The unit of Bandgap is in electron volt. when doing calculations, you have to either use SI units or CGS. Stick to one unit. it does not matter what you use. Make sure you convert all the variables to same unit.
hey good day to find absorbance value from transmittance data the formula is 2-log(T) right Can i know the value of T shld be in percentage (%) or we need to divide with 100 then find absorbance ?
Hey Tivendren, As far as I know, %T = antilog (2 - absorbance). I don't think you have to divide by 100. There is a table here from Sigma that might help you clear that:www.sigmaaldrich.com/US/en/support/calculators-and-apps/absorbance-transmittance-conversion
You can cite this: pubs.acs.org/doi/10.1021/acs.jpclett.8b02892 for the method. Specific papers for ZnO: link.springer.com/article/10.1007/s13391-019-00173-4 or this: onlinelibrary.wiley.com/doi/pdf/10.1002/pssb.201700393
@@proprocessengineer7231 respected I have calculated data using material studio simulation software and now I want to calculate the optical band gap and don't know how I have to calculate?
@@ShakeelAhmad-tf6lz I would suggest to compare your simulated values with experimental data. If they are similar, you can use the same process to calculate bandgap.
god job, I have a question, is this technique used only for smooth films? (sputering, PVD,...) or does it also work for rough surfaces? (dip coating, spin coating,...)
Thanks. While I am not an expert in optics, I can say that if your sample's roughness is small, it shouldn't matter. The reason is the wavelength range of light used will be bigger than roughness and will not be able to resolve it. For instance, in my samples, I have nanowires 50 nm apart and are not continuous and the smallest wavelength of light used during the UV-Vis scan is 250 nm, which can not resolve 50 nm. So, in this case, you can assume your film to be continuous or smooth. BTW, my films were made by spin coating. I have tried similar for samples made by PVD and it works. But keep in mind that ZnO derived by PVD and spin coating will have different quality (crystalinity and orientation), so their bandgap might not be the same. I hope this helps.
To understand why K (proportionality constant) is popped out, you will have to refer to the derivation of the equation. You can probably start off from here and find older publications from the references: pubs.acs.org/doi/10.1021/acs.jpclett.8b02892. BTW, it does not matter what the value of K is, since the x-intercept will not change. X-intercept is what we need to determine the bandgap. There is a way to determine K (also known as Urbach energy) can be calculated. You can read these papers for more info: 1) www.sciencedirect.com/science/article/pii/S0749603615302615?via%3Dihub 2) www.sciencedirect.com/science/article/pii/S0925838815303352?via%3Dihub
@@proprocessengineer7231 thanks for your reply. Nevertheless, this approach is the same that you are applying just the differences is that they are using f(R) instead of alpha. And they are also skipping the B (in our case K) value. I understand that you need the value to measure de band gap but as I said before if you do the algebra you will notice it is necessary that constant because you the constant if you graph the y and x axis as you present (and conventionally is presented) I am not sure if I not getting something else here. What I mean, what are the considerations to pop k out? It approaches to 1 in this circumstance? Also in the first paper they substitute the kubelka-munk function for alpha. But this function is equal to the absorption coefficient over the scattering coefficient. So another way to see this is that this scattering coefficient be equal to the constant K. So that what you and other research do in their research is correct. If that doesn't happen it doesn't makes sense. So if you have any information regarding this I would really appreciate it. I have been trying to understand what happen to this constant but I haven't got into the right answer yet. Thanks in advance!
@@davidyong5928 David, K is not something popped out. It is a proportionality constant added after the proortionality is defined. This is probably in the seminal work by Davis and Mott. Should I happen to find it, I will share it with you. Thank you for bringing this up. I created this channel for having such critical discussions that can provide clarity to everyone.
Thanks for your great explanations
Clear explanation. Thank you
very helpful
helpful, txs
Which is more appropriate, considering the thickness in cm or meter?
The unit of Bandgap is in electron volt. when doing calculations, you have to either use SI units or CGS. Stick to one unit. it does not matter what you use. Make sure you convert all the variables to same unit.
hey good day
to find absorbance value from transmittance data the formula is 2-log(T) right
Can i know the value of T shld be in percentage (%) or we need to divide with 100 then find absorbance ?
Hey Tivendren,
As far as I know, %T = antilog (2 - absorbance). I don't think you have to divide by 100.
There is a table here from Sigma that might help you clear that:www.sigmaaldrich.com/US/en/support/calculators-and-apps/absorbance-transmittance-conversion
This is only for thin film or nanoparticles also possible sir
MOst likely suitable for thin films
How to calculate alpha if we have liquid uv
Please check if this helps: www.ncbi.nlm.nih.gov/pmc/articles/PMC6716051/
thank you, very helpful. there is any reference for the citation?
You can cite this: pubs.acs.org/doi/10.1021/acs.jpclett.8b02892 for the method.
Specific papers for ZnO: link.springer.com/article/10.1007/s13391-019-00173-4
or this: onlinelibrary.wiley.com/doi/pdf/10.1002/pssb.201700393
hi. very helpful. how did you know the thickness of the thin film is 50 nm?
I gre the film using ALD. I later measured the film thickness using ellipsometry.
can I just use other formula if I didn't measure the thickness of the film?
@@rotuafetricia8238 Yes, the formula will remain the say. You can use an expected or approximate thickness.
Can we compute optical band gap using large machine-based data anyway?
I am not sure I understand your question. Is your data from experiment? What do you mean by machine-based data?
@@proprocessengineer7231 respected I have calculated data using material studio simulation software and now I want to calculate the optical band gap and don't know how I have to calculate?
@@ShakeelAhmad-tf6lz I would suggest to compare your simulated values with experimental data. If they are similar, you can use the same process to calculate bandgap.
how to calculate alpha for (liquid) dispersed
solution of nanoparticles??
I am not very familiar working with liquids. I suggest reading this paper : www.ncbi.nlm.nih.gov/pmc/articles/PMC6716051/
@@proprocessengineer7231 thank you
god job, I have a question, is this technique used only for smooth films? (sputering, PVD,...) or does it also work for rough surfaces? (dip coating, spin coating,...)
Thanks. While I am not an expert in optics, I can say that if your sample's roughness is small, it shouldn't matter. The reason is the wavelength range of light used will be bigger than roughness and will not be able to resolve it. For instance, in my samples, I have nanowires 50 nm apart and are not continuous and the smallest wavelength of light used during the UV-Vis scan is 250 nm, which can not resolve 50 nm. So, in this case, you can assume your film to be continuous or smooth. BTW, my films were made by spin coating. I have tried similar for samples made by PVD and it works. But keep in mind that ZnO derived by PVD and spin coating will have different quality (crystalinity and orientation), so their bandgap might not be the same. I hope this helps.
what does 50E-9 means?
50 nm or 50 * 10^-9 m.
@@proprocessengineer7231 is this 10^8 or 10^9?
@@salmanriaz7998 50 * 10^-9 m
What about the K constant? why are you just popping it out? the slope of that is not Eg, is Eg times K (K*Eg) would you mind explaining this?
To understand why K (proportionality constant) is popped out, you will have to refer to the derivation of the equation. You can probably start off from here and find older publications from the references: pubs.acs.org/doi/10.1021/acs.jpclett.8b02892.
BTW, it does not matter what the value of K is, since the x-intercept will not change. X-intercept is what we need to determine the bandgap.
There is a way to determine K (also known as Urbach energy) can be calculated. You can read these papers for more info:
1) www.sciencedirect.com/science/article/pii/S0749603615302615?via%3Dihub
2) www.sciencedirect.com/science/article/pii/S0925838815303352?via%3Dihub
@@proprocessengineer7231 thanks for your reply. Nevertheless, this approach is the same that you are applying just the differences is that they are using f(R) instead of alpha. And they are also skipping the B (in our case K) value. I understand that you need the value to measure de band gap but as I said before if you do the algebra you will notice it is necessary that constant because you the constant if you graph the y and x axis as you present (and conventionally is presented) I am not sure if I not getting something else here. What I mean, what are the considerations to pop k out? It approaches to 1 in this circumstance?
Also in the first paper they substitute the kubelka-munk function for alpha. But this function is equal to the absorption coefficient over the scattering coefficient. So another way to see this is that this scattering coefficient be equal to the constant K. So that what you and other research do in their research is correct. If that doesn't happen it doesn't makes sense. So if you have any information regarding this I would really appreciate it. I have been trying to understand what happen to this constant but I haven't got into the right answer yet. Thanks in advance!
@@davidyong5928 David, K is not something popped out. It is a proportionality constant added after the proortionality is defined. This is probably in the seminal work by Davis and Mott. Should I happen to find it, I will share it with you.
Thank you for bringing this up. I created this channel for having such critical discussions that can provide clarity to everyone.
@@proprocessengineer7231 looking forward to it. Thanks a lot!