How to calculate crystallite size from XRD data using origin

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  • čas přidán 11. 09. 2024

Komentáře • 302

  • @MuhammadIsmail-sp6uj
    @MuhammadIsmail-sp6uj Před 3 lety +10

    Thanks sir explained in nutshell...

  • @basavajyothi89
    @basavajyothi89 Před rokem +4

    As a beginner in research field ur videos are very helpful for me, thank u for all your videos 😊

    • @SAYPhysics
      @SAYPhysics  Před rokem +1

      Thanks for the appreciation dear. You may share them with others 🙂. Thanks

  • @aishatuhabib2724
    @aishatuhabib2724 Před 10 měsíci +2

    For the past 5 months, i have been trying to figure out how to analyse my results on characterization of nanoparticles but indeed am so excited with this video. God bless you sir.

    • @SAYPhysics
      @SAYPhysics  Před 10 měsíci

      I'm happy for you 😊. Thanks for the appreciation dear

  • @bultiroy978
    @bultiroy978 Před 2 lety +2

    Thank you so much sir. You are a social worker. Thanks for your contribution to our nation.

  • @omsri-science
    @omsri-science Před 2 lety +4

    Precisely explained, thank you so much. Wishing the best for you, sir! :)

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +1

      Thanks for the appreciation dear

  • @kryptic9056
    @kryptic9056 Před 2 lety +2

    thank you so much. You're helping me more than my supervisor.

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for the appreciation dear ☺️

  • @kvraoy
    @kvraoy Před 3 lety +1

    Straightway explanation and very good explanation thank you

    • @SAYPhysics
      @SAYPhysics  Před 3 lety

      Glad to know you liked it. Thanks

  • @miftahulkhair4416
    @miftahulkhair4416 Před rokem +1

    thank you Sir for the clear, simple, nice explanation of this important topic.

    • @SAYPhysics
      @SAYPhysics  Před rokem

      Thanks for the appreciation dear 😊

  • @faragayad4513
    @faragayad4513 Před 2 lety +1

    Realy the explanation is very helpful , thank so much

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for the appreciation dear 😊

  • @brilliantedutech1629
    @brilliantedutech1629 Před rokem +1

    Super clearly presented

    • @SAYPhysics
      @SAYPhysics  Před rokem

      Thanks for the appreciation dear 😊

  • @mhusain6185
    @mhusain6185 Před 3 lety +1

    BY FAR THE BEST VIDEO

  • @preciousekwere1508
    @preciousekwere1508 Před 2 lety +1

    Thank you sir for the excel. God bless you

  • @hope2633cu
    @hope2633cu Před rokem +1

    excellent really appreciate you i use it and learn one main point todayyy woww

    • @SAYPhysics
      @SAYPhysics  Před rokem

      Thanks for the appreciation dear 😊

  • @analist-3000
    @analist-3000 Před 7 měsíci +1

    Thanks for information and files

  • @brhaneamha
    @brhaneamha Před 3 lety +1

    Great ; This topic is not my major But I loved it and watched it, man !!! Keep it up !!

    • @SAYPhysics
      @SAYPhysics  Před 3 lety

      I’m glad and honoured to know. Thanks Brhane

  • @SaadANawaz
    @SaadANawaz Před 2 lety +1

    Mashallah, what a great explanation. Thank you sirr

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for the appreciation dear 😊

  • @mujeebrahman5848
    @mujeebrahman5848 Před 2 lety +1

    Great explanation sir

  • @learnwithfaixoo7570
    @learnwithfaixoo7570 Před rokem +1

    Finally i am on right video to determine size from xrd

    • @SAYPhysics
      @SAYPhysics  Před rokem +1

      Thanks for the appreciation dear 😊. Share it please with others too.

  • @user-dc1bz4qu8p
    @user-dc1bz4qu8p Před 4 měsíci +1

    Thank you so much, sir.
    I have two question about crystallite size.
    1. Why scherrer equation is valid for crystallite size up to 200 nm??
    2. If crystallite size larger than 200 nm, how to calculate crystallite size??

    • @SAYPhysics
      @SAYPhysics  Před 4 měsíci

      Thank you for watching the tutorial! Great questions.
      1. The Scherrer equation is often considered valid up to 200 nm because it assumes a uniform distribution of crystallites and does not account for factors like strain or defects. Beyond 200 nm, other factors start to become significant, potentially affecting the accuracy of the Scherrer equation.
      2. For crystallite sizes larger than 200 nm, alternative methods such as electron microscopy or small-angle X-ray scattering (SAXS) can be more appropriate. These techniques offer higher resolution and can provide more accurate measurements for larger crystallites. Feel free to explore these options based on the size range you're working with.

  • @pinkblossomsky
    @pinkblossomsky Před rokem +1

    Thanks! You're the best!

    • @SAYPhysics
      @SAYPhysics  Před rokem +1

      Thanks for the appreciation 🙂 dear

  • @chamlaphysics2625
    @chamlaphysics2625 Před 8 měsíci +1

    Thanks Dr Sb.Well Explained

  • @j0unce772
    @j0unce772 Před 2 lety +1

    Wow that was a great explanation!

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for the appreciation dear 😊

  • @syedshaheenshah2482
    @syedshaheenshah2482 Před 3 lety +1

    Interesting, thank you Sir for the nice explanation of this important topic.

  • @imydz466
    @imydz466 Před rokem +1

    Thank you. It is well explained

  • @LearnMathwithZain
    @LearnMathwithZain Před 3 lety +1

    Great work Sir
    #LearnMathwithZain

  • @keilanecruz5009
    @keilanecruz5009 Před rokem +1

    thank you soooo much

  • @user-ty2pw8yf1n
    @user-ty2pw8yf1n Před 6 měsíci +1

    감사합니다. 당신이 내 구세주입니다.

  • @aarondanielrodriguezrodrig2442

    Thank you very much!

  • @usmanali-t4b7w
    @usmanali-t4b7w Před rokem +1

    awesome sir well explainned

  • @khizarhussain632
    @khizarhussain632 Před rokem +1

    😮😮 it's an amazing video

    • @SAYPhysics
      @SAYPhysics  Před rokem +1

      Thanks for the appreciation dear 😊

  • @learningwithmajid850
    @learningwithmajid850 Před 3 lety +1

    amazing sir jazakALLAH,,keep it sir and thnx

  • @hinaimtiaz2332
    @hinaimtiaz2332 Před 2 lety +1

    Thank You so much MAy ALLAH bless u Ameen

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for the appreciation dear 😊
      Aameen

  • @SNTOfficial_786
    @SNTOfficial_786 Před rokem +1

    Sir, our Xc value is 30.21472
    and FWHM is 0.93755
    And divided xc/2 than cos is 15.107 and Cos(15.107) = -0.82
    What should do now?? It's giving a negative value

    • @SAYPhysics
      @SAYPhysics  Před rokem

      Please use the Excel template which is provided in the video description. It will do all the calculations for you. Thanks

  • @anciyasantha9779
    @anciyasantha9779 Před 9 měsíci +1

    Very nice presentation sir

    • @SAYPhysics
      @SAYPhysics  Před 9 měsíci

      Thanks for the appreciation dear 😊

  • @TheTedgwada
    @TheTedgwada Před 6 měsíci +1

    Thanks for the Video. Can you explain me why are you ignoring the broadening of the instrument ?

    • @SAYPhysics
      @SAYPhysics  Před 6 měsíci

      Thanks for watching the tutorial! In that specific video, I focused on crystallite size calculation. If you're interested in instrument broadening, check out the complete series at [czcams.com/video/0PUTB0IhlWw/video.html&pp=iAQB]. It covers various aspects of XRD data analysis.

  • @1971mn
    @1971mn Před 2 lety +1

    Thank you for your lecture..

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for the appreciation dear 😊

  • @chellapparajamani876
    @chellapparajamani876 Před 2 lety +1

    Very useful in my ph. D sir. Put up the video vickers microhardness testing. How I calculated meyers index n value.

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      There's a formula for the Vicker's hardness in which you use the diagonal lengths of the microindentation. I'll try to include it in the tutorial series. Thanks

  • @josejuangarciagarcia2807

    Amazing, your video helps me a lot.

  • @arthurgonzaga1027
    @arthurgonzaga1027 Před 2 lety +1

    @SAYPhysics why do you use 𝜽 Bragg’s angle in degrees? 𝜽 Bragg’s angle in Scherrer formule should be in radians as beta?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      You're right for β to be in radian, while 𝜽 is to be in degrees. I've confirmed it through literature as well as the modern software (like Highscore etc.) are using the same to calculate the crystallite size. Thanks

  • @ANILKUMAR-bs8gz
    @ANILKUMAR-bs8gz Před 9 měsíci +1

    Thanks alot sir

  • @aqibaziz1792
    @aqibaziz1792 Před 2 lety +2

    Thanku sir.. but I have a question that how crystallite size affect the properties of a material and which one is better, a small crystallite size or large? Thank you

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +2

      I don't know which properties you're interested in. Let me say few things generally. The smaller the crystallite size, the more efficient the sintering (lower possible sintering temperature). However, the drying and firing shrinkage will be very high. The hardness of a material increases as the crystallite size decreases. Thanks

  • @sumbulz768
    @sumbulz768 Před 2 lety +1

    Excellent video and excellent explanation..one question though ..why didnt u take xray wavelength in nm in sherrer equation ..ur answer was in nm but xray wavelength was in angstrom

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +1

      Thanks for the appreciation dear. Answer will remain the same. If you convert it first from angstrom to nm or later. From the machine specs, we get it in angstrom. Thanks

  • @abhishekkj9664
    @abhishekkj9664 Před 2 lety +2

    How do you assign error valuation for the average crystallite size ?
    Assuming contribution from instrument and strain be ignored, what is the error contribution from FWHM ?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      As you know, different peaks have been taken into account instead of the very intense peak only. The average crystallite size covers many contributing factors. Thanks

  • @kalaipriyaaesthetics4042
    @kalaipriyaaesthetics4042 Před 2 lety +1

    Thank you sir for clarifications on grain vs particle. I wat to know why we took Beta is in radians and cos theta in degrees?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +1

      Thanks for the appreciation dear. Theta is the goniometer angle, plotted as 2theta on the x-axis, which is in degrees as we move it around the sample. While beta is the measure of fwhm, which by its equation in radians. These are as per literature, whose references are provided in the video description. Thanks

  • @dailyquotesforeveryone
    @dailyquotesforeveryone Před rokem +1

    i have origin 2022 pro and when i click on ok i do not see this complete data sheet as you got in the video instead i am getting that small grey box only

    • @SAYPhysics
      @SAYPhysics  Před rokem

      I can't say anything blindly. You may send the screen recorded video at sayphysics@gmail.com. I'll respond to that. Thanks

  • @yousafafridi8931
    @yousafafridi8931 Před 3 lety +1

    Thank you sir its too helpFul

  • @aleksandralaska3767
    @aleksandralaska3767 Před 2 lety +2

    Thank you for this video! I have a question - is it possible that if the grains become smaller, the crystallites grow? In my studies, during the processing of aluminium sheets, the grains are much smaller after the process, but the crystallites are larger. Is it possible?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +1

      Grain is either a single crystalline or polycrystalline material, and is present either in bulk or thin film form. During the processing, smaller crystallites come closer and grow to become larger due to kinetics. Therefore, in the most likely scenario, the grain is larger than a crystallite. Thanks

  • @JorgeOliveira-li9lr
    @JorgeOliveira-li9lr Před 2 lety +2

    Nice explanation! Shouldn't we consider the weighted arithmetic average? As the peaks have different intensities, they contribute differently in the average.

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +3

      Somehow you're right Oliveira. Let me elaborate what do you mean actually. The weighted arithmetic mean is similar to an ordinary arithmetic mean (the most common type of average), except that instead of each of the data points contributing equally to the final average, some data points contribute more than others.
      Now, if all the weights are equal, then the weighted mean is the same as the arithmetic mean. The facts is, as different peaks represent the same crystallites but their different planar orientations, so, both the averages will give the same result. Thanks

  • @izazahmad5058
    @izazahmad5058 Před 2 lety +1

    Many Thanks sir

  • @the_rare_soul
    @the_rare_soul Před rokem +1

    Sir, when we take cos theta...do we need to take half of 2 theta..but in the video, you directly take 2theta value to the following step.

    • @the_rare_soul
      @the_rare_soul Před rokem +1

      Also , rad to degree conversion is incorrect ..need to multiply with pi and divide with 180

    • @SAYPhysics
      @SAYPhysics  Před rokem

      In the vIdeo description, I have attached an Excel file, where all calculations are being done. Thanks

    • @SAYPhysics
      @SAYPhysics  Před rokem

      Here, Origin gives the value in radian and in the Scherrer equation we need it in degrees,, so in Excel template, which is available in the video description , we convert it into degrees. Whatever calculations you are saying, Excel is doing it itself. Use the provided template for calculations, please. Thanks

  • @physicswithghani7126
    @physicswithghani7126 Před 3 lety +1

    many many thanks sir very intersting

  • @Sree_madness
    @Sree_madness Před rokem +1

    As per me, you are considering the doublet as a single reflection and collecting the FWHM (for example at video time 6:20 min). Is this the correct way to estimate the crystallite size, I don't think so you are getting accurate crystallite sizes.

    • @SAYPhysics
      @SAYPhysics  Před rokem

      You are right. In general, it is not recommended to consider a doublet as a single reflection for calculating the crystallite size from XRD data, as this can lead to inaccuracies in the results. This is because a doublet typically indicates the presence of multiple crystallographic domains or twinning, and treating it as a single reflection can result in an overestimation of the crystallite size. However, it is always better to consider all the peaks and then finally take the average of all the calculated crystallite sizes, as explained at 07:50 of the video. As a cross-check, we can compare the calculations from a single reflection and then of the average to confirm these calculations in the video. Thanks for your input.

    • @Sree_madness
      @Sree_madness Před rokem +1

      @@SAYPhysics taking average can help you when consider particular hkl planes , per example 00l. Every reflection indicates/arising from particular hkl plane. Insuch case you can not consider doublet as a singlet. If you want to get quatitateve size then you have to use siglet only. I can not agree singlet can giver over estimation of size. And one more thing origin program not use for quantitative data it's just for rough idea.
      For more basics kindly read any crystallography text.

    • @SAYPhysics
      @SAYPhysics  Před rokem

      Right. Thanks for your input.

  • @ishfaqchohan7297
    @ishfaqchohan7297 Před 2 lety +1

    @SAYPhysics Sir very helpfull video. Thanks very much.... I have a short question.... Sir in your Excle formula you devided the whole formula 10 that is not existed in equation? can u elaborate for me plz

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for the appreciation dear. Division by 10 is just the conversion to nm. Thanks

    • @ishfaqchohan7297
      @ishfaqchohan7297 Před 2 lety +1

      Thanks....your highness

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks dear

  • @anisahsajidahsaud1327
    @anisahsajidahsaud1327 Před 3 lety +1

    Hi, is the excel calculation u show based on the the peak in earlier video? They were 5 peaks there but ur excel is showing 6 values. Another question is, can we just select the 3 or 4 intense peak for calculation? Thanks

    • @SAYPhysics
      @SAYPhysics  Před 3 lety +3

      Dear Anisah, This excel file is for the calculations in this video only. You can select peaks of your choice. There's no restriction on the number. Sometimes, some peaks cause a problem in fitting etc., they may be ignored. Thanks

  • @friendlydango
    @friendlydango Před 2 lety +2

    I have a question. Does the peak heights represent the relative amount of those size crystallites? If so, then shouldn't taking a _weighted_ average yield a more accurate answer?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +2

      If all the weights are equal, then the weighted mean is the same as the arithmetic mean. The facts is, as different peaks represent the same crystallites but their different planar orientations, so, both the averages will give the same result. Thanks

  • @seeminomi5413
    @seeminomi5413 Před 3 lety +1

    Very helpful video

  • @muhammadsahil6437
    @muhammadsahil6437 Před 2 lety +1

    thank you sir for this video.. but I have one objection.
    you wrongly explained the grain.
    Grain size can not be calculated through XRD.
    Grain size is the pattern size that we can observe through TEM or SEM. we assume that pattern as Grain. and then we compare this grain size with crystallite size. And consider this pattern approximately equal to crystallite size. Conclusion: crystallite size calculates through XRD and grain size calculates through TEM OR SEM study.

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for your input dear. I haven't calculated grain size but the crystallite size. I have explained the difference between crystallite and grain size clearly. I'll appreciate to watch the tutorial carefully and thoroughly instead of jumping to a wrong conclusion. Thanks

  • @ziyadukyli2382
    @ziyadukyli2382 Před 2 lety +1

    thanks a lot Sir

  • @ph_ahmedali2207
    @ph_ahmedali2207 Před měsícem +1

    Please sir we see that from TEM mean particle size have low magnitude compared with the value of crystal size … so does particles size smaller than crystal size or inverse please shown me this problem

    • @SAYPhysics
      @SAYPhysics  Před měsícem +1

      The smaller mean particle size from TEM compared to crystal size could be due to differences in measurement techniques, where TEM measures actual particle size, and XRD measures crystallite size. Factors like particle aggregation, polycrystallinity, surface effects, and differences in sample preparation or calibration can also contribute to this discrepancy. If you have more details on the measurements, I can offer more specific insights. Thanks

    • @ph_ahmedali2207
      @ph_ahmedali2207 Před měsícem

      ⁠ I used a pulsed laser in 5 ml of distilled water and deposited the colloidal solution on small pieces of silicon for the purpose of performing an XRD test. I performed a TEM of the solution and the results were as I indicated. There is a difference between the TEM and XRD values ​​with respect to the crystal size and particle size, noting that the targets were silver and gold. @@SAYPhysics

  • @didinsahidin3574
    @didinsahidin3574 Před 2 lety +1

    tq for explenation

  • @vitalidoroshenko9134
    @vitalidoroshenko9134 Před rokem +1

    Hello! Thank you for your video. At the beginning you said that this method is valid for crystallites up to 200 nm. Could you please indicate the source where this is stated?

    • @SAYPhysics
      @SAYPhysics  Před rokem

      Thanks for the appreciation dear 😊
      In the video description, you will find some relevant references.

    • @vitalidoroshenko9134
      @vitalidoroshenko9134 Před rokem +1

      @@SAYPhysics Thank you for the reply! Do you know of any book or article that states this?

    • @SAYPhysics
      @SAYPhysics  Před rokem

      Thank you for watching my video and for your question regarding the validity of the method for calculating crystallite size using XRD data in Origin.
      The information about the method's validity for crystallites up to 200 nm comes from various references, for example, as mentioned below.
      www.nature.com/articles/nnano.2011.145
      www.ncbi.nlm.nih.gov/pmc/articles/PMC6459976/
      www.sciencedirect.com/topics/chemistry/scherrer-equation
      prism.mit.edu/XRAY/oldsite/CrystalSizeAnalysis.pdf

  • @arsalanadil14912
    @arsalanadil14912 Před 3 lety +1

    Great video sir

  • @yufu7122
    @yufu7122 Před 2 lety +1

    Thank you for your good explanation! Well, I have a question on my research topic: can we calculate the diameter of nanowires (NWs) or thickness of nanoplates(NPls) directly on the Scherrer equation? How can we define K and can the result after calculation represents the real thick/ diameter of NWs/NPls? Looking forward to your reply!

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for the appreciation dear. Yes, but change in symmetry is reflected in the graph shape. These calculations are for simple symmetry. While, when there will be another symmetry, calculations will slightly modify. Thanks

  • @ZaidLHadi
    @ZaidLHadi Před 2 lety +1

    Genius

  • @fazalurrehman9446
    @fazalurrehman9446 Před rokem +1

    well explain but plz also tell me about that how you convert beta value which is in radian into another value

    • @SAYPhysics
      @SAYPhysics  Před rokem

      The value of beta is provided by origin in radian which I have converted into degrees as per requirements of the equation. Please check in the video description the Excel file in which I have done the calculations you can see in the file that when you take radian of a radian value then it is converted into degrees. Thanks

  • @hidayahidris5167
    @hidayahidris5167 Před 2 lety +1

    This video helps me a lot, really appreciate it. Thanks very much 🙂

  • @louizaguerguer5369
    @louizaguerguer5369 Před 2 lety +1

    Thank you sir,I have a questions,What is the accuracy of XRD measurment? How to determine the reliable accuracy of the crystallite size.

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      XRD, being a technique, is very accurate and reliable. However, calculation of crystallite size depends on accurate measurement of internal parameters like fwhm etc. Thanks for the appreciation dear 😊

    • @louizaguerguer5369
      @louizaguerguer5369 Před 2 lety +1

      Thank you very much,sir.

  • @wafazina4225
    @wafazina4225 Před 3 lety +1

    This video is perfect, so in amorphous material we can't clculate the crystalline size , because it has not grain and boundary grain, is it right or no,
    Thank you so much

    • @SAYPhysics
      @SAYPhysics  Před 3 lety

      Thanks for the appreciation. Yes, in amorphous, we don't have long range order, so crystallite or grain size calculations are not possible. You're welcome Wafa....

  • @peterndungu3615
    @peterndungu3615 Před 2 lety +1

    Awesome video. Where can I download winXpow software?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for the appreciation dear. You'll have to purchase it or some universities provide it.

    • @peterndungu3615
      @peterndungu3615 Před 2 lety +1

      @@SAYPhysics Thank you for the information.

  • @aymenfarhi1425
    @aymenfarhi1425 Před 4 měsíci +1

    salam alaykom, I don't get the two tables in left of page, how can get them to read FWHM please ?

    • @SAYPhysics
      @SAYPhysics  Před 4 měsíci +1

      In the data worksheet, there comes two extra sheets after fitting, click the first to see the tables. Thanks

    • @aymenfarhi1425
      @aymenfarhi1425 Před 4 měsíci +1

      @@SAYPhysics Thank you brother

    • @SAYPhysics
      @SAYPhysics  Před 4 měsíci

      @aymenfarhi1425 you're welcome dear

  • @SNTOfficial_786
    @SNTOfficial_786 Před rokem +1

    What would be K for tetragonal particle??

    • @SAYPhysics
      @SAYPhysics  Před rokem

      k is a dimensionless shape factor (typically around 0.9). The value of k in the Scherrer equation is not dependent on the type of Bravais lattice. It is a constant that depends on the shape of the crystalline domains and the instrument used. In practice, k is often determined experimentally by measuring the size of well-characterized standard samples with known crystallite sizes. Thanks

  • @rihzel
    @rihzel Před 2 lety +1

    Thanks for this explanation sir. I have a question please: for metals, is Crystallite size = Grain size (the same thing)?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +1

      For a single crystal, they're the same only. Thanks

    • @rihzel
      @rihzel Před 2 lety +1

      @@SAYPhysics an other question please, when the crystallite size increase ,the grain size also increase? And vice versa, that it's true or not?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +1

      Principally, yes. Thanks

    • @rihzel
      @rihzel Před 2 lety

      @@SAYPhysics thanks sir

  • @user-mm9hl3bb6m
    @user-mm9hl3bb6m Před 2 lety +1

    something wrong with your excel. equation of D is gone. can you fix it, thank you.

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      I rechecked. It's OK. If you mention specifically what is wrong with it, I'll address that. Thanks

  • @md.joherulalamchanchal63
    @md.joherulalamchanchal63 Před 2 lety +1

    Thanks for the video. In my case, it is showing "Fit did not converge. Maximum iteration setting of 400 was reached" COD9R^2 =n 0.38 only. what could be the reason?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Thanks for the appreciation dear. It seems that you have fitting issues with your peak(s). I'll suggest you to watch the videos in the playlist to best fit your peaks. Thanks
      czcams.com/play/PLeWSImvDbplc5W0ru6UNKcq6tEJsYxPmY.html

  • @sunandobanerjee7665
    @sunandobanerjee7665 Před rokem +1

    WHY THR FWHM VALUE IS NOT SHOWING, OTHER VALUES LIKE Y X W ARE SHOWING BUT NOT FWHM, I FOLLOWED SIMILIAR PROCEDURE.

    • @SAYPhysics
      @SAYPhysics  Před rokem

      After fitting, it generates the table and an extra sheet to the data in which you have all the parameters including FWHM. Thanks

  • @XXKASRXX
    @XXKASRXX Před 2 lety +1

    If I have a composite material, the x-ray wavelength will be the sum of the components at that material??

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +1

      The incident wavelength of X-ray is of the machine, independent of the nature of a material being characterized. Thanks

  • @mohamedsadik7708
    @mohamedsadik7708 Před 3 lety +2

    How do you convert the FWHM value to radian on the calculator?

    • @SAYPhysics
      @SAYPhysics  Před 3 lety +1

      One radian = 180/ PI degrees and one degree = PI /180 radians. Therefore to convert a certain number of degrees in to radians, multiply the number of degrees by PI /180 (for example, 90º = 90 × PI /180 radians = PI /2). Thanks

  • @colanifakude9501
    @colanifakude9501 Před 2 lety +1

    Good explanation. I am just worried about the Angstrongs ie 1.5406 when the final answer is in nm. Why didn't you use 0.15406 nm wavelength instead?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +1

      Thanks for the appreciation. You're right. We have the value from machine specs in the Angstroms, but you can convert as per your ease.

    • @colanifakude9501
      @colanifakude9501 Před 2 lety +1

      @@SAYPhysics Thanks man. Keep up the good work.

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Sure. Thanks

  • @somtirthadas6610
    @somtirthadas6610 Před 4 měsíci +1

    Sir if lambda is in angstrom the crystallite size is also in angstrom isnt it?

    • @SAYPhysics
      @SAYPhysics  Před 4 měsíci

      It doesn't matter if we use Angstrom or nanometer, provided the dimensions are accurate. Thanks 😊

    • @somtirthadas6610
      @somtirthadas6610 Před 4 měsíci +1

      @@SAYPhysics thank you sir

    • @SAYPhysics
      @SAYPhysics  Před 4 měsíci

      You're welcome dear

  • @Andresordojim
    @Andresordojim Před 3 lety +1

    Thanks for the explanation, but I wonder why Cos(theta) is in degrees and not radians are used

    • @SAYPhysics
      @SAYPhysics  Před 3 lety

      As per formula, beta should be in rad and cos must be degrees to make the equation dimensionally correct. Thanks

  • @hichembenz1608
    @hichembenz1608 Před 2 lety +1

    Hello Sir, I sow other videos they radians not degree for bragg position , wich one is correct

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      The one that I did is correct as I confirmed it from an extensive literature. Thanks

  • @siddigmohieldein7285
    @siddigmohieldein7285 Před 2 lety +1

    If my particle in range of 100 micro can I use it to calculate the crystal size? Also if I have 2 or 3 compounds in one material can I calculate the size for all together?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety +2

      I have explained that crystallite, grain and particle size are different things. Yes, you can calculate the crystallite size. Thanks

  • @tamizhchemist
    @tamizhchemist Před 2 lety +1

    Sir how do you get radian value

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      The Excel template is provided in the video description where all calculations are clearly mentioned. Thanks

  • @Hafizsajjad77
    @Hafizsajjad77 Před rokem +1

    Assalamo allikum sir! kindly provide the link of paper you mention at 2:06 .

    • @SAYPhysics
      @SAYPhysics  Před rokem +1

      This is the doi of the article. If you can't download it, let me know at sayphysics@gmail.com. Thanks dx.doi.org/10.1107/s0021889878012844

  • @saltanatauyez611
    @saltanatauyez611 Před 2 lety +1

    I don't understand, how you converted radian to number? i mean could you explain how radian(0,23779) is (0,004150218)
    for example, my data is radian(0,21594) how much it will be like number?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      In the video description, I have attached an Excel file in which all these calculations are being done. Please note that Excel calculate the value by default in radian, so to have it in degrees, we take its radian again. Thanks

    • @saltanatauyez611
      @saltanatauyez611 Před 2 lety +1

      @@SAYPhysics Thank you very much!

  • @arjunkrishna4815
    @arjunkrishna4815 Před 3 lety +2

    sir can you define what is microstrain in simple words?

    • @SAYPhysics
      @SAYPhysics  Před 3 lety

      Here, I have explained microstrain in details. Thanks czcams.com/video/oUwJpGtwOVk/video.html

  • @sachinkrao3010
    @sachinkrao3010 Před 2 lety +1

    Sir, where did you get the XRD data from?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      I didn't get your question what exactly you want to ask. Thanks

  • @sanianaseer2522
    @sanianaseer2522 Před 2 lety +1

    Good explaination. where can i get similar sort of excel sheet for hexagonal system?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      In the description of the video you can find the Excel file. For a hexagonal structure, you may find structural equations in the literature. Thanks for the appreciation dear.

  • @zijunryanmeng3921
    @zijunryanmeng3921 Před 3 lety +1

    8:27 the first peak is quite an outlier it seems, should it be removed from average or is it common.

    • @SAYPhysics
      @SAYPhysics  Před 3 lety

      In my case, it is not an outlier. However, sample to sample it may not be the case. Thanks

  • @zahidsarfraz4361
    @zahidsarfraz4361 Před rokem +1

    Very nice

  • @michipiscitelli
    @michipiscitelli Před rokem +1

    Thank you so much. I would like to know how to claculate size around 0.5 nm, Do you have a paper with that information? Thank for that

    • @SAYPhysics
      @SAYPhysics  Před rokem

      Thanks for the appreciation dear. There's an upper limit to the Scherrer's equation only. I'm sorry, I don't have any ready reference where calculations have been made for 0.5 nm or less. If I came across one, I'll let you know. Thanks

  • @karimjaved9649
    @karimjaved9649 Před rokem +2

    Sir how we find limit of detection and limit of quantification in orign....

    • @SAYPhysics
      @SAYPhysics  Před rokem +1

      To find the limit of detection (LOD) and limit of quantification (LOQ) in OriginLab, you can follow these steps:
      Prepare your data: Ensure that you have experimental data with a series of known concentrations of your analyte or target compound.
      Create a calibration curve: Plot the signal response (e.g., peak area, intensity, absorbance) against the corresponding concentrations of your standards. You can use regression analysis to fit a curve to your data points. OriginLab provides various regression models such as linear, polynomial, exponential, etc., which you can choose based on your data.
      Determine the LOD: The LOD is typically defined as the concentration at which the signal-to-noise ratio (S/N) is above a certain threshold (commonly 3:1 or 2:1). To calculate the LOD, you need to estimate the noise level of your data. You can select a region of the baseline or noise in your data and determine its standard deviation (SD) or root mean square (RMS) value.
      a. Select a region of the baseline or noise in your calibration curve.
      b. Calculate the SD or RMS value of the selected region.
      c. Determine the LOD using the formula: LOD = (SD or RMS of noise) × (S/N threshold).
      Determine the LOQ: The LOQ is typically defined as the concentration at which the S/N is higher than a certain threshold (commonly 10:1 or 3:1). To calculate the LOQ, you can follow similar steps as for LOD:
      a. Select a region of the baseline or noise in your calibration curve.
      b. Calculate the SD or RMS value of the selected region.
      c. Determine the LOQ using the formula: LOQ = (SD or RMS of noise) × (S/N threshold).
      OriginLab provides various analysis tools to calculate the regression parameters, determine the noise level, and perform the calculations mentioned above. You can refer to the OriginLab documentation or help resources specific to your version of the software for detailed instructions on using these tools and functions. Thanks

    • @karimjaved9649
      @karimjaved9649 Před rokem +2

      Sir how we find slope ,R² value , standard deviation in orign.....

    • @SAYPhysics
      @SAYPhysics  Před rokem

      The following tutorial is explaining about your queries. There are even more in the playlist. Thanks
      czcams.com/video/QESi2dDRnrw/video.html

  • @joshuacarranza349
    @joshuacarranza349 Před 2 lety +1

    gracias!

  • @qaneetayounas5090
    @qaneetayounas5090 Před 2 lety +1

    sir please share the video how to calculate the intrumental broadning of XRD?

    • @SAYPhysics
      @SAYPhysics  Před 2 lety

      Here are the videos
      czcams.com/video/tO_frVLxmsg/video.html
      czcams.com/video/238yM0OgaH8/video.html

  • @traveltreetaotao
    @traveltreetaotao Před rokem +1

    Hi Sir, thanks for your video, but how to get cos(44.95887/2) to 0.924016? I calculated cos(44.95887/2) with excel is -0.883138 ? can you explain ? thank you.

    • @SAYPhysics
      @SAYPhysics  Před rokem +1

      It seems there might be a misunderstanding in the way Excel is handling the angle measurement. In trigonometry, the angle is typically measured in radians, not degrees. When you calculate cos(44.95887/2) in Excel, it is interpreting the angle as degrees unless you explicitly specify it to be in radians.
      To get the correct result, you should convert the angle to radians before calculating the cosine. Here's the correct way to do it:
      Convert 44.95887 degrees to radians:
      Angle in radians = 44.95887 * (π / 180)
      Angle in radians ≈ 0.78539816339 radians
      Now, calculate the cosine of half the angle:
      cos(44.95887/2) ≈ cos(0.78539816339)
      Using a calculator or Excel (making sure the mode is set to radians), you should get:
      cos(44.95887/2) ≈ 0.924016 (rounded to 6 decimal places)
      So, the correct value for cos(44.95887/2) is approximately 0.924016, not -0.883138. The negative value you obtained might have been a result of using degrees instead of radians.
      Thanks

  • @ishach835
    @ishach835 Před 3 lety +1

    I was searching for schrerrers const value for mgcr2O4 and sae your comment on research gate .

  • @siskabudi459
    @siskabudi459 Před 3 lety +1

    I want to ask sir, Is Deby Scherer crystal size 1 dimensional? because it has units of nm

    • @SAYPhysics
      @SAYPhysics  Před 3 lety

      It's a diameter of a crystallite not length. Thanks

  • @chidinmajudith7679
    @chidinmajudith7679 Před 3 lety +1

    Please Sir after fitting and saying okay. My origin only gives the other values but no FWHM, instead it gives R-Square (COD) and Adj. R-Square. Please how can I get the FWHM

  • @tanyaramsail9349
    @tanyaramsail9349 Před rokem +1

    How to select the particular peak?

    • @SAYPhysics
      @SAYPhysics  Před rokem

      At time stamp 05:55 in this video, I have explained how to select a given peak. Thanks

  • @sumitdubal7887
    @sumitdubal7887 Před 2 lety

    Thank you sir...👏