Part-6+(rev).+Unkowns+Ordering++Shape(2018)+CP.ppt

H k l mixed 4f zn i f s 16f zn 2 f s 2 h k l unmixed

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h, k, l   mixed 4(f Zn - i f S 16{(f Zn ) 2 + (f S ) 2 } h, k, l   unmixed  and  h+k+l = odd 4(f Zn -f S 16{(f Zn ) - 2f Zn f + (f S ) 2 } h, k, l   unmixed   and (h+k+l)/2 = odd 4(f Z +f S 16{(f Zn ) +2f Zn f + (f S ) 2 } h, k, l   unmixed   and (h+k+l)/2 = even
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1 2 3 4 5 6 7 8 9 32.5 37.5 54.2 65.0 67.5 80.0 88.8 92.0 104.0 0.078 0.103 0.2075 0.2887 0.30865 0.41317 0.48952 0.5174 0.6209 1 2 3 4 5 6 8 9 10        2 4 6 8 10 12 14 16        3 4 8 11 12 16 19 20        0.078 0.0515 0.0692 0.072 0.061 0.0688 0.0611 0.057 0.0629 0.039 0.0258 0.0346 0.0360 0.0308 0.0344 0.0349 0.0323 0.0260 0.0258 0.0259 0.0262 0.0257 0.0258 0.0257 0.0259 4.77 4.796 4.786 4.759 4.805 4.796 4.805 4.786 111 200 220 311 222 400 331 420 Ave=4.788 Å   =1.54056 Diffraction equation 2d hkl sin = 2a h 2 + k 2 + l 2 sin = sin 2 h 2 + k 2 + l 2 = 4a 2 2 S=h 2 +k 2 +l 2
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1 2 3 4 5 6 7 8 9 =1.54056 Diffraction equation 2d hkl sin = 2a h 2 + k 2 + l 2 sin = sin 2 h 2 + k 2 + l 2 = 4a 2 2 S=h 2 +k 2 +l 2 What is next? You now find the lattice constant(a) and the structure (FCC).  Look at the data sheet, together with other information of this material, Find what it is. (some speculation is involved, but more the data, better.) You are able to compute the intensity of each diffraction peak. Intensity   (|F hkl | 2 )•(L-P factor) • (Mult) ) Followed by the normalization You then plot the intensity with 2 , And compare with the spectrum  given. 32.5 37.5 54.2 65.0 67.5 80.0 88.8 92.0 104.0 0.078 0.103 0.2075 0.2887 0.30865 0.41317 0.48952 0.5174 0.6209 1 2 3 4 5 6 8 9        2 4 6 8 10 12 14 16        3 4 8 11 12 16 19 20        0.078 0.0515 0.0692 0.072 0.061 0.0688 0.0611 0.057 0.0629 0.039 0.0258 0.0346 0.0360 0.0308 0.0344 0.0349 0.0323 0.0260 0.0258 0.0259 0.0262 0.0257 0.0258 0.0257 0.0259 Ave=4.788 Å   4.77 4.796 4.786 4.759 4.805 4.796 4.805 4.786 111 200 220 311 222 400 331 420
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Now your turn Taken by Cu-K  line  It is a single element metal   Cubic Shiny Line 2 Sin SC S           sin 2 /S BCC S           sin 2 /S FCC S           sin 2 /S Diamong S           sin 2 /S A(A) (hkl)
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Sharp distinct peak Broader peaks Two overlapping peaks K 1 K 2
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Single peak @ low 2 Mixed peaks @ mid 2 Double peaks   @ high 2 Diffraction equation 2d hkl sin = = sin -1 2d hkl K 1 =1.540562Å vs K 2 =1.544390Å 2 ( K 1 ) = 2sin -1 1.540562 2•3.4966 =25.538˚ 2 ( K 2 ) = 2sin -1 1.544390 2•3.4966 =25.508˚ 2 ( K 1 ) = 2sin -1 1.540562 2•1.3356 =70.442˚ 2 ( K 2 ) = 2sin -1 1.544390 2•1.3356 =70.064˚
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CuAu Cu 3 Au Order-Disorder Transformation   • Often occurs in most substitutional solid solutions,  when two kinds of atoms    A and B are arranged more or less at random on the atomic sites of the lattice.  • Good example:    Cu-Au alloys      For high T,   phase is always FCC.    (Why?) • Three different  ordered structures       Cu 3 Au CuAu      CuAu 3   
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  • Fall '12
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