
#1
Mar2408, 11:22 AM

P: 25

Hello,
I came to this forum looking for an answer to a pretty simply question about the Scherrer equation for peak broadening in the XRD pattern. I understand that the fullwidthhalfmaximum of the peak must be in radians, but when I measure the FWHM, it is in 2 theta. Do I need to half it to theta before plugging it into the equation? This is important because doing so doubles the answer you get for crystal size. Thanks for any help. 



#2
Mar2408, 12:59 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

Yes, you need to write the peak width in units of theta, not 2theta.




#3
Mar2408, 01:01 PM

P: 25

Great! Thanks for that info!
So basically when I measure the peak width to be something like 2.5 (2theta) it is 1.25 (theta) and then convert that to radians so I can plug and chug? I really appreciate the help! 



#4
Mar2408, 01:33 PM

Emeritus
Sci Advisor
PF Gold
P: 11,154

Scherrer Equation Question (XRD)
mesogen,
I may have spoken in haste. I will need to see your Scherrer forumla before I can say whether or not to divide by 2. If I recall correctly, and it's been a while, if you have [itex]dcos\theta=K\lambda/w[/itex], w has units of 2theta, and if you have [itex]dcos\theta=K\lambda/2w[/itex], then w has units of theta. Let me confirm that with a text when I get a chance today. 



#5
Mar2608, 08:51 PM

P: 25

I'm using the form that goes:
L = K*lambda / B*cos(theta) where L = average crystal size (sorta) B = the full width half maximum of the peak But I was looking around online and nothing would tell me, but I think this page says that B is 2(delta)(theta) http://www.eng.uc.edu/~gbeaucag/Clas...herrerEqn.html so I'm thinking maybe I should just take the degrees 2theta value and convert it to radians and then plug that into the equation. I don't know. An answer like 20 nm seems just as reasonable as 40 nm to me. 



#6
Oct1408, 07:17 AM

P: 1

Hi,
if you are using the formula as L= k*lamda/B*cos(theta) your theta should be half of 2 theta, thete should be in degree (you may try rad as well, but not much different in the result you will get). e.g. a Bragg peak at 2theta 40, then theta is 20 degree. Instead, B (usually you will take FWHM) must be in rad. The k constant varies from 0.8 to 0.98 depends on the crystalline shape. Hope this will help. Best regards 



#7
Jul2809, 09:45 AM

P: 3

Hi,
I also use L= k*lamda/B*cos(theta) to calculate crystallite size and I convert B and theta in rad. The results I obtain are the same with those obtained using the XRD crystallite size calculator from the following link: http://www.d.umn.edu/~bhar0022/dpcalculator/index.php I think this is the right way to calculate L. By converting both B and theta in to rad only the SI units are used Best regards! P.S Sorry about my poor english 



#8
Jul3109, 11:09 AM

P: 29

Could you kindly let us know where your reference that you used to quote the numbers for for k? 



#9
Aug509, 07:14 AM

P: 3

Hi,
From the courses I took during the M.S.C program I know that the value of k depends whether I take B=B1/2 or B=Bi where B1/2 is FWHM and Bi is the integral width of the diffraction line. When using B=B1/2, k=0.9 and k=1.05 for B=Bi. As far as I know Bi is more convenient for theoretical calculations. Try the following reference: http://books.google.ro/books?id=HuF...age&q=&f=false 



#10
Aug609, 05:37 AM

P: 9

i have 2th = 35.546 deg, d spacing 2.5236 A, fwhm 2 th  0.148 deg, now can sombody put these values to get size of crystallite? i want to know the steps
thanks 



#11
Aug1009, 08:06 AM

P: 3

Hello everybody,
I think I found a good paper that contains a good discussion about the values of K. My problem is that I don't have access to this paper and neither other persons I know. I give you the link of the paper: http://www3.interscience.wiley.com/s...&pages=102113 Hope that somebody from this forum could download this paper and if it is possible I would also like a copy for me Thanks in advance! 



#12
Aug2510, 09:00 AM

P: 1

I don't know how I can get B .




#13
Nov1510, 06:55 AM

P: 1

Hello,
From this thread I managed to answer all my questions, except one. What is FWHM and how do I get the value for it? Thanks 



#14
Apr213, 06:44 AM

P: 5

FWHM is full width at half maxima. So, first you find the maximum intensity of the peak (absolute value of max. intensity minus the base line intensity). Make half of this value. Find out the 2theta positions of the two points on the peak with this halfintensity value. The difference between these two 2theta positions (not theta positions) is your FWHM. It will be in degree. Convert this deg value in radian, for using Scherrer equation.
In Scherrer equation, there is cos theta at the denominator. This theta is the value of the 2theta position of max intensity divided by 2. Deg or radian does not matter since cos theta will be same. 



#15
Apr213, 06:57 AM

P: 5

In Scherrer equation, B is the measure of the broadening of the peak. B can be expressed in many ways. Simplest is FWHM. More accurate is Integral Breadth since it is calculated from the area of the peak, considering all the intensity points of the peak w.r.t base line.
There are two components in B measured in XRD peak. One is the broadening due to sample characteristics (crystallite size, lattice strain and defects). Another is Instrumental broadening which comes from the experimental parameters of Xray Diffractometer. You can estimate B(Inst.) by measuring the broadening of a peak at similar 2theta position of a standard sample provided with the diffractometer. This effect can be subtracted from measured B by various analytical ways. You can find this from published literature. 



#16
May1513, 02:00 AM

P: 5

I have few publications in this field. If anyone needs some ref., I can provide mine. That will help.



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