32.7 Some Important Crystal lographic Formu as 1. Zone and zone equation All Planes parellel to common axis u v w constitute a (crystal) zone, the [u v w] zone If (h k i) belongs to zone [u v wl, then hu++ lw=0 2. If hu-tlyrly=0, then [u v wl lies in(h n 3. For cubic crystals, [h()
§2.7 Some Important Crystallographic Formulas 1. Zone and zone equation All Planes parellel to common axis [u v w] constitute a (crystal) zone, the [u v w] zone. If (h k l) belongs to zone [u v w], then hu＋kv＋lw＝0 2. If hu＋kv＋lw＝0，then [u v w] lies in (h k l) 3. For cubic crystals, [h k l] ⊥(h k l)
4. The zone [u v w] contains two planes(h, k li) and (, k, l), then :v: (k1l2-k241):(4h2-l2h):(h1k2-h2k1) 5. The plane(hk l) belongs to two zones [u vI wIl and [u2 v2 w2l if h:k:l= (12-v21):(w2-w21):(u12-21)
4. The zone [u v w] contains two planes (h1 k1 l1 ) and (h2 k2 l2 ),then ( ):( ):( ) 1 2 2 1 1 2 2 1 1 2 2 1 k l −k l l h −l h h k −h k 5. The plane (h k l) belongs to two zones [u1 v1 w1 ] and [u2 v2 w2 ] if ( ):( ):( ) 1 2 2 1 1 2 2 1 1 2 2 1 v w −v w wu −w u u v −u v h: k :l = u : v :w=
6. The distance between adjacent plane interplanar distance) dhk)f(a, b,c a, B, y h,k,l a For cubic: d h2+k2+ 2 1 h k For orthor homb ic: b For hexagona crystals 1 4 h2+hk tk
6. The distance between adjacent plane ( interplanar distance ) d(hkl) ＝f(a,b,c α,β,γ h,k,l) For orthorhombic: 2 2 2 2 2 2 2 1 c l b k a h d = + + For cubic: 2 2 2 h k l a d + + = For hexagonal crystals: 2 2 2 2 2 2 3 1 4 c l a h hk k d + + + =
7. The length of [u v w LUmg =vua2+(vb)+(w2)2+2vwbccos a + 2uwac cos B+2uvabcos y For cubic: Lz 1m=ay=2+y2+v2 8. The angle op between(h, k li) and (h, k, l)
7. The length of [u v w] ( ) ( ) ( ) 2 cos 2 2 2 [ ] L u v w vwbc uvw = a + b + c + + 2uwac cos + 2uvabcos For cubic: 2 2 2 L[uvw] = a = u + v + w 8. The angle Ф between (h1 k1 l1 ) and (h2 k2 l2 )
For cubic h1h2+k1k2+l12 cos pp= 2 (h12+k12+l1)(h2+k2+2 For orthorhombic h,h,, kk2 l1l2 2 b COS(= h k )2+(4)2]×( h2\2,k2 b C
For orthorhombic: [( ) ( ) ( ) ] [( ) ( ) ( ) ] cos 1 2 1 2 1 2 2 2 2 2 2 2 2 1 2 2 1 2 2 1 2 c l b k a h c l b k a h c l l b k k a h h + + + + + + = For cubic: ( )( ) cos 2 2 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 2 h k l h k l h h k k l l + + + + + + =
For hexagonal: CoS pp 3.a h2h2+k1k2+ 12+(h1k2+h2k1) 2+2+3)2+2 3.a +k2+()22+h2k2 4 C 4 9. The angle o between [ui vI wil and [a2v2w2]
For hexagonal: cos = + + + + + + + + + + 2 2 2 2 2 2 2 2 1 1 2 2 1 2 2 1 2 1 1 2 1 2 2 1 2 1 2 1 2 2 ( ) 4 3 ( ) 4 3 ( ) 2 1 ( ) 4 3 l h k c a l h k h k c a h k l l h k h k c a h h k k 9. The angle between [u1 v1 w1 ] and [u2 v2 w2 ]
For cubic uiu,++W1W cos pp 2 1+V1+w 1V2+n2+1 For orthorhombic uia+v,vob+wwoc cos pp (4a)2+(vb)2+(c)2y( (l2a)2+(2b)2+(2c)
For orthorhombic: cos = 2 2 2 2 2 2 2 1 2 1 2 1 2 1 2 2 1 2 2 1 2 (u a) (v b) (w c) (u a) (v b) (w c) u u a v v b w w c + + + + + + For cubic: cos = 2 2 2 2 2 2 2 1 2 1 2 1 1 2 1 2 1 2 u v w u v w u u v v w w + + + + + +
For hexagonal: cos 9 l4l2+y1V2+1w2()2-(4V2+a21) a C 1+V1-+W uiV L2)+12+1 a
For hexagonal: 2 2 2 2 2 2 2 2 1 1 2 2 2 1 2 1 2 1 1 2 2 1 2 1 2 1 2 1 2 ( ) ( ) ( ) 2 1 ( ) u v a c u v u v w a c u v w u v u v a c u u v v w w + + − + + − + + − + cos =
10. The volume of unit cells y V=abcv1-cos a-cos y+2 cos a cos B cosy
10. The volume of unit cells V V = 1 cos cos cos 2cos cos cos 2 2 2 abc − − − +
32.8 Stacking Mode of Crystals I. A crystal can be cons i dered as the result of stack ing the atom ic l ayers, say (hk1), one over another in a specific sequence For simple cubic (001)aaa (110) abab shift =1101 along 1 10 This sequence is cal led the stacking order
§2.8 Stacking Mode of Crystals ◆ Ⅰ.A crystal can be considered as the result of stacking the atomic layers, say (h k l ), one over another in a specific sequence. For simple cubic (001) aaaa…… (110) abab…… [110], along [110] 2 1 shift = This sequence is called the stacking order