Reciprocal Space 2: Condition for Diffraction

[@elmamathew2588]

Sir, Thank you so much for your wonderful lectures. It is hard to find people like you who teach the concepts so much in detail in the UG/PG level. Your sincerity and dedication will definitely pay off.

[@nidhinsathyan6904]

This person has enough potential to write a renowned solid state physics book.

[@chacho0001]

+Nidhin Sathyan He wrote ir already. Physics of materials: Essentials concepts of solid-state physics. Dr. Prathap Haridoss

[@akshitasharma5340]

Best! Nptel lectures are much better than any other online platforms.

[@tumeshkumarsahu6817]

the way prof. has explain its too good any body can understand thanks a lot prof who presented this topic in very simple manner

[@WLVBS]

Maharshi Ray I think the lectures are excellent, they helped me understand the subject. If you want you can always watch them on vlc player and speed them up to 1.5x or 2x ;)

[@ChaineTa3lim]

you can do that (x1.5) here

[@danvergelmasiong3041]

I really appreciate your lecture sir. It helps me a lot to understand the concepts of solid state physics which is my major subject this term.

[@MuhammadAli-px1ob]

Nicely Explained! I appreciate the effort!

[@03kingh]

This is amazing!

[@123yogees]

Thanks a lot. I was looking for a basic lecture like this.

[@Ayushmishra-gz4pg]

for p=3 q=5 r=7 if we take h=0.5, k= 0.5, l= Any integer , we will get hp+kq+lr = integer

[@kuldipkunghadkar6802]

Yes !.. Good question.. how then the hkl values must be always intergers?

[@rafshantik]

41:41 why should h, k, l be integers only? For example, if p, q, r are all even numbers, and h=k=l=0.5, then also we get the sum as integer value.

[@ahmednesartahsinchoudhury2628]

Because hp + kq + lr should be an integer regardless of the values you choose for p, q, and r. Suppose you choose h = k = l = 0.5 and p = q = r = 2, but I can now choose p = 1, q = 2, and r = 2 and hp + kq + lr won't be an integer anymore. In order to ensure that hp + kq + lr is an integer regardless of the values chosen as p, q, and r, the values of h, k, and l should be integers.

[@karansavantable]

Very helpful Lectures.....Thankyou

[@snaseer9279]

awesome'

[@gobin961]

owesome thank u

[@LM53180]

I LOVE YOU!!!

[@namitachhibba]

Thank You :)

[@telelight]

It doesn't sense that at 41:30, where he says h,k,l must be integers for the sum (hp+kq+lr) to be integer where p,q,r are integers. This is not a true statement because for all integers greater than 1 let h=1/p, k=1/q, l=1/r then we have (hp+kq+lr)= (1+1+1)=3 which is at integer. Yet clearly we have chosen non-integer values for h,k,l and the statement remains true. So his conjecture is incorrect, h,k,l do no have to be integers for the sum (hp+kq+lr) to be an integer.

[@srikanthmnb]

Quite late here but it might be useful for others watching the video. sum(hp+kq+lr) is an integer for ANY integral values of p, q and r, not just one set of (p,q,r). Thus, each of (h, h, l) have to be integers.

[@alokkumar3703]

@@srikanthmnb thanks bro,great job

[@Ayushmishra-gz4pg]

two problems 1. Path difference has dimensions of area 2. if 2a vector is taken in real space its position in real space will be 1/(2a) in a direction perpendicular to the a vector so I will have a non integer multiple of b vector (which has a magnitude of 1/a in reciprocal space

[@rituyadav5808]

Sir the lecture is amazing but video quality is very poor 😓

[@Ayushmishra-gz4pg]

in fcc arrangement for the atoms at the middle of the face of the cell the values of p,q and r are not integers

[@akashbashir2153]

@Auysh mishra in FCC or bcc lattice we define primitive vectors for primitive cell which are formed of either lattice points of faces(in case of FCC) or the centre points ( in case of bcc) .

[@yisehakg5039]

Good. But I had to play the whole video on X1.25speed,...... too slow.

[@hrs7305]

man you are too slow you are repeating a lot of times

[@MaharshiRay]

tooooooooo slow ! please avoid repeating the same arguments over n over again !

[@sidraowais3453]

Kindly improve ur english literally Fed up toooo slow