ࡱ> a )jbjbdd $dee%.......dl6l6l6l6\6ld#H@78888888FFFFFFF,IRoKh G.88888 G;..88G;;;8.8.8F;BV....8F;;rDT..E47 l6::EE G0#HHEK;KE;.dddd1. How does Fhkl change for centrosymmetric structures? Show your derivation step by step. Can the x-ray reflection intensities for centrosymmetric structures be used to directly calculate the electron density distribution? 2. The diffraction pattern below was obtained from a common cubic material using CuKa radiation. Determine the Miller indices for each reflection ("index the pattern"), and thereby determine the centering type for the lattice from an inspection of the systematic absences. Calculate the lattice parameter, a. To index the pattern, guess the (hkl)'s for the first two or three reflections for P, I, and F cubic, and calculate the corresponding a values for these reflections. Since this cubic material has a single lattice parameter, the assignment of indices which gives a values which are very close to each other is the correct one. You will run into an additional problem here; in attempting to solve the problem, do not use h,k,l values higher than 5. In indexing the pattern, remember that the objective is to find some set of indices (hkl) that gives essentially the same lattice parameter for all reflections. d() (hkl) for P (hkl) for I (hkl) for F  3.157 1.931 1.647 1.366 1.253 1.1150 1.0512 0.9657 0.9233 0.9105 0.8637 0.8330 The most precise values of lattice parameters are those obtained for reflections which occur at high angles. In fact, all errors can be eliminated at q = 90, and lattice parameter values generally reported are those obtained from an extrapolation of the data to q = 90 against some function intended to make the lattice parameter variation linear. Two such functions, obtained through an extensive evaluation of systematic experimental errors, are the Bradley and Jay function, cos2q (frequently used for diffractometer data, and truly linear only over the range q = 60-90) and the Nelson-Riley function (cos2q/sinq + cos2q/q, linear over the range 30-90q, and used for other instrument geometries). Here are some observed data for a-iron. Determine the best lattice parameter to 5 decimal places by extrapolating the a values versus cos2q in a graph. lCuKa1 = 1.5405981 . To index the pattern, remember that a-iron is I cubic. q d (hkl) a cos2q  22.3360 32.5114 41.1675 49.4742 58.1948 68.5881 4. For the data below, determine the weight fractions of rutile and anatase present in a mixture. Pure rutile: Reflection Intensity Background  1 102309 1508 2 47564 1453 Rutile in the mixture: Reflection Intensity Background  1 36987 1486 2 17408 1439 5. A rock specimen was analyzed for a-quartz. To correct for the unknown absorption of the specimen, an internal standard was used. The internal standard was chosen as KCl because of its excellent crystallinity, relatively simple diffraction pattern, and a strong reflection which is relatively close to the strong a-quartz reflection. Three samples were measured: pure KCl, pure quartz and a mixture of the rock specimen with KCl. The mixture was prepared by intimately mixing 200 mg of finely ground KCl with 1000 mg of the rock, also finely ground. The intensities of the 3.15 KCl reflection and the 3.34 quartz reflection were measured using CuKa radiation. The data are given below. Phase d () mo Pure compound Mixture  quartz 3.34 35.0 48360 cps 2648 cps KCl 3.15 124.0 19072 6160 For all intensities, the background was 240 cps. The dead time for the counting system was 1.0 msec. The correction for the dead time, t, makes use of the following equation: true counts/sec =  EQ \f(observed counts/sec, 1 - tobserved counts/sec)  Calculate the concentration of quartz in the rock specimen. Note that:  EQ \f(Ia, I\o(o,a)) = \f(m\o(o,a),m) Xa  where: I and Ia are the intensities for the pure compound and the mixture for phase a, respectively. m EQ \o(o,a) and m are the mass attenuation coefficients for the pure compound and the mixture, respectively. Xa is the weight fraction of phase a in the mixture. An equation of this type can be written for both quartz and KCl. Combining these two equations to eliminate m, the unknown mass attenuation coefficient for the mixture: Xq = XKCl  EQ \f(I\o(K,o)Cl,I\o(o,q)) \f(m\o(o,K)Cl,m\o(o,q) ) \f(Iq, IKCl)  First calculate the weight fraction Xq of quartz in the diluted specimen. Then, from the value of Xq and the dilution factor, calculate the weight fraction of quartz in the original mixture. 6. MoB is I41/amd with a = 3.110, c = 16.95 , and Mo and B atoms in 8e, z = 0.196 and 0.352, respectively. Calculate F001 and F004. F001 = ______ Fhkl =  EQ \S( N,S,j=1) fj exp(2i(hxj + kyj + lzj)). F004 = ? What must be calculated? d004 = _______ (sin q)/l = ________ 0.0 0.1 Mo 38.2 32.6 3.5 2.4 Mo = _______ B = _______ 8e: (00z) (0, EQ \f(1,2) , EQ \f(1,4) +z) ( EQ \f(1,2) ,0, EQ \f(3,4) -z) ( EQ \f(1,2) , EQ \f(1,2) , EQ \f(1,2) -z) F004 = _______ 7. b-Np is P4212 with a = 4.897, c = 3.388 , and Np in 2a ((000) and ( EQ \f(1,2)\f(1,2) 0)) and 2c ((0 EQ \f(1,2) z) and ( EQ \f(1,2) 0 EQ \o(z,_) )), z = 0.375. Find I101 for CuKa radiation.  calculation: (sin q)/l 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7  93 87 78 69 60 53 48 44 F calculation: What is p? 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How does F ((hkl <( c-@)1h-)"ange-D) -)for-@)[ -)centro-@)s-)ymmetric-D(  -) structures?-@({ -) Show-D) -)your-@w){ -(v#derivation step by step. Can the x(9-)ray reflection intensities-D(  -)for-@)[ -(Qv centrosymmetric structures be us-@(Q3e-) d-D)# -)to-@)< -)directly-D) -) calculate-@) -)the-D)` -)electron-@) -(v density di-D(s-) tribution?-@3( (v -(L2.)4 The) diffraction pattern below wa-@(L,s-) obtained-D(LM -)from-@w) -)a-) -)c-@)o-)mmon-D) -)c-@)u-)bic (vmaterial using CuK(va) radiat-@)i-)on.-D)N -) Determine-@( -)the-D)` -)Miller-@) -)indices-D) -)for-@)[ -)ea-@)@c-)h (vreflection ("index-D(f -)the-@)` -) pattern"),-D( -)an-D)>d-)$ thereby-@) -) determine-D( -)the-@w)` -) centering-( " -(Gv8type for the lattice from an inspection of the systemati(G c absences. -@(G  -(v!Calculate the lattice parameter, ( a)!.-) -@3( -+TBTo index the pattern, guess the (hkl)'s for the first two or three-@(B  -(vreflections for P, I 2(, <) and F cubic, an-@(Id-)" calculate-D(u -)the-@)` -) corresponding-D(  -)a)! (v0values for these reflections. 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How does F ((hkl <( c-@)1h-)"ange-D) -)for-@)[ -)centro-@)s-)ymmetric-D(  -) structures?-@({ -) Show-D) -)your-@w){ -(v#derivation step by step. Can the x(9-)ray reflection intensities-D(  -)for-@)[ -(Qv centrosymmetric structures be us-@(Q3e-) d-D)# -)to-@)< -)directly-D) -) calculate-@) -)the-D)` -)electron-@) -(v density di-D(s-) tribution?-@3( (v -(L2.)4 The diffraction patte(L rn below wa-@(L,s-) obtained-D(LM -)from-@w) -)a-) -)c-@)o-)mmon-D) -)c-@)u-)bic (vmaterial using CuK(va) radiat-@)i-)on.-D)N -) Determine-@( -)the-D)` -)Miller-@) -)indices-D) -)for-@)[ -)ea-@)@c-)h (vreflection ("index-D(f -)the-@)` -) pattern"),-D( -)an-D)>d-)$ thereby-@) -) determine-D( -)the-@w)` -) centering-( " -(GvDtype for the lattice from an inspection of the systematic absences. -@(G  -(vCalcu)late the lattice parameter, ( a)!.-) -@3( -+TBTo index the pattern, guess the (hkl)'s for the first two or three-@(B  -(vreflections for P, I 2(, <) and F cubic, an-@(Id-)" calculate-D(u -)the-@)` -) corresponding-D(  -)a)! (v0values for these reflections. Since this cubic -@(m-).aterial-D) -)has-@w)_ -)a-) -)single-@w) -(=vlattice paramete-@(=Hr-),) the assignment -@(=?o-)f-D) -)indices-@) -)which-D) -)gives-@) -)a)! values (vwhich -@)a-)re-D)> -)very-@)z -)close-D) -)to-@)< e-)2 ach other-D(= -@)i-)s-D) -)the-@)` -) correct one.-D(U -) You-@)w -)will-D)Z -)run-@)[ -(vinto an additional problem h-@(}e-) re;-D)P -)in-@)0 a-)0 ttempting-D(s -)to-@)< -)solve-D) -)the-@)` -)problem,-D) -@)d-)"o (7v$not use h,k,l values higher than 5. -@(70 I-)1n-D) -)indexing-@) -)the-D)` -)pattern,-@) -)remem)ber-@)b -(vthat the objective-@({ -)is-D). -)to-@)< -)find-D)q -)some-@) -)set-D)Z -)of-@)> -)indices-D) -)(hkl)-@) -)that-D)z -)gives-@) -(v essentially-D( -)the-@)` -)same-D) -)lattice-@) -)par-@)\a-)meter-D) -)for-@)[ -)all-D)? -) reflections.-@3( (2 -2*T -)d()Y-Y\){ -(y (hkl) for P-~!( -) (hkl) for  2)I <j-iD) -)| (hkl) for F-@w(  x  - @3(  1  x --U*T -)3.157-) -@3(  -0*T -)1.931-) -@3( , -0*S -)1.647-) -@3(  -0*T -)1.366-) -@3( z -0*T -)1.253-) -@3( ! -0*T -)1.1150-) -@3(  -0*S -)1.0512-) -@3( p --U*S -)0.9-@)Y6-)$57-)J  !x!x ! ! w!w ! ! v!v ! ! u!u ! ! t!t ! ! s!s ! ! r!r ! ! q!q ! ! p!p ! ! o !o ! ! n !n ! ! <@< NormalCJaJmH sH tH DA@D Default Paragraph FontRi@R  Table Normal4 l4a (k@(No ListPd!zP        | } ~     ' ( ) | } Y Z    7 S T U WXY9:+,:;tu  $%&'(89:;<=>?@OR00ʀ00000000000000000000000000000000 0 00О0̀0̀00͠0͠0͠0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0О0О0ʀ00О0О0О0О00О0О0О00О0О0О0О0О0О0О0О0О0О0О0О000О00О0О0О0О0О0О0О00О0О0О00О0О0О0О0О0О0О0О0О0О0О0О0О0О0О0О0О00О000О00О0О0О0О0О0О0О0О0О0О0@0О@0О@0О0@0О@0О@0О@0О@0О0@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О@0О0О0О0О0000000000000000000000000000000000000000000000        | } ~     ' ( ) | } Y Z    7 S T U WXY9:+,:;tu  $%&'(89:;<=>?@OR0̀0̀0̀0̀0̀0̀0̀0 0͠0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀0̀ 0̀000000 0 0 000000000000008080 8080 8080 8080 8080 8080 0  0  8080 8080 8080 8080 8080 8080 8080 8080 8080 0   0   8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 0   8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080 8080  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0I9I9I9I9I9I9@0@0@0@0@0@0@0@0@0@0@0@0@0@0@0@0I9I9I9"@08@0Y@0Y@0Y@0Y@0Y`0Y@0Y@0YIQG IQG IQG :@^{PZ8@^{PZ8@^{PZ8@^{PZ8@^{PZ8@^{PZ8@^{PZ08@^{PZ8@^{PZ8@^{PZ@0C3 @080000000000000000000000 000000`0 0  0 0@ 0, T"$@&H'V(zd&'()+-.0 X, jU"$L(() !"#$%*,/1)U;ICU')6=JN[boq~&35BP1111111111111111-9=AD7:#z}bevzS Y ^ e % ( x { 8 ; !=@acgimoaeRv~'/]nNSuv4 D   Y]{~KMrR:::::::::::::::::::)^)`)b)d)f)h)j)l)n)))^gd:  0]0^gd: ":p: / =!"#$% Ignore absorption and temperature factor.