Mark Farnworth - Pilkington European Technical Centre

  1. Use the structure factor equation to calculate the F2 (intensity) values for (i) the 200-reflection, and (ii) the 100-reflection from sodium chloride. You have already been given a sample calculation for the 111-reflection, Calculating the Intensity of Diffraction Using the Structure Factor Equation, and all the information that you need is contained within this example. In the case of the 100-reflection there is something odd about the answer and you should try to explain this.

    [17 Marks]

    2. i) The 200 reflection

    The formula for d for cubic crystals is:
                      dhkl = a/Ö(h2 + k2 + l2 )
    For h,k,l = 2,0,0 we have
                      d111 = 5.638/Ö(22 + 02 + 02 ) = 5.638/Ö4 = 2.819 Å
    From Bragg's law
                         l = 2d sinq
    so
                    sin q/l = 1/2d = 1/(2 ´ 2.819) = 0.177 Å-1
    The atomic scattering factors for Na+ and Cl- ions can be obtained from the extract of the International Crystallographic Tables provided in the notes. The table below shows my calculated values for sin q /l = 0.177 Å -1 (weighted between 0.15 and 0.20)

    sin q/l 0.15 0.177 0.20
    fNa+9.035 8.698.39
    fCl- 14.12 13.0812.20

    Now

                      F(S)  =   
    å
    n    		 fn cos {360 (hx + ky + lz)} + i 
    å
    n    		 fn sin {360 (hx + ky + lz)} (in degrees),

    Working out the cosine part of F200

     
    å
    n    		 fn cos{360 (hx + ky + lz)} =   8.69 cos {360 (2´0 + 0´0 + 0´0)}
          + 8.69 cos {360 (2´0 + 0´1/2 + 0´1/2 )}
          + 8.69 cos {360 (2´1/2 + 0´0 + 0´1/2 )}
          + 8.69 cos {360 (2´1/2 + 0´1/2 + 0´0)}
          + 13.08 cos {360 (2´1/2 + 0´0 + 0´0)}
          + 13.08 cos {360 (2´0 + 0´1/2 + 0´0)}
          + 13.08 cos {360 (2´0 + 0´0 + 0´1/2 )}
          + 13.08 cos {360 (2´1/2 + 0´1/2 + 0´1/2 )}
        =   8.69 cos 0 + 8.69 cos 0 + 8.69 cos 360 + 8.69 cos 360
          + 13.08 cos 360 + 13.08 cos 0 + 13.08 cos 0 + 13.08 cos 360
        =   8.69 + 8.69 + 8.69 + 8.69 + 13.08 + 13.08 + 13.08 + 13.08
        = 87.08

    Working out the sine part of F200

     
    å
    n    		 fn sin{360 (hx + ky + lz)} =   8.69 sin {360 (2´0 + 0´0 + 0´0)}
          + 8.69 sin {360 (2´0 + 0´1/2 + 0´1/2 )}
          + 8.69 sin {360 (2´1/2 + 0´0 + 0´1/2 )}
          + 8.69 sin {360 (2´1/2 + 0´1/2 + 0´0)}
          + 13.08 sin {360 (2´1/2 + 0´0 + 0´0)}
          + 13.08 sin {360 (2´0 + 0´1/2 + 0´0)}
          + 13.08 sin {360 (2´0 + 0´0 + 0´1/2 )}
          + 13.08 sin {360 (2´1/2 + 0´1/2 + 0´1/2 )}
        =   8.69 sin 0 + 8.69 sin 0 + 8.69 sin 360 + 8.69 sin 360
          + 13.08 sin 360 + 13.08 sin 0 + 13.08 sin 0 + 13.08 sin 360
        =   0

    Combining the cosine and sine parts of F111

                      Fhkl2 = {cosine part}2 + {sine part}2
                      F2002 = (87.08)2 + 02 = 7583 (Intensity)

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