There are many ways but I will explain few of them. 1. Just meet up with your seniors and ask that you wish to contribute and become the part of the research group. 2. Visit the relevant Professors at your universities. Also visits other universities Professors nearby and ask for the collaboration. Here you have to ask like this.... I am very interested in your research field and would like to become the member or contribute in any research activities or simply ask them to give you any topic to work on it. 3. Just write to foreign professor of your relevant fields but at this stage target low profile professors... Those whose citations are less than 500 4. For successful collaboration, you must need to built trust. If you win here then you will have a successful collaboration. This is why People published a lot of papers because of collaboration... Individually you hardly published few papers... *Collaboration is like a TEAM work!* 5. Before approaching or trying the aforementioned options, I strongly recommend to work on some topic and at least gathered some good data. So you can show them that you know how to write or do analysis etc.... You can also make a PPT for presentation in case they ask you. Let me know if you need any other tips and guidance Wish you good luck!

@qamarwali thank you sir you explain very well.

*First reason:* This simply means that your two sample images have NOT taken from the same exact area. This is why I am always stressed that everyone should sit with the SEM operator. *Second reason:* Your sample may be composed of different size NPs because usually it is quite difficult to get uniform NP distribution. Generally, 80 nm and 120 nm are not far away, and the small differences should be there . In the research papers, we usually report like range such as ~ 80 to 120 nm or 80 - 120 nm ....

Sure, you can write. But I would recommend taking any action during your study with the permission of your supervisor. Later, you can do whatever you want, but at least during your student, take great care. Simply ask him or her that you wish to write a review paper. I am sure they will support you.

Better you ask the experts in your field of interest. Because they will know much more that what are the hot areas to work in the horticulture. You can also write in Goolge for the same purpose. If you have access to Scopus, then it would be great to provide with much great information. I can tell you the key areas need to work in Materials Science because I have expertise in this field

@qamarwali thankyou sir

You welcome 🙏

Look that kind of information we get from *XRD* not from EDX. In XRD analysis, we identify the phases (crystal structure) of the materials. For instance, SnTiO2 shows tetragonal structure. Fine! Now if these elements re-arranged from its position so the XRD will not give us the PEAKS and other parameters like before. This simply means that the structure is distorted. In the case of EDX, it only reveals the concentration of each element in term of *wt%* and *atm%* How much atoms of each element exist and what is the weight of that element. Therefore, for clarification that you exactly synthesized the SnTiO2 structure, we have to perform BOTH *XRD* and *EDX* analysis. I hope it is clear now. If you still have any questions please do ask here.

@@qamarwali thanks for your cordial response. Actually, I'm working on a double perovskite material which hasn't been reported before. but the problem arises when it comes to determine the space group, which was supposed to be fm3m but experimentally I got simple cubic pm3m(from XRD peak). however, if I consider it a single perovskite according to XRD & EDX results, nonetheless how can I be sure about atomic position of each materials and crystal identification, like: which material is in A site, which is in B site and which is doped? Sadly, I don't have access to ICDD, and its too expensive as well. hope you would reply.

You can simply drop your questions here in the comment, and I will respond to you ASAP.

I just made a video for the information. Better you search it by visiting the main website.

Yeah, I have synthesized and chararcterized single well-ordered crystalline materials.

Just write the sign in Web of Science and it will take you to the main page. Then, just follow the steps involved

No, I have not. I just drew the data in Origin Pro. The data was obtained from the BET machine.

You can direct copy figures from the other research papers and use it in your review paper, but you have to *add *references* of that paper. Usually, we write sentences like ... the figure has taken with permission of (journal or publisher name)

Thanks for your nice words 👍

Sure, I will make more interesting videos.

I am happy it helps you. Wish you good luck 👍

Can I write review paper without any support from professor during my post graduation.?

@@neon_chemistry Sure, you can write. But I would suggest that you must do everything with the permission of your supervisor during the study. Simply discuss with him or her that you wish to write a review paper.

I don't understand what you mean by photo? We set the scale to cm or inch and then measure the diameter of the sample

You are welcome 😊. Sure, I will keep uploading more interesting videos. Wish you good luck 👍

You are right, Raman spectroscopy is a very interesting technique. I'll try to make videos about it in the future. *I only made posts about this technique* which are available on my channel but videos not yet. This is the link to the posts if you are interested kzbin.info/door/lu9cyYyOT4-7nkoVTjPONgcommunity Many thanks

I am glad you found the video helpful. 😊

Many thanks for your comment. You are welcome 😊

Thanks for the comment. Could you write it here that question, please

@@qamarwaliSure sir! It’s with different x ray sources, the calculated binding energy of auger electrons will change.Please explain this a little bit,and it will be really helpful for me.

*Let's first discuss the Auger Peaks:* These peaks are produced when the core electron vacancy filled by valence electrons and the subsequent emission of electron from the same level. There are four main Auger series observable in XPS namely KLL, LMM, MNN and NOO series, identified by specifying the initial and final vacancies in the Auger transition, where K, L, M, N, and O are shells. For instance, the KLL peak means that the initial vacancy occurs in the K-shell, the vacancy is filled by the L-shell electron and subsequent emission of electron from the L-shell. There are also similar notation like KVV, where V for valence level, which means that the final vacancies (double vacancies) are in the valence level. The KE of Auger electrons are KEAuger ~BE(K) - BE(L1) - BE(L3) The KE of the Auger electron depends on the BE of specific orbitals in the atom from which it is originated. The BE of these orbitals are independent of the types of x-rays sources used so the KE of the Auger electrons is also independent on the type of x-rays source used. Therefore, with different x-ray sources, the calculated BE of the Auger electrons will changed.

@ Thank you,sir!However,I just still don’t get that “the calculated BE of the auger electrons will changed as the type of x-ray changes”.Mentioned that KE of the auger electrons is independent on the type of the x-ray,is it the BE of the auger electrons will not change as the type of X-ray changes?

We agree with this statement "The KE of the Auger electron depends on the BE of specific orbitals in the atom from which it is originated. The BE of these orbitals are independent of the types of x-rays sources used so the KE of the Auger electrons is also independent on the type of x-rays source used" using KE(Auger) ~BE(K) - BE(L1) - BE(L3). Let's come to the basic XPS (photoelectric) equation. hv =BE+ KE + WF(spec), this is also a energy conservation equation. The LHS x-rays photon energy distributed among the BE, KE and WF(spec). KE =hv- BE - WF(spec) BE =hv- KE - WF(spec), so here the KE of the Auger electrons independent on the X-rays energy (hv) as we have proved this in the top most paragraph. The WF(spec) is a constant value for the spectrometer. This implies that the BE of the Auger electrons changes if we change the photon energy (hv)..... directly proportional There are two types of x-rays sources in XPS Al K(alpha) ~1486.6 eV and Mg K(alpha) line ~ 1253.6)

Usually amorphous materials shows BROAD XRD peaks. This is due to the reasons that there is no long range order of the atoms in the amorphous materials. The x-rays scatter from the atoms in different directions and we get diffused spectra (less intense and broad peaks) The Crystalline sample give Sharp XRD Peaks. This is due to the fact that atoms arranged in the crystal acts like a plane (mirror). I always give this example to clarify the concepts. *Crystalline materials:* We we shine visible light on the mirror, the reflection of the light is intense (sharp). *Amorphous materials:* When you break the mirror into pieces and the pieces are still in the frame and then shine light, the reflections is NOT sharp but instead diffuse Why the peaks are getting less intense at higher 2theta and with the lower angle, the peaks getting smaller. I strongly recommend watching this full playlist on XRD analysis kzbin.info/aero/PLWgqLpcPbMgckHs0e677v8_GUh9-itYoS

@ thank you sir 😍

We use EDS to determine the contents or compositions of each element in the sample. For instance, your synthesized TiO2, so the EDS will shows the contents of Ti and O in the form of weight% and atomic%. Weight% is simple that how much is Ti and how much is O in the sample you prepared i.e., lets net weight of the sample comes from the 60% of Ti and 40% of O. Atomic% means that how many atoms of Ti and how many atoms of O exist in the sample. Because, the size of the atoms for Ti and O are different....... We get the value of weight% and atomic% in the form of TABLE from the SEM (EDS) operator. We are not going to find these values. We can also guess from the PEAKS in the EDS graphs that what is the % of each element exist (roughly). If you are making composite of TiO2- ZnO, so you can add one component more than the other so this you can increase the quantity of one element versus the other. I hope it will clear your doubt. If not please do comment for further questions

Thank you sir

Many thanks for the nice words. Wish you good luck!

Let's explain it here. In XRD analysis, the more atoms, the more scattering, higher the peaks intensity and vice versa. This is why some peaks are low intensity while others are higher in intensity. Because the peaks in XRD plot are due to the x-rays scattering from the atoms exist in the crystal. The same I explained in the description of the video. We consider two unit cells (shown in the attached images), i.e., Base centered orthorhombic & Body centered orthorhombic. Here in both cases, TWO atoms per unit cell, one atom from the corner, and other atoms are the additional atoms at the base or body position. But in both cases, the extra Base & Body atom is located in different positions. Either unit cell is derivable from the other by a simple shift of one atom by the vector ½ c. Consider diffraction or reflection from (001) plane, for the base centered lattice, it is showing that Bragg’s law (λ = 2dsinθ) satisfied for the particular values of λ and θ, which means that the path difference ABC between rays 1´ and 2´ is equal to λ. Therefore, rays 1´ and 2´ are in phase, and X-rays diffraction occurs, and we get peak in the XRD pattern. The same is true for the body centered lattice for reflected rays 1and 2 i.e., these both rays meet constructively, and diffraction occurs, which are in phase and the path difference ABC is equal of λ . However, in the case of body centered lattice, there is another plane of atoms midway between the (001) plane, and the path difference DEF between reflected rays 1 and 3 is ½ λ .Therefore, rays 1´ and 3´ are completely out of phase, and no diffraction occurs. That’s why there is no (001) reflection from the body-centered lattice. We have proved that the position of atoms matters a lot, and the intensity of the XRD spectrum depends on the atom’s position. X-rays scatter in three ways from the crystal: 1) - Scattering from electrons : Polarization factor 2) - Scattering from an atom : Atomic scattering factor (f) 3) - Scattering from a unit cell : Structure factor (F) I strongly recommend watching this full playlist and also read the description with each video kzbin.info/aero/PLWgqLpcPbMgckHs0e677v8_GUh9-itYoS I also strongly recommend to read the post on my channle kzbin.info/door/lu9cyYyOT4-7nkoVTjPONgcommunity

It is written on the article first page, like *review paper* or *research paper* Also, from the title, you can clearly understand that the article is a review or research paper.

Size of what? The shift shows the size as I explained in the video. By simply increasing or decreasing the size of the quantum dots, the peak shift towards one or other direction on the wavelength line

@@qamarwali Yes, but if you know the wavelength of a QD sample, can you quantify what the size of those QDs are

Sure. You can find the size of the QD from the emission wavelength. We can also estimate the size of the QD from the absorption peaks (absorption wavelength as it usually occur at lower value than the emission wavelength) You can also find size of the QD TEM analysis instead if you don't from the absorption or emission spectra. As it is very understood that the size of the QD affect the Band gap (E) and then this energy affect the wavelength i.e., E = hc/lambda Read the following article will help you Analyst, 2011,136, 2391-2396

Thanks 😊

I am immensely grateful for your nice words. Please do comments if you need any kind of help in writing a review paper.

K, L, and M basically show the shells where electrons are residing. K shell is the closest to the nucleus.

I am happy that you appreciate my efforts. Wish you good luck 👍 I recommend you explore the whole playlist on XRD. kzbin.info/aero/PLWgqLpcPbMgckHs0e677v8_GUh9-itYoS&si=ca1gsYndzHs0_b2B Also, read the comments with every video. This way, you will gather a lot of information

If I want to compare the same material before and after a test only calculating of peak shift, broadening are enough? *Response:* What do you mean by before and after the test. You can compare the peaks after the heat treatment, after the doping, after increasing one constituents in the case of composite, changing the parameters like humidity, temperature, speed etc of the sample during synthesis....... In such scenarios you can compare the peaks and see the affects. if I find a new peak can it lead to new structure? *Response:* New peaks simply means that there is another phase in the crystal structure. The constituent of a component is so much large that it form another phases in the crystal structure or the crystal texture is oriented in some other preferred directions.... I recommend you to watch all videos in the playlist. kzbin.info/aero/PLWgqLpcPbMgckHs0e677v8_GUh9-itYoS