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I cracked KCl and made two samples. One is a fine powder, another is a coarse powder. Using these samples, diffraction intensity was measured. The results were as followed.

[peak number, relative intensity (fine), relative intensity (coarse)]
1: 100, 100
2: 52, 35
3: 16, 10
4: 19, 14
5: 27, 12
6: 19, 8
7: 6, 3
8: 13, 6

By the way, relative intensity can be theoretically calculated, using

$I = |F(h,k,l,\vec{K})|^2 \cdot p \cdot L(\theta) $ (eq.1) .

I wrote a simple program and evaluated its values.

[peak number, relative intensity]
1: 100
2: 67.1
3: 21.9
4: 9.7
5: 25.7
6: 18.7
7: 6.1
8: 11.0

Now I have two questions.

  1. What is the difference between the fine powder and the coarse one? (eq.1) doesn't include how the powder was cracked, so it's seemed that there is no difference. But the relative intensities are in fact different. I've heard the size of the powder is related to extinction effect, but don't know how to add the size factor to (eq.1).

  2. Why are the theoretical data and the experimental data different? When thinking about relative-intensity(fine) and that of theoretical, the values of the peak of 1,5,6,7,8 are similar, and 2,3,4 are far from similar. It's seemed so strange to me.

Any hints will be appreciated.

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How did you measure the intensities? A powder pattern consists of concentric rings, which are the sums of the spot patterns of randomly oriented small single crystals.

In the old days, one used to have a strip of photograph film on which the circle segments looked like lines. Theses were then analyzed by a densitometer. Now, there is often electronic detection.

If the powder is coarse, some larger crystals in the sampling volume will dominate. The circles will be non-uniform in intensity. If one is only sampling a small part of the diffraction circles, the intensities will deviate from the average over all of the circle.

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  • $\begingroup$ Thank you. I've read Cullity's book and I'm now coming up with the reason for the difference, putting together knowledge I got from the book and information you thankfully gave me. $\endgroup$ – ynn Dec 17 '17 at 7:16

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