If a peak shifts towards the left by 0.4 degrees (2 thetas) while the calculated crystallite size is also lower. Then, what is the relation between peak shift & d-spacing?
Dear Abdul Moyez ,
according your 'Skills and expertise' from your RG homepage one should expect that you are familiar with the Bragg law*):
n*lambda= 2 * d * sin(theta)
Differentiating this relation will give you:
0 = 2* delta_d*sin(theta) + 2*d*cos(theta)*delta_theta
Here you have the relation between the peak shift delta _theta and the d-shift delta_d.
You may solve for delta_d by yourself.
Please pay attention to the signs:
left shift of the peak (lower theta) will give higher d-spacing... and vice versa
*) https://en.wikipedia.org/wiki/Bragg%27s_law
Best regards
G.M.
Dear Gerhard Martens as theta will decrease d value will increase. For exact calculation you can solve it by bragg's law.
The best answer is given by Professor Gerhard Martens.
Thank you a lot for good experience and valuable answering.
Dear Nitin Kumar ,
you may accidently have addressed me instead of Abdul Moyez .
I myself know how to deal with the Bragg law as you can see from my answer above.
Additional remark:
The crystallite size/peak width (as mentioned in the question above) has no impact onto the d theta relation (Bragg law).
By Bragg's law
nλ = 2d sinθ
d = nλ/2 sinθ
where n = 1 for any materails
therefore d = λ/2 sinθ
here θ = mean peak position (half of 2θ)
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