Question

. What is the condition for systematic absences in Face Centred Cubic Packing?

1. a) h k l either all odd or all even

2. b) h k l all odd

3. c) k k l all even

4. d) h+k+l even

5. e) h+k+l odd

can you explain to me the answer please

Answer 1

Atom at (0,0,0) & (2, V2, 0) and equivalent positions Face Centred Cubic {[27(hx', +k y',+1 z;]| , %, 0), (2, 0, %2), (0, %2, Y) |F = f, h+k i[2T( 2 k+1. i[2T( 2 1+h i[2T 2 F = feT27(0)] +e +e +e 022 Real 002 iz(1+h) ir(k+l) +e іл(h+k) fll+ +e 202, 222 ССР ("FCC сrystal") becomes "BCC crystal" in reciprocal space 111 020 000 in(k+) +e in (l+h) - F f[1ei(h+k) 220 +e 200 |F = 4f F2 16f-| (111), (200), (220), (333), (420) (h, k, 1) unmixed h,k,lall even or all odd F = 0 F 2 o (100), (211); (210), (032), (033) (h, k, 1) mixed Two odd and one even (e.g. 112); two even and one odd (e.g. 122) Continued

Face Centred Cubic Mixed indices Two odd and one even (e.g. 112); two even and one odd (e.g. 122) Mixed indices 1 CASE A O O e В O e ino) +e'#{©] = [1+1-1-1]= 0| -en()]= [1-1+1-1] 0 л(е) CASE A: [1+e CASE B: [1+er(°) +e 2 F =0 F=0 e.g. (100), (211); (210), (032), (033) (h, k, 1) mixed All odd (e.g. 111); all even (e.g. 222) Unmixed indices Unmixed indicesCASE k A O В e in (e) in(e) in(e) +e +e T?]=[1+1+1+1] CASE A: [1e = 4 This implies that in FCC only h,k,1 unmixed'reflections are present. e= [1+1+1+1] 4 i(e) CASE B [1er(e) +e'r(e) F = 4f F 2 (h, k, 1) unmixed e.g.(111), (200), (220), (333), (420)

Na* at (0,0,0) +Face Centering Translations -» (/, V, 0), (%, 0, ), (0, , 2) E Cl at (2, 0, 0) FCT -> (0, 2, 0), (0, 0, 2), (2, 2, 2) NaCl: FCC lattice k+l h+k i[27 2 1+h! i[27 2 (^) e i[27 e 2 |F = fe127(O)] +e h+k+l. i[27( 2 12 k i[27() е i[27(^] e far 2 2 2 +e +e ,in(k+l) іл(h- іn (1+h) F = f [1+e(h+k) 1 +e Na oin(k) e e iz(l) -eih+kl)1 in fler(h - [1+e (+*k) +e# (k+1) + e!#(i+h)]+ F = f Na fer ei +k+)[e# (-k-) F = [5Na* e'(-1-h) +e*(-n-k) +1I i + iz(1+. ein(l+h) elh+k+!)j1+e!r(h+k) + eiz(k+l) f +

NaCl: FCC lattice F =[fxet + for er(h+k+!)][1+ e'#(h+k) + e'#(k+!) + @}#(l+h) ]; Na Cl |F = [Term -1][Term-2 Zero for mixed indices Mixed indices CASE h | Mixed indices A O B e e e]= [1+1-1-1= 0| en] = [1-1+1-1] 0| in(e) oin(0) in CASEA Term 2 = [1+e" іл(о) e іл(е) +e CASE B: Term- 2 =[1+ F2 F = 0 0(100), (211); (210), (032), (033) (h, k, 1) mixed Unmixed indices | CASE h k A Unmixed indices O В e e T(e) +e e ©] = [1 +1+1+1] i T(e) CASE A : Term=2 = [1 - 4 - iT(e) + + e e]= [1+1+1+1]= 4 CASE B: Term-2 =[1 - 1o#(e) Continued

NaCl: FCC lattice h+k+l) F = 4[fN +fer e(h |(111), (222); (133), (244) (h, k, 1) unmixed h,klall even or all odd -F2 =16[f 4[fA F = + CI If (h + k + 1) is even-* Na Na CI (222),(244) |F = 4fNa = Jer- 2 -far] = I = 16[ f, If (h + k 1) is oddF Na CIT (111), (133) bj 002 022 202 222 11 UC in reciprocal space Superlattice reflection l020 Fundamental reflection *NaCl gives B2 in reciprocal space !! 220 200