Kittel,Charles IntroductionTo Solid State Physics 8Th Edition Solution Manual

Introduction to Solid State Physics

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C HAPTER 1 1. The vectors ˆˆˆ ++ x y z and −− + x y z are in the directions of two body diagonals of a cube. If θ is the angle between them, their scalar product gives cos θ = –1/3, whence . 1 cos 1/ 3 90 19 28' 109 28' θ= = °+ ° = ° 2. The plane (100) is normal to the x axis. It intercepts the a' axis at and the c' axis at ; therefore the indices referred to the primitive axes are (101). Similarly, the plane (001) will have indices (011) when referred to primitive axes. 2a' 2c' 3. The central dot of the four is at distance cos 60 a ctn 60 cos30 3 aa ° = ° from each of the other three dots, as projected onto the basal plane. If the (unprojected) dots are at the center of spheres in contact, then 2 2 2 ac a, 2 3 ⎛⎞ ⎛⎞ =+ ⎜⎟ ⎝⎠ or 22 21 c 8 a c ; 1.633. 34 a 3 == 1-1
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CHAPTER 2 1. The crystal plane with Miller indices hk is a plane defined by the points a A 1 /h, a 2 /k, and . (a) Two vectors that lie in the plane may be taken as a 3 / A a 1 /h – a 2 /k and 13 /h / A aa . But each of these vectors gives zero as its scalar product with 12 hk 3 = ++ A Gaa a , so that G must be perpendicular to the plane . (b) If is the unit normal to the plane, the interplanar spacing is hk A ˆ n 1 ˆ /h na . But , whence . (c) For a simple cubic lattice ˆ /| | = nGG 1 d(hk ) G / h| | 2 / | G| =⋅ A aG ˆˆˆ (2 /a)(h k ) = π+ + A Gx y z , whence 22 2 2 2 1G hk . d4 a == π A 12 3 11 3a a 0 2. (a) Cell volume 3a a 0 00 ⋅×= aa a c 2 1 3a c. 2 = 23 1 2 ˆˆ 41 1 (b) 2 3a a 0 || 2 2 3a c 21 ( ), and similarly for , . a 3 ×π = ⋅× π =+ x ˆ c y z b xy bb (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 33 31 3 3 C (a a ) (a a ) (a a ) ) ( 2 ) / | ( a a a ) | |(a a a ) | /V . BZ V( 2 (2 ) × ⋅××× 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
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2 2 1 2 2 1 2 1 exp[ iM(a k)] 1 exp[iM(a k)] |F| 1 exp[ i(a k)] 1 exp[i(a k)] sin M(a k) 1 cosM(a k) . 1 cos(a k) sin (a k) −− ⋅ ∆ − =⋅ −− ⋅∆ −⋅ == (b) The first zero in 1 sin M 2 ε occurs for ε = 2 π /M. That this is the correct consideration follows from 1 zero, as Mh is an integer 11 sin M( h ) sin Mh cos M cos Mh sin M . 22 ± π+ ε= π ε+ π ε ±²³ ²´ 1 2 5. j1 j2 j3 2i ( xv+ yv+ zv) 123 S (v v v ) f e j −π Referred to an fcc lattice, the basis of diamond is 111 000; . 444 Thus in the product S(v v v ) S(fcc lattice) S (basis) = × , we take the lattice structure factor from (48), and for the basis 1 i (v v v ). 2 S (basis) 1 e −π+ + =+ 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 v 1 + v 2 + v 3 = 4n, where n is an integer. For example, for the reflection (222) we have S(basis) = 1 + e –i3 π = 0, and this reflection is forbidden.
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Kittel,Charles IntroductionTo Solid State Physics 8Th Edition Solution Manual

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