# XRD Crystallite Size Calculator (Scherrer Equation)

## XRD Crystallite Size Calculator (Scherrer Equation)

#### 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).
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**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"

## Post a Comment

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

ReplyDeleteIt is very good web.

ReplyDeleteThis does not take into consideration instrumental widths right?

ReplyDeleteHow to do correction in fwhm values. While calculating fwhm for clays

ReplyDeleteThank you very much. It is useful.

ReplyDeleteit's good idea to help researchers

ReplyDeleteTHANK YOU..BUT K=0.94 IS FOR COPPER RADIATION

ReplyDeleteI THINK YOU HAVE TO ADD OTHER CONSTANT LIKE K=0.89 FOR COBALT RADIATION..

YOURS

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

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

ReplyDeleteVery good, but shape faktor for different crystals is vary.

ReplyDeleteThat has got a necessity for new section which can insert shape factor.

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

ReplyDeleteI agree with your details , fantastic post.

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

ReplyDeleteThanks for this helpful site to find crystal size.

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

ReplyDeleteVery useful

ReplyDeleteOne of our guests not long ago advised the following website.

ReplyDeleteVery useful.

ReplyDeleteThank you for easiness, but have considered the shape factor?

ReplyDeleteWith regard

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

ReplyDeleteif we express theta values and FWHM in radians then crystallite size is different instead of express values in degrees?

ReplyDeleteWhy it is so?

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

DeleteNot wordy, nice explanation and easy to grasp so

ReplyDeleteSmart work..its very useful

ReplyDeleteThis is really helpful. Thanks for saving our time!

ReplyDeletedo you includes contributions of instrumental error?

ReplyDeletenope

DeleteIf both beta and theta in degree then which one have to convert in radians

ReplyDelete