# XRD Crystallite Size Calculator (Scherrer Equation)

## XRD Crystallite Size Calculator (Scherrer Equation)

Peak Position (2θ)

FWHM (2θ)

X-Ray Wavelength

Results
nm

#### Calculation Tutorial:

STEP1: Open the XRD graph of the material, which is obtained from the instrument. STEP2: Now zoom on the area for which you want to calculate the crystallite size and note down the angle at which peak is shown and peak Full Width at Half Maximum (FWHM). Raw HTML STEP3: Now enter the measured Peak Position (i.e. 31.8) and peak FWHM (i.e. 0.5) in desire columns of the calculator. You should get the calculated results of the crystallite size in the "Calculated Result" field.

NOTE: Default value of wavelength of LASER is set is 0.15418 (Cu K-alpha), which is mostly used in the instruments.

Theory Behind Calculations: X-Rays are having wavelength between 0.01nm to 10nm. Hence X-Rays can penetrate inside the crystal structure of any material very easily; and tells us the properties of material while coming out from that material. Which is why X-Ray spectroscopy is very useful technique for characterization of different types of materials. We can easily calculate the size of particles from Scherrer formula given:

Scherrer Formula:

Dp = (0.94 Χ λ) / (β Χ Cosθ)

Where, Dp = Average Crystallite size, β = Line broadening in radians, θ = Bragg angle, λ = X-Ray wavelength

NOTE: Please don't worry about the β(in radians), All the calculations are made such that you can enter β (i.e. Full Width at Half Maximum) value directly in degree as shown in "Calculation Tutorial"

1. Step one: Prepare a suitable sample, one with an appropriate size range and prepared to avoid preferred orientation.

2. It is very good web.

3. This does not take into consideration instrumental widths right?

4. How to do correction in fwhm values. While calculating fwhm for clays

5. Thank you very much. It is useful.

6. it's good idea to help researchers

7. Sedrati Hichem29 May 2018 at 00:05

THANK YOU..BUT K=0.94 IS FOR COPPER RADIATION
I THINK YOU HAVE TO ADD OTHER CONSTANT LIKE K=0.89 FOR COBALT RADIATION..
YOURS

1. The K is a material related, not source related, constant. Is true that they could give the option to give the K for your material, but is easy to fix dividing the outcome by 0.94 and multiplying for your desire K

8. why The crystallite size of Ca-substituted in YBCO at Ba site samples was decreased for high concentration and increased for low concentration

9. Iroda Yuldashova21 June 2018 at 20:03

Very good, but shape faktor for different crystals is vary.
That has got a necessity for new section which can insert shape factor.

10. komal prasad malla23 July 2018 at 14:56

Thank you so much it is a easier way to calculate particle size of nano materials.

11. LaquandaGissler14055@yahoo.com21 August 2018 at 15:39

I agree with your details , fantastic post.

12. what if we have multiple peaks as in the case of silver nanoparticles? which peak should we select?

13. Thanks for this helpful site to find crystal size.

14. Harshwardhan Wadikar20 November 2018 at 11:38

K depends on the crystal shape factor, K~0.9 for spherical crystallites.

15. Very useful

16. One of our guests not long ago advised the following website.

17. How can I calculate FWHM12 February 2019 at 13:53

Very useful.

18. Thank you for easiness, but have considered the shape factor?
With regard

19. Different peaks in a XRD pattern gives different values of the particle size . How to evaluate the exact particle size then?

20. Habib Ur Rehman1 March 2019 at 22:51

if we express theta values and FWHM in radians then crystallite size is different instead of express values in degrees?
Why it is so?

1. Esto es porque los radianes solo están relacionados a una medida lineal -radio- (necesaria para conocer, por ejemplo, un parámetro de red), mientras que los grados se relacionan además con la inclinación de incidencia de los rayos X. Al expresar la FWHM en radianes, se esta realizando la conversión de unidades necesaria para concer la medida lineal del parámetro de red

21. Not wordy, nice explanation and easy to grasp so

22. Smart work..its very useful

23. Niyitanga Theophile12 April 2019 at 17:57

This is really helpful. Thanks for saving our time!

24. do you includes contributions of instrumental error?

1. 25. If both beta and theta in degree then which one have to convert in radians