I'll request to follow the steps carefully and it must be done. When you accept its working, modify your comment as a decent one. Thanks

Hi Namratha, thank you so much for your kind words! I'm glad to hear my videos have been helpful for your research. Regarding your question about calculating absorbance from reflectance, you're right that different sources provide slightly different approaches. When dealing with reflection (R) and transmission (T), the general formula is (A = 1 - (R + T)). However, in cases like UV-Vis DRS where you don’t have transmittance data, the common approximation is (A = 1 - R), especially for opaque or highly scattering samples. This is widely accepted for such cases. If you are only working with reflectance data, you can apply this approximation, assuming the transmittance is negligible. For further reading, I suggest looking into papers that cover Kubelka-Munk theory, which is often used in DRS analysis. It provides a good explanation of how reflectance can be related to absorbance. Feel free to ask more questions if needed, and best of luck with your PhD!

This channel have Physics content. I don't have any WhatsApp group. Thanks

Welcome dear 😊

Thanks dear 😊

Thanks for the appreciation dear 😊. You're right, sharing it with others will make their life easy too.

Thank you dear 😊

Yes. Here's a complete playlist for the calculation of crystallite size and other parameters from XRD data. Thanks XRD Data Plotting and Analysis: kzbin.info/aero/PLeWSImvDbpldse8hqfvZX7MasFSkTwIIT

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In this specific integration, there are interdependencies between variables, so this is the only way to solve it. Thanks

Thanks dear

Introduction to Electrodynamics, D. J. Griffiths, 3rd ed. Thanks

Thanks dear

Many thanks for the appreciation dear 😊

@@SAYPhysics Professor, could you please help me to draw 2nd heat curves of DSC data from TA universal analysis software to origin? Really I need it. I sent you an email already.

@mitamunsi1991 right dear. I've received your email, will respond you soon IA

@@SAYPhysics Thanks a lot, I am waiting for the reply as it is urgent, please.

Please synchronize the K and k with the book. In this topic, both K (involved in quantization and e-iKx) and k (the k-space and energy) have been used and it is hard to differentiate while writing (although I have tried my best to differentiate). Thanks

@@SAYPhysicsThanks, got it professor

@Respect-shorts75 you're welcome dear

You're right dear. However, this will limit our audience. Thanks 👍

Thanks for your question! Negative microstrain values can sometimes result from instrumental or sample broadening, incorrect baseline corrections, or errors in peak fitting. Make sure that any instrumental broadening is accounted for and that baseline corrections are applied properly during the calculation process. For a more accurate approach, I suggest trying the Williamson-Hall (WH) plot method. In my other video tutorial, I demonstrate how to calculate both microstrain and crystallite size, including considerations for peak broadening due to strain. You can check it out at kzbin.info/www/bejne/n6G7ZJyLf5qJmbc.

Thanks for your question! You can fit the BCS model equation for heat capacity in Origin by customizing the fitting function. If the BCS equation isn’t pre-defined in Origin, you can manually input it and perform the fit. I recommend watching my video on how to customize fitting equations in Origin for a step-by-step guide. It will help you set up the BCS model specifically. You can find the video at kzbin.info/www/bejne/jJqslXiiodpopqM

@@SAYPhysics I tried fitting it but it's gives me a value of 1.83E-295 for any value of X, my aim to fit C/T vs T^(2) and be able to obtain T_c from the BCS model fit.

@mlungisimasondo3613 it's definitely not a valid value. Try refining your parameters. Thanks

Thanks dear

Glad to know dear. Thanks

Thanks dear

Again I will say that modern software like Xpert HighScore often uses integral breadth for its calculations, not FWHM. What your suggestions based on established literature? Thanks

Thank you for pointing that out! One key difference is that modern software like Xpert HighScore often uses integral breadth for its calculations, not FWHM. This can lead to some variation in the results compared to methods that rely on FWHM, such as those done manually or in Origin. However, difference of exactly two times is to think over and revisit the formulae.

Thank you for your detailed comment! You're right that different methods, like Scherrer and the Williamson-Hall (WH) plot, can yield varying results when calculating microstrain and crystallite size. The Scherrer equation typically gives an average crystallite size, while the WH method accounts for strain by plotting FW(s)×cosθ on the y-axis and sinθ on the x-axis. The slope provides the strain, and the intercept gives the crystallite size. In my video on the WH method, I’ve also explained when to use which relation for more accurate results. I recommend checking that out if you haven’t already-it might help clarify the differences between the approaches. Thanks again for sharing your experience! kzbin.info/www/bejne/n6G7ZJyLf5qJmbc

Welcome dear ☺️

This is good to hear. Thanks for the appreciation dear 😊

Thank you for your understanding and respect. While I have no problem delivering a lecture in Urdu, doing so would restrict my audience primarily to the subcontinent, and I want to reach a broader, more diverse audience. Our accent is clear and easy to understand, and since the textbooks and teaching materials are in English, we maintain English as the language of instruction. I appreciate your support and understanding.

Yes, you're right. Thanks

Regarding your custom fitting function, if the peaks have broad tails, you may need to extend the range beyond the FWHM to account for the additional intensity in those regions. Some methods account for peak asymmetry by using skewed Gaussian or Lorentzian functions, which might help better fit peaks with tails. The key is to ensure that the area considered for each peak reflects the contribution to crystallinity while minimizing the noise or background influence. Let me know if you'd like further details on implementing this in code! Thanks

@@SAYPhysics Thanks for your reply.. Would you please elaborate more?

Thanks for your follow-up! When selecting the symmetric area for CI calculation, consider extending beyond the FWHM for peaks with broad tails to capture all relevant contributions. Using skewed Gaussian or Lorentzian functions can help fit asymmetric peaks better. Start with the FWHM, then gradually extend the range until the intensity stabilizes, minimizing background noise. In your code, calculate the FWHM, define the extended area based on the peak profile, and integrate the area for CI. Testing with known standards will help refine your approach. Let me know if you have any specific questions or need further assistance!

In the start of the few minutes of this tutorial, I have explained about the thickness calculation. Let me know if it needs further clarification. Thanks

Glad you enjoyed it! Thanks for the appreciation. Share it with others too.

You should receive the key via an official (institutional) email. Personal emails are not accepted for this purpose. If your email is institutional but Origin Corporation does not recognize your institution, they may request that you wait for verification. In this situation, you can reply by providing your institution's website address. Thank you.

The units of eV·cm⁻¹ may have caused some confusion. When you divide 1240 by the wavelength in nm, the result is in eV, not eV·cm⁻¹. The factor 1240 comes from converting between energy in eV and wavelength in nm (since 1 eV ≈ 1240 nm for photons). The term "cm⁻¹" refers to the wave number, which is the inverse of the wavelength in cm. If you're calculating the band gap in eV, you don't include the "cm⁻¹" unless you're specifically converting to wave number. Thanks

@@SAYPhysics Thank you very much doctor, but then, why in the video in the bandgap graph the y-axis has units of ev*cm-1, if the conversion was not made? 🥹

@antonioculebro8038 because on the y-axis, we don't have energy but alpha hv. The cm-1 comes there due to the absorption coefficient alpha. Thanks

Thanks for the appreciation dear.

Both coefficients (√3/4 and √1/4) can exist in the same row of the Clebsch-Gordan table, but they correspond to different total angular momentum states. The choice depends on the quantum numbers you are using to describe the system (like j1, j2, M, and the resulting coupled states). Thanks

@@SAYPhysics thanks professor

@Respect-shorts75 welcome dear 😊

The following video will guide you how to insert such symbols. Thanks kzbin.info/www/bejne/Y53HmpqVqLudrtE

After a month IA. Thanks

Glad to know dear 😊. Please share it with others. Same love from the city of SRK, Peshawar, Pakistan. Thanks

Glad it was helpful! Thanks for the appreciation. Share it with others too.

Yes, you're right. I have missed x in the exponential, and there should be z instead of y in the last part, as sin(m pi z/b), not sin(m pi y/b). Thanks for the correction.

Thanks for the acknowledgement dear 😊. Share it with others too as a token of appreciation.

Yes, you're right. <Sx>=0, it is the <Sy> which is equal to -12/25hbar. Thanks for the correction.

Right dear. It will be continued IA. Please share it with others too. Thanks

I know sem and tem will give me proper result... But any rough idea about size of nanoparticles using XRD data

I'll respond to this in your next question. Thanks

XRD can only give you the crystallite size that can be related with the TEM. The size of NPs can be estimated from uv vis absorption data in case of metal or metal oxide NPs. The following videos can help you estimate the sizes. Thanks 😊 kzbin.info/www/bejne/oaW7apieetuLjbM kzbin.info/www/bejne/nIK3hmBrnNl2ibM

@@SAYPhysics thank you sir

@lakhichetry3733 you're welcome dear

You're welcome dear

Thanks for the appreciation dear 😊. Your acknowledgement means a lot to me. Yes, sure. I'll cover all of the remaining QM. Stay tuned.

Magnitude has nothing to do with being a positive or negative, it is the reference point which makes it negative or positive. Thanks

@@SAYPhysics we always write magnitude in Modulus to get positive value of magnitude.. any how it's really good information for me that it's not have any thing with positive or negative

When a ball is thrown upwards, the sign of \( \frac{dx}{dt} \) (velocity) depends on the chosen positive direction. If upward is positive, the velocity is positive while the ball rises but decreases due to gravity, becoming negative as it falls. If downward is positive, the velocity is negative while rising, slows down to zero at the peak, and becomes positive as the ball falls. The choice of reference frame affects whether velocity and gravity are positive or negative, but the motion remains the same in both cases. Thanks

@@SAYPhysics مننه سر