Sir, your google search was just amazing... this shows that you are not stuck just to the syllabus of the course (as the other professors do) you are interested in knowledge..thanks a lot.
Your are great, mi new students never respond the ask: which one is the corresponding Bravais lattice for HCP structure? your video support very well my course, TNX.
Sir, I am confused as in what are the boundaries of lattice, How do we see hexagonal symmetry in this crystal structure, which object is symmetric w.r.t rotation about a six fold axis (where is this axis) ?
Well, by the definition, i.e., by the requirement of translational symmetry, a lattice is infinite. So it has no boundaries. We look at the symmetry by bringing the lattice points (in case of a lattice) or atoms (in case of a crysta) intro self coincidence after the symmetry operation. The hexagonal symmetry axis is parallel to the c-axis (or perpendicular to the basal plane) and passes through the C sites of the ABAB packing. In fact, it is not a pure 6-fold axis but a 6_3 screw rotation axis. In this course, I have not discussed the screw rotation axes.
Prof, could you please verify my 2nd approach to why there isn't a HCP lattice: First of all, we won't get a "CLOSE PACKING" if there are no atoms. We came to a concept of HCP, after stacking atoms layers on one another, carefully such that we get maximum packing efficiency. Now if we don't have atoms, there is no concept of "Close Packing" and hence, there is no concept of "Hexagonal close packing Lattice".
You are right. Using 'close-packed' for a lattice does not make sense as a lattice can never be close-packed. The interesting question to ask is whether the centres of atoms in an HCP structure form a lattice or not.
sir.....in hcp you said it can be described as a lattice + a 2 atom motif........but hcp is not a lattice......so sir this means if a motif has multiple atoms then all motif atoms must have translation symmetry only then the structure will be called as lattice otherwise it will not be called as lattice but can be described as a combination of lattice and motif..............is this so sir? also sir question 2......Crystal is lattice + motif......if hcp is not a lattice....then why do we call it hcp crystal???.......it must be simply hcp.
Only the lattice must have symmetries, not the motif. HCP is a crystal structure, it is based on a hexagonal p lattice, but it is not a lattice on it's own.
@@justinchow1644 The total length of Y axis is assumed to be b, so 1/3 b is just assumed as the length of the b atoms Y axial position. Actually from the sketch the professor drew it looks like the measurement is 1/3 b. Tha'y why I think
I think firstly we should understand 'Centeroid' concept and secondly and mainly there is 'Equal Intercept Theorem' of parallel lines, which helps us relate the maths he is explaining. also as its about 3d arrangement of atoms a,b and c are vectors. he is repersenting 3d array of atoms in 2d. This is my assumption. I may be wrong.
Sir, your google search was just amazing...
this shows that you are not stuck just to the syllabus of the course (as the other professors do) you are interested in knowledge..thanks a lot.
Sir you are amazing! awesome way of teaching Physics. Grounded and philosophical as well as rigorous. Thanks a lot.
Very useful video. Prof. explained the concepts slowly but clearly.
thank you professor
also thank you for saying " happy Norooz"
with love from Iran
Your are great, mi new students never respond the ask: which one is the corresponding Bravais lattice for HCP structure? your video support very well my course, TNX.
Useful
Sir, I am confused as in what are the boundaries of lattice, How do we see hexagonal symmetry in this crystal structure, which object is symmetric w.r.t rotation about a six fold axis (where is this axis) ?
Well, by the definition, i.e., by the requirement of translational symmetry, a lattice is infinite. So it has no boundaries. We look at the symmetry by bringing the lattice points (in case of a lattice) or atoms (in case of a crysta) intro self coincidence after the symmetry operation.
The hexagonal symmetry axis is parallel to the c-axis (or perpendicular to the basal plane) and passes through the C sites of the ABAB packing. In fact, it is not a pure 6-fold axis but a 6_3 screw rotation axis. In this course, I have not discussed the screw rotation axes.
@@introductiontomaterialsscience no problem with that sir, i am only going with GATE requirements.
Thank you.
Ok Sir I'll always use hexagonal lattice. Thank you so much
Prof, could you please verify my 2nd approach to why there isn't a HCP lattice:
First of all, we won't get a "CLOSE PACKING" if there are no atoms. We came to a concept of HCP, after stacking atoms layers on one another, carefully such that we get maximum packing efficiency. Now if we don't have atoms, there is no concept of "Close Packing" and hence, there is no concept of "Hexagonal close packing Lattice".
You are right. Using 'close-packed' for a lattice does not make sense as a lattice can never be close-packed. The interesting question to ask is whether the centres of atoms in an HCP structure form a lattice or not.
sir.....in hcp you said it can be described as a lattice + a 2 atom motif........but hcp is not a lattice......so sir this means if a motif has multiple atoms then all motif atoms must have translation symmetry only then the structure will be called as lattice otherwise it will not be called as lattice but can be described as a combination of lattice and motif..............is this so sir?
also sir question 2......Crystal is lattice + motif......if hcp is not a lattice....then why do we call it hcp crystal???.......it must be simply hcp.
Only the lattice must have symmetries, not the motif.
HCP is a crystal structure, it is based on a hexagonal p lattice, but it is not a lattice on it's own.
Sir can you explain Miller indices and slip planes of hcp
th-cam.com/video/toRd5L7S--A/w-d-xo.html
Thank you sir
can anyone explain to me how it became 1/3 b?
nevermind, thanks
@@justinchow1644 The total length of Y axis is assumed to be b, so 1/3 b is just assumed as the length of the b atoms Y axial position. Actually from the sketch the professor drew it looks like the measurement is 1/3 b. Tha'y why I think
İ didnt getit too
I think firstly we should understand 'Centeroid' concept and secondly and mainly there is
'Equal Intercept Theorem' of parallel lines, which helps us relate the maths he is explaining. also as its about 3d arrangement of atoms a,b and c are vectors. he is repersenting 3d array of atoms in 2d. This is my assumption. I may be wrong.
🙏👍👍👍