Assignment 3
Zhanhui Yuan

Answer:

1.
Structure Factor equation:

                  F(S)  =   
å
n		 fn e2p i (hx + ky + lz)
                  F(S)  =   
å
n		 fn cos {2p(hx + ky + lz)} + i  
å
n		 fn sin {2p(hx + ky + lz)} (in radians)
or
                  F(S)  =   
å
n		 fn cos {360 (hx + ky + lz)} + i 
å
n		 fn sin {360 (hx + ky + lz)} (in degrees),

Na: 0, 0, 0; 0, 1/2, 1/2; 1/2 , 0, 1/2; 1/2, 1/2, 0.
Cl: 1/2, 0, 0; 0, 1/2, 0; 0, 0, 1/2; 1/2, 1/2, 1/2.

d = 5.638Å

(I)
(1). Using crystallography and Bragg's law
The formula for d for cubic crystals is:
                  dhkl = a/Ö(h2 + k2 + l2 )

For hkl = 200 we have
                  d200 = 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

(2) Working out what atomic scattering factors (fn):

From the International Crystallographic Tables we get the following f values:

sin q/l 0.150.20
fNa+ 9.0358.39
fCl- 14.1212.20

Using an intermediate value weighted produces the following result(needed values in bold):

sin q/l 0.150.1770.20
fNa+9.035 8.698.39
fCl- 14.1213.0812.20

(3) 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

(4) Working out the sine part of F200

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

(5) Combining the cosine and sine parts of F200

                  F200 = 87.08 + i ´0

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

(II)
(1). Using crystallography and Bragg's law
The formula for d for cubic crystals is:
                  dhkl = a/Ö(h2 +k2 + l2 )

For hkl = 100 we have
                  d100 = 5.638/Ö(12 + 02 + 02 ) = 5.638/Ö1 = 5.638 Å
From Bragg's law
                    l = 2d sinq
so
                sin q/l = 1/2d = 1/(2 ´ 5.638) = 0.0887 Å-1

(2) Working out what atomic scattering factors (fn):

From the International Crystallographic Tables we get the following f values:

sin q/l 0.150.20
fNa+ 9.0358.39
fCl- 14.1212.20

Using an intermediate value weighted produces the following result(needed values in bold):

sin q/l0.0887 0.150.20
fNa+9.839.0358.39
fCl- 16.4714.1212.20

(3) Working out the cosine part of F100:

 
å
n		 fn cos{360 (hx + ky + lz)} =  9.83 cos {360 (1 ´ 0 + 0 ´ 0 + 0 ´ 0)}
   + 9.83 cos {360 (1 ´ 0 + 0 ´ 1/2 + 0 ´ 1/2)}
    + 9.83 cos {360 (1 ´ 1/2 + 0 ´ 0 + 0 ´ 1/2)}
   + 9.83 cos {360 (1 ´ 1/2 + 0 ´ 1/2 + 0 ´ 0)}
   + 16.47 cos {360 (1 ´ 1/2 + 0 ´ 0 + 0 ´0)}
   + 16.47 cos {360 (1 ´ 0 + 0 ´ 1/2 + 0 ´ 0)}
   + 16.47 cos {360 (1 ´ 0 + 0 ´ 0 + 0 ´ 1/2)}
   + 16.47 cos {360 (1 ´ 1/2 + 0 ´ 1/2 + 0 ´ 1/2)}
  =  9.38 cos 0 + 9.38 cos 0 + 9.38 cos 180 + 9.38 cos 180
   + 16.47 cos 180 + 16.47 cos 0 + 16.47 cos 0 + 16.47 cos 180
  =  9.38 + 9.38 + (-9.38) + (-9.38) + (-16.47) + 16.47 + 16.47 + (-16.47)
  =0

(4) Working out the sine part of F100

 
å
n		 fn sin {2p (hx + ky + lz)}=  9.83 sin {2p (1 ´ 0 + 0 ´ 0 + 0 ´ 0)}
   + 9.83 sin {2p (1 ´ 0 + 0 ´ 1/2 + 0 ´ 1/2)}
   + 9.83 sin {2p(1 ´ 1/2 + 0 ´ 0 + 0 ´ 1/2)}
   + 9.83 sin {2p (1 ´ 1/2 + 0 ´ 1/2 + 0 ´ 0)}
   + 16.47 sin {2p (1 ´ 1/2 + 0 ´ 0 + 0 ´ 0)}
    +16.47 sin {2p (1 ´ 0 + 0 ´ 1/2 + 0 ´ 0)}
   + 16.47 sin {2p (1 ´ 0 + 0 ´ 0 + 0 ´ 1/2)}
   + 16.47 sin {2p (1 ´ 1/2 + 0 ´ 1/2 + 0 ´ 1/2)}
  =  9.83 sin 0 + 9.83 sin 0 + 9.83 sin p + 9.83 sin p
   + 16.47 sin p + 16.47 sin 0 + 16.47 sin 0 + 16.47 sin p
   = 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0
  = 0

(5) Combining the cosine and sine parts of F100

                  F100 = 0 + i 0

so
                  Fhkl2 = {cosine part}2 + {sine part}2
                  F1002 = (0)2 + 02 = 0
(III)

When two X-rays start off in phase, they will reach the detector and have different distance to travel. So we would say that the two X-ray photons are out of phase by the time they reach the detector. It is more usual to express this phase difference in terms of an angle (degrees or radians) whereby one angular revolution corresponds to a phase difference of one wavelength. If the phase difference is half a wavelength in distance then destructive interference occurs. In above case, the two waves end up being about half wavelengths, i.e. the phase difference is l/2 in distance. So the intensities of the two X-rays are cancelled each other and the detector eventually gets nothing. Th is is the reason why the value of intensity for 100-reflection of NaCl cubic is 0.

2.

The number +0 is identical in every respect to the number -0, so there are in fact two equivalent sets of planes to 200 (i.e.200 and -200), all with the same d spacing. So we would say that the multiplicity j200 is 2 in this NaCl cubic case.

3.

(1)Working out the angles q, 2q :

First we remind you how to calculate the d spacing in cubic crystals:
                  dhkl = a/(h2 + k2 + l2 )1/2
                  d200 = 5.638/(22 + 02 + 02 )1/2 = 5.638/Ö4 = 2.814 Å

For Copper Ka radiation the wavelength is about 1.54 Å, and so inserting into Bragg's law gives us:
                  1.54 = 2 ´ 2.814 ´ sinq
                  sinq = 0.273
                  q = sin-1(0.273) = 15.84°
                  2q = 31.68°

(2) Calculating P200 :

For the laboratory powder diffractometer we use the formula of Case (3):
2q = 31.68°
cos (2q) = 0.849
cos2 (2q) = 0.721
1 + cos2 (2q) = 1.721
P200 = (1 + cos2 (2q ))/2 = 0.8605
So polarisation produces a reduction in intensity to 86.05% in the case of the 200-reflection.

4.

Lorentz Factor, L

Again taking the NaCl 200 reflection, and using the value for theta already obtained in the polarisation calculation:
            q = 15.84°
            sin q = 0.273
            sin2q= 0.0745
            cos q= 0.962
            sin2qcos q = 0.0717
            L = 1/sin2q cos q= 13.95

5.

The formula previously given, relating the measured intensity to the structure factor, Ihkl = c jhkl Phkl Lhkl A Fhkl2, can now be used with respect to all parameters except A (absorption) which we will ignore for the purposes of this exercise. The various values that have been calculated in this section for the case of 200 diffraction are:

j200= 2
P200= 0.8605
L200= 13.95
F2002= 7583

Multiplying these (and ignoring absorption) gives:

 
 I200= 182052 c units.
The units are not important since the usual approach in Crystallography is to compare two values by measuring them under, as far as possible, identical conditions. In a later assignment you will be asked to make such a comparison of I111/ I200 . Obviously if the experimental measurements of I111 and I200 were identical (other than j, P, L) and we ignore differences in the absorption, then the constant c will disappear on division:

I111
I200		 =  j111 P111L111 F1112
j200 P200 L200 F2002
 = 0.288

6.

(1). Temperature Effects:Temperature will have an effect on the intensities of the powder diffraction pattern. The equation of modified atomic scattering factor fT is: fT = f exp(-B sinq2/l)
(2). Preferred Orientation of the Sample:The intensities of sample in different mounting ways are quite different, such as in capillary, flat-plate.
(3). Sample Texture: The sample texture causes that in the sample there are insufficient crystallites to provide a true powder diffraction pattern.
(4). Transpareny of Sample:For the Bragg-Brentano geometry, poor data may be obtained when the sample is too transparent for X-rays.
(5). A bad X-ray filters: With a bad metal foil filter, the X-ray pattern has high energy background radiation.
(6).A bad X-ray source: Not pure monochromatic beam causes varied pattern.
(7).Magnetic effect:Strong magnetic field effects on the X-ray so as to bad X-ray pattern comes out.
(8).Orientation of X-ray instrument:With bad orientation of X-ray instrument, X-ray running in vibration condition causes bad result of X-ray pattern.
(9).The holder of sample:Different shape and quality of the holder of sample produce different background of X-ray pattern.
(10).Count Times:The setup of count times is important to obtain perfected X-ray pattern.
(11).Slit effect: The good slit gives a less asymmetric and narrower peak shape of X-ray pattern, especially at low scattering angles.
(12).Calibration of X-ray Detectors:The position sensitive detectors require careful calibration for both wire position and efficiency so that both scattering angles and intensities can be reliably determined.
(13).Choice of X-ray Target:With bad choice of X-ray target, a powder diffraction yields the data with a very high background and very little signal.
(14).Gaseous environment:The intensity is changed while the sample is under gaseous environment.
(15).Pressure of environment:The intensity of a sample under different pressure is changed.

yuan@img.cryst.bbk.ac.uk ©Zhanhui Yuan