A2
    = = .
    d 2 4π 2 a2

    1 1
    3a a 0
    2 2
    1 1
    2. (a) Cell volume a1 ⋅ a 2 × a3 = − 3a a 0
    2 2
    0 0 c

    1
    = 3 a 2 c.
    2

    xˆ yˆ zˆ
    a 2 × a3 4π 1 1
    (b) b1 = 2π = − 3a a 0
    | a1 ⋅ a 2 × a3 | 2
    3a c 2 2
    0 0 c
    2π 1
    = ( xˆ + yˆ ), and similarly for b 2 , b3 .
    a 3

    (c) Six vectors in the reciprocal lattice are shown as solid lines. The broken
    lines are the perpendicular bisectors at the midpoints. The inscribed hexagon
    forms the first Brillouin Zone.

    3. By definition of the primitive reciprocal lattice vectors

    (a 2 × a 3 ) ⋅ (a 3 × a1 ) × (a1 × a 2 )
    VBZ = (2π)3 = (2π)3 / | (a1 ⋅ a 2 × a 3 ) |
    | (a1 ⋅ a 2 × a 3 ) |
    3

    = (2π)3 / VC .

    For the vector identity, see G. A. Korn and T. M. Korn, Mathematical handbook for scientists and
    engineers, McGraw-Hill, 1961, p. 147.

    4. (a) This follows by forming

    2-1
    1 − exp[−iM(a ⋅ ∆k)] 1 − exp[iM(a ⋅ ∆k)]
    |F|2 = ⋅
    1 − exp[−i(a ⋅ ∆k)] 1 − exp[i(a ⋅ ∆k)]
    1 − cos M(a ⋅ ∆k) sin 12 M(a ⋅ ∆k)
    2
    = = .
    1 − cos(a ⋅ ∆k) sin 2 12 (a ⋅ ∆k)

    1
    (b) The first zero in sin Mε occurs for ε = 2π/M. That this is the correct consideration follows from
    2

    1 1 1
    sin M(πh + ε) = sin πMh cos Mε + cos
     πMh

    sin Mε.
    2 
    2 2
    zero, ± 1
    as Mh is
    an integer

    −2πi(x j v1 +y j v 2 +z j v3 )
    5. S (v1 v 2 v 3 ) = f Σ e
    j

    1 1 1
    Referred to an fcc lattice, the basis of diamond is 000; . Thus in the product
    4 4 4

    S(v1v 2 v3 ) = S(fcc lattice) × S (basis) ,

    we take the lattice structure factor from (48), and for the basis

    1
    −i π (v1 + v 2 + v3 ).
    S (basis) = 1 + e 2

    Now S(fcc) = 0 only if all indices are even or all indices are odd. If all indices are even the structure factor
    of the basis vanishes unless v1 + v2 + v3 = 4n, where n is an integer. For example, for the reflection (222)
    we have S(basis) = 1 + e–i3π = 0, and this reflection is forbidden.

    ∞
    f G = ∫ 4πr 2 (πa 0 Gr)−1 sin Gr exp ( −2r a 0 ) dr
    3
    6.
    0

    = (4 G 3a 0 ) ∫ dx x sin x exp ( −2x Ga 0 )


    3

    = (4 G 3a 0 ) (4 Ga 0 ) (1 + r G 2 a 0 ) 2
    3 2

    16 (4 + G 2 a 0 ) 2 .
    2

    The integral is not difficult; it is given as Dwight 860.81. Observe that f = 1 for G = 0 and f ∝ 1/G4 for
    Ga 0 >> 1.

    1
    7. (a) The basis has one atom A at the origin and one atom B at a. The single Laue equation
    2
    a ⋅ ∆k = 2π × (integer) defines a set of parallel planes in Fourier space. Intersections with a sphere are
    a set of circles, so that the diffracted beams lie on a set of cones. (b) S(n) = fA + fB e–iπn. For n odd, S = fA –

    2-2
    fB; for n even, S = fA + fB. (c) If fA = fB the atoms diffract identically, as if the primitive translation vector
    1 1
    were a and the diffraction condition ( a ⋅ ∆k ) = 2π × (integer).
    2 2

    2-3

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