the crystal structure

is something that you know ah in fact you have probably learnt off from high school

days and certainly maybe early college days you have heard of course the structure you

have heard a lot of descriptions of it and so on. so in in those descriptions you would

have heard of crystal directions in other words axis ah and crystal planes

so there are conventions based on which we ah designate ah the directions and planes

and so on ah and on that basis you can have lot of the crystal structures that ah we have

in the books okay. so ah so these are there ah normally these are ah in dimensions of

length we basically a certainly ha crystal axis when we unit vectors we say with respect

to, so we we define unit vectors dimensions of length okay.

so this is how it is done, we write xyz coordinate system we say you know ah and if you take

a real ah crystal structure you may have specific dimensions in the in the form of say ah of

the order of say 2 angstroms inter atomic spacing is of the order of 2 angstroms say

1.5 angstroms 1.8 angstroms, angstroms is 10 power -10 meters, so it is still in the

dimensions of length okay. so when you normally put such information

together okay, so we would call this as real space. a real space is what we are conventionally

used to ah which where we write an xyz coordinate system and within the framework of that xyz

coordinate system we define dimensions, we define unit vectors, we define directions,

we define ah the planes and spacing between planes and so on. so this is real space okay,

what we are going to do today is define something that we will call reciprocal space. you will

define something called reciprocal space and we will begin to see some of it’s properties

okay. ah reciprocal space in fact takes ah the basic

ah umm approach here is that the same crystal structure information that you have here or

ah and many of its important key features can be represented in another format known

as reciprocal space okay. so we are not actually changing anything in a in a fundamental sense,

because you are still talking of the same material, you are still talking of the same

kind of ah relationships between the planes between those materials, the atoms between

those present within the materials and so on. so we are not fundamentally changing anything.

all we are doing is we are representing the same information using a different set of

ah a different framework. if you want to call it that it is a different framework which

is called reciprocal space, here the dimensions of the vectors we will use will be 1 by length.

so dimensions used to be 1 by length or length power -1 is the dimension that we will use.

ah in sort of a trivial sense you can see ah that this can be related to our k vector

because k vector is already in the dimension of 1 by length okay.

so therefore ah it may be it may make it easier for us to relate certain things associated

with the wave vectors of electrons which are running across the crystal structure. to the

crystal structure itself okay, if you use the same ah ah same framework within which

you are describing both of them. the framework here we are using is 1 by length 1 by lambda.

so it would help, if we also define the crystal structure in 1 by length dimensions okay.

so that is maybe ah a little bit of a trivial way of saying it, but it will ah it is it

it suddenly conveys the immediate ah link between what we have just done up until now

and what we are planning to do okay. in reality actually it is much more than that ah it turns

out that ah ah many of the information you can represent in 1 by length in this in this

reciprocal space. ah actually conveys certain details of how interactions occur between

between a crystal structure and ah waves that are present much more elegantly. then ah ah

is then is done in real space okay. so in a broader sense this is, ah ah this

conveys certain information much more elegantly reciprocal space and actually is able to highlight

specific details a much better than the real space way of representing information does

and that is the real, that is the ah ah fundamental reason why actually if you get into ah a diffraction.

if you get into meti i mean ah diffraction as a means as a tools of as a tool for looking

at structure and such information characterization of materials. ah you will find a lot of heavy

usage of reciprocal space. ah the first time we encounter reciprocal space

it is not as intuitive and it may seem like you know we are unnecessarily complicating

ah ah the issue so to speak okay. so it looked like why go to all the ah troubles of creating

everything in 1 by length when real space is already there for us. it is just that if

you use it enough you find that there is lot of ah ah information that is much more elegantly

and clearly represented in in when you use this notation than when you use real space

notation. so it is from that perspective that this is

really caught on and it traces itself back to a person by name ewald, who actually ah

worked on this ah in around years around the year 1920 okay. so around that time frame

was when this ah work was put together. and so there is a there is notation which is which

is named after him also in in this relationship okay, so ah in this context.

so we will see that as ah ah i mentioned you know we have sort of indicated why you may

need to go in for ah ah1 by length dimensions but that is not a hard-and-fast that is not

the best way of indicating, but you can see the link that we have already got to pi by

lambda 4k wave vector and we want to see the interaction between the electron waves and

the crystal structure ah and therefore it would help if we or if you present at all

the information in the same framework. so ah you can think of that as a loose link

for what why we are doing what we are doing. as we progress forward we will see ah much

more ah a better understanding of how this system works out okay. so now we will look

at ah umm ah in the next two or three classes we will actually focus ah exclusively on reciprocal

space and we will build its relationship to the reveal real space because that is something

we need to understand. ah and to some degree this discussion may

seem a little disconnected from what it is that we have discussed so far ah but we will

need to build this framework so that we can connect it up link it up to our discussion

earlier and see what ah ah what benefits we can gain from the process okay. so but temporarily

for at least this class and in most of next class this will be a discussion exclusively

on reciprocal lattice and on how diffraction can ah as a phenomenon can occur in the reciprocal

lattice. after that only we will make a link back to

real lattice, a real space and also take up in the extend this idea the diffraction occurs

in reciprocal space and it has a certain way of being conveyed and see what is the consequence

of that discussion, on how the electrons are interacting with a crystal lattice. so that

that step is going to be 2 or that is at least 2 classes away before we get there but now

we will have to build the framework which will enable us to handle that discussion ok

so that is what we would do at this point. so ah when you look at real lattice real space,

we say that you know we are usually defining it by unit vectors

and typically we would use the notation a 1 a 2 and a 3 okay. so a 1 a 2 and a 3 are

unit vectors in real space. ah so we have a crystal structure in real space represented

by these unit vectors a 1 a 2 and a 3 okay. now we will define a reciprocal space in other words we will define a space, we

will define a coordinate system. so we are just going to define it up front

this definition is at this stage may look somewhat arbitrary, we will just accept it

as an arbitrary definition. the definition the way it is given will give the space a

lot of useful properties which we can use later on. ah how arbitrary or otherwise this

definition is we will see a little later but at the moment it is just a definition we will

just accept the definition and we will work with it, so we are defining it okay.

so this is a choice we are making, so we are defining it we are defining a reciprocal space

to consist of three vectors again unit vectors there b 1, b 2 and b 3 but these are not arbitrary

ah vectors they are being defined with respect to real, i mean real space vectors in a certain

way in other words there is in the in the process of this definition we are already

making a link between these vectors and these vectors okay and what is that link it is it

is written like this so b 1 is a 2 cross a 3 by v.

what this v is we will see in just a moment. it is a volume actually volume of this unit

cell ah of consisting of a 1 a 2 and a 3, ah b 2 is defined as a 3 cross a 1 by v and

b 3 is similarly is a 1 cross a 2 by v. so these are cross

product, these are vector cross products so that is what they are so ah so these are all

vector ah quantities b 1, b 2 and b 3 are vector quantities. so they have ah a magnitude

as well as a direction, a 1, a 2 a 3 are also vector quantities they all also have magnitude

and direction and this is a cross product. and so it is defined that way b 1 is defined

as a 2 cross a 3 by v, b 2 is defined as a 3 cross a 1 by v and b 3 is defined as a 1

cross a 2 by v okay. so this is the way they are defined. ah by defining it like this certain

properties ah become ah certain properties ah arrive for ah i mean end up being available

for b 1 b 2 and b 3 which becomes convenient for our utilization later on okay.

so now we will see what immediately based on this definition itself simply because we

have defined it like this what is the meaning of what is the consequence or what ah what

is the relationship between b b 1 and this ah reciprocal, so these are called reciprocal

lattice vectors okay. these are unit vectors in reciprocal space they are also called reciprocal

lattice vectors. these are real lattice vectors and this is real space. so by simply using this definition what is

it that we have created as a relationship between what is the consequence in in the

relationship between b 1 b 2 and b 3 with respect to what we have in real space. so

that is the first thing that we will examine okay. so to do that let us take an arbitrary

say a triclinic cell. a triclinic cell is one where by definition the a 1 a 2 and a

3 vectors need not have the same length. so a 1 need not be equal to a 2 need not be

equal to a 3 and the angles between them so alpha beta and gamma which are the three angles

that exist ah in the system need not be the same okay. so that is the so in that sense

it is of it is like a very general cell we have placing a really placing no restrictions

on it. ah now ah we are defining we will we will write b 3 again here fine a 1 cross a

2 by v. so now by definition ah see volume does not

have any is not a vector volume is just volume it is not a vector it is a scalar quantity.

so you have a 1 cross a 2 by definition of cross product if you have a cross product

the ah result is perpendicular to both a 1 and a 2 okay. so that is the meaning of a

cross product. so be 3 therefore by definition is perpendicular to a 1 and a 2. so on this

scale if you want to draw b 3 it will show up something like this.

it will show up somewhere in this direction it would be perpendicular to the plane being

described by a 1 and a 2 okay. so the planes a 1 and i mean the vectors a 1 and a 2 define

a plane which is which we are now treating as the horizontal plane on this board a sort

of in this representation and b 3 would now appear in this direction perpendicular to

a 1 and a 2. so that is the ah definition by way of definition

but what about its actual magnitude if you look at a 1 cross a 2, ah if you ah if you

see the by the standard definition of a 1 cross a 2 that is the area of this parallelogram

okay. so if you write this as the origin write this as oa ah let us say this is b and let

us say this is c, so the ah area of this parallelogram oacb is what is being given by a 1 cross a

2 okay so oacb area of parallel a 1 cross a 2 okay oacb the parallelogram.

so that is what this area is ah if you look at the ah volume of the unit cell this v here

is the volume of the unit cell in real space okay. so that is the volume that is that v.

so which is that volume if you take this a 3 vector and you complete this solid okay.

so that solid, so ah this unit cell that we have drawn here i mean if you draw it properly

you will get it appropriately this volume of this unit cell is this volume

v that we have written here fine. so that is the volume that we are looking at. so what

is that volume in terms of geometrical terms it is simply the area of this base times ah

the distance between these planes or the height of this unit cell okay. so that is all the

volume is the volume is the area of the base times the height of that structure. so that

is what we are looking at, so the height is simply this the whatever this is abc so this

is d let’s call this d od, so od is the height of this ah unit cell that we have drawn

and it is simply the ah ah projection of a 3 on this on this axis on this axis.

which is perpendicular to a 1 and a 2 right, so that is the height, so ah we will call

that ah so the height is od ok so the height is od so therefore volume is the area of oacb

ah into the height od okay. so it is a product of the area times the height that is the volume

okay. so therefore if you look at it that way b 3 okay or the modulus of b 3 is area

oacb by area oacb into od. so that is what the modulus of b 3 is, so these two will cancel

out, so it is simply one by od, so it is simply one by od okay.

so that is the ah magnitude of ah b 3, so b 3 is in this direction and the magnitude

of b 3 is simply 1 by od, what is 1 by od if you look at the conventional planes that

we are looking at if you look at the way we define planes this is the 1 0 0 play okay.

so if you look at 1 0 0 plane this is a d 1 1, d 1 0 0 plane it is spacing between 1

0 0 planes. if you go back to your elementary crystallography the 1 0 0 planes are defined

this way and then these ah are 0 0 1 plates d 0 0 1 this is perpendicular to ah a 3 axis

okay. so d 0 0 1 if you want to call it 0 0 1.

it is the spacing between 0 0 1 planes so spacing between 0 0 1 planes is what we have

now looked at. so od is the spacing between 0 0 1 plane, so d 0 0 1 okay, so b 3 is therefore

spacing. ah this is the standard notation for crystallography, so if you go and look

up crystallographic from your elemental crystallography this is d 0 0 1 is the spacing between 0 0

1 planes and modulus of b 3 is simply 1 by d 0 0 1 okay. so so therefore we find and

by analogy in fact i mean. in fact if you extend this argument the same

thing would hold for the other ones to b 1 is 1 by d 1 0 0, b 2 1 by d 0 1 0 and b 3

which we just did is 1 by d 0 0 1. so b 1 b 2 and b 3 are ah inversely proportional

to the spacing of those planes ah 1 0 0 , 0 1 0 and 0 0 1 planes. so that is the by way

of our definition we have created the situation okay. so it is not arbitrarily occur since

we defined it this way this is the way it has occurred okay. ah also if you look at

the way we have defined it we will also see that if you just take a product say ah b 3

cross a 2 or b 3 dot a 2 okay. if you just take this dot product between

b 3 and a 2, i have just arbitrarily pick these two vectors we find that since b 3 is

already perpendicular to a 1 and a 2 this dot product is 0 okay. so by definition it

will be there will be a cos 90 degrees which was up here therefore this is 0. similarly

b 3 dot a 3 sorry a 1 equals 0, because b 3 is perpendicular to both a 2 and a 1 simply

based on how we have defined it that is all it is okay. so therefore these two are 0,

if you look at b 3 dot a 3 okay. based on our definition if you go back to

our the picture we have drawn here b 3 has the is this distance od, i am sorry this is

1 by the distance od, so ah which we just saw here okay 1 by od b 3 is modulus of b

3 is 1 by od, so this is simply equal to 1 by od times the projection of a 3 on b 3 okay.

when you do a dot product that is basically what it is it is one vector times the projection

of the other vector on itself okay. so what is the projection of a 3 on b 3, projection

of a 3 on b 3 is od, a 3 on b 3 the projection of a 3 on b 3 is od as shown in this diagram

fine. so therefore the dot o dot a 3 that we are having in that dot product is simply

works out od, so to speak so so this projection will become od okay. so therefore this is

equal to 1. so we find the relationships between those vectors the the reciprocal lattice vectors

and the real lattice vectors based on how we have defined those vectors okay.

based on how we have defined those vectors creates the situation where b 3 dot a 2 is

0 b 3 dot a 1 is 0 and b 3 dot a 3 is 1 so more generally if you have b i dot a j, then

this is equal to zero when i not equal to j and is equal to one, when i=j okay. so

that is the notation that we have and that is the consequence of what the wave in which

we have defined these vectors okay. ah now we have already seen, ah if you have taken

a specific case actually where we are saying that if you have a particular vector.

so d b 1 in this case is a particular vector and we found that the way we have defined

it, it works out to be perpendicular to the two vectors a 2 and a 3 and it is equal to

the in magnitude it is equal to 1 by the spacing d 1 0 0 okay. in reciprocal lattice i we can

actually generalize this much more. we will generalize it as for us we can write

any vector ah ah h h k, h is the notation that is given for a reciprocal a general,

a generic reciprocal lattice vector okay. so where we it could be anything, so h is

a general reciprocal lattice vector we will give it subscripts h k l okay. so if this

is a reciprocal lattice vector and the unit vectors in the reciprocal space are b 1 b

2 and b 3 then this is simply equal to h b 1 + k b 2 + l b 3.

this is simply based on our definition and vectorial, standard vectorial notation, standard

vectorial notation b 1, b 2, b 3 are unit vectors and h k andl are the ah ah specific

distances we are travelling along those unit vector directions okay. so h h k l is a vector

in in reciprocal space and it is therefore equal to h be 1 + k b 2 + l b 3, those h k

l are the amounts that we are travelling on those respective dimensions okay so that is

what it is. we say that when we define when unit when

reciprocal space is defined the way we have just defined it, then when you take a general

vector h h k l okay, ah in the reciprocal space we can there are some relationships

it has to real space vectors and dimensions in real space just the way b 1, b 2, b 3 themselves

have relationships to the a 1, a 2 and a 3 okay. we just saw a relationship between b

1, b 2, b 3 and a 1, a 2, a 3. similarly we there is a very general relationship

between any h k l vector in reciprocal space and certain quantities in real space. what

is that relationship it is basically that h h k l is perpendicular

to the plane which has the miller indices h k l and modulus of h h k l=1 by d h k

l. please note in both these cases we are relating something in reciprocal space to

something in real space. this is not complicated because we just did

that already when we looked at a 1, a 2, a 3 and b 1, b 2, b 3 and we made relationships

between them when i said that know b 1 is 1 by d 1 0 0, b 2 is 1 by d 0 1 0 and b 3

is 1 by d 0 0 1, there b 3 is a reciprocal lattice quantity and d 0 0 1 is a real lattice

quantity. we found that you know it’s simply because of the way we have defined it, when

in our definition itself we have linked real space and reciprocal space

so they are not arbitrary quantities okay, so they are already linked by definition therefore

some quantity in real space can relate to something in reciprocal space within the framework

of the definition. so that is basically all we are doing here this is a ah ah reciprocal

lattice vector and it is found that it is perpendicular to a plane in the real lattice.

which has that the ah ah miller indices h k and l and the modulus of this reciprocal

lattice vector is 1 by d h k l. these are properties that the reciprocal or

in other words one by the spacing of h k l planes. so these are two properties that any

ah vector in the reciprocal space has as a result of the definition of the reciprocal

lattice okay. so right now i have just stated it in the in the next few minutes we will

prove these two once we prove these two we have a good understanding of what holds in

reciprocal space and how it relates to real space and later we can use those two okay. so we will now try an attempt to prove this,

so to do that let us actually draw a general plane in the real space. so we will say that

we have, so this is a 1 this is a 2 this is a 3 okay, so these are unit vectors in real

space and we will draw the h k l plane here okay. so this is h k l plane fine, so ah ah

now we will just say that ah in in reciprocal space we have the h h k l we at this moment

i am just arbitrarily denoting it denoting it here.

this is an arbitrary de notion at the mean i have arbitrarily denoted it this way i have

indicated it here in this figure this way ah simply for the sake of convenience to show

it in the same figure ah what relationship it actually has to h h k l is not forced upon

it simply because of how i have drawn it we will show that in fact it does have some appropriate

relationship. so that relationship we will just see we are actually going to prove it.

now by definition of h k l plane its intercept along a 1, a 2 and a 3 are simply ah a 1 by

h, a 2 by k and a 3 by l. so these are the intercepts okay, only because the plane h

k l happens to intercept a 1, a 2 and a 3 at those locations such that its intercepts

are at a 1 by h, a 2 by k and a 3 by l, that is the reason why we even call it the h k

l plane okay. so therefore if you see if you take this ah vector here, this vector here

is a 1 by h this is a 1, a 1 by h is this vector that is why you get this h notation

in the h k l plane. similarly this vector here will be a 2 by

k that is why you will get the k notation. this vector here from here to here is a 3

by l that is why you get that l in the h k l. so this is a 1 by h. i will also named

this locations where this plane intercepts a 1, a 2 and a 3 as a, b and c okay. so ah

oa is a 1 by h and ob, so this is origin o, oa ob and oc. so oa is a 1 by h, ob is a 2

by k fine, so this is what we have. so ah simply by vectorial notation if you write

ah oa + ab you should get ob right. so oa simply because of standard vectorial

notation oa these are vectors by the way oa + ab should equal ob, that is easy to see,

if you go from here to here and then you go from here to here it is the same as going

from here to here that is all it is oa + ab is ob. so this is what we have a 2 okay, so

now that we know these are the vectors we can replace them oa is a 1 by h and ob is

a 2 by k. so therefore a 1 by h + ab, vector ab should equal a 2 by k fine. so this is

what we have, we can rearrange this a little bit. so we wrote a 1 by h + ab=a 2 by k, rearranging

this we get ab is a 2 by k + sorry – a 1 by h. so we now have in terms of the a 1, a 2

vectors we have the vector ab given to us in terms of the unit vectors a 1 and a 1 and

a 2 which we have already defined for a real space. we already said that if you take a

vector h h k l, we are defining it based on this notation as h b 1 + k b 2 + l b 3.

so the h k l, h k and l are simply integers okay, so this is some vector in reciprocal

space so h k and l are just integers, so as long as they are just integers you can use

them in whichever space you wish they are just integers b 1, b 2 and b 3 are reciprocal

lattice vectors therefore h b 1 + k b 2 + l b 3 is now a reciprocal lattice vector.

whereas when you use the h k and l which are just integers in the real space a 2 by k and

a 1 by h. if you take this difference which is ab it

is a real lattice vector because these are just integers where which you are using in

the real space. but they are the same ah ah integers at this time ah we have the same

h k and l being used in two different places. the h k l plane is this results in this relationship

and the same h k l values, we have now used for this definition although we have got no

significance for it. yet there is the same we are enforcing the

same h k l in this ah definition for h h k l. so if you now take a dot product of h h

k l and ab, what will you get? so if you do h h k l dot ab okay. so we have ah b 1 here

h b 1 okay, we already saw that if you have a i dot ah b j then this is equal to 0, if

i is not equal to j and this is=1, if i=j and since it is a dot product you can

have at a i dot b j is the same as b j dot a i, if the order in which you do this dot

product is not important to us because either way you will get the same thing.

so if you look at this dot product here you have h b 1, h b 1 dot a 2, so these are just

the h is just an integer b 1 dot a 2 is zero because it is 1 and 2 here. so that is 0,

b 1 dot a 1 is 1 and you have a h b 1 here and a 1 by h here so h and h will cancel you

get -1 okay. so h b 1 times or dot product of h b 1 and a 1 by h will give us – 1 right.

if you take b 2 here k b 2 times a 2 by k, the k and k will cancel, you will have b 2

dot product of b 2 and a 2 which is 1 + 1 and this b 2 dot a 1 is going to be 0 because

it is a subscript 2 and that is a subscript 1 we already saw that by definition

so that is a plus 0, so that term will become 0 and then b 3, b 3 dot a 2 is going to be

0, b 3 dot a 1 is going to be zero. so this term does not contribute in any way it becomes

all 0, so this becomes 0, so this becomes 0 the product with this gives us a +1 the

dot product with this gives us a – 1, so this is equal to 0. so we have a situation

where a vector in reciprocal space h h k l dot a vector in real space is equal to 0.

so in other words we have a dot product between two vectors which is zero which simply implies

that h h k l is perpendicular to this vector ab. therefore h h k l is perpendicular okay.

so if you go back to our figure it means that this h h k l is perpendicular to this ab okay.

so this ab is there the way this h h k l is defined it is perpendicular to ab. using exactly

the same ah ah derivation that we have done instead of we started with this this location

being a 1 by h and this being a 2 by k. we can do the same thing with a 2 by k and

a 3 by l okay, in which case we will find that h h k l will become perpendicular to

bc. similarly we can also do it with a 3 by l and a 1 by h and exactly the same calculations

we will do we will find that h h k l is perpendicular to ac. so we find that h h k l is perpendicular

to ab it is perpendicular to bc and it is perpendicular to ca based on the same derivation

that we have done. you simply have to select the other two axis, you will come to the exact

same conclusion. the same math will the mathematics is work

out in exactly the same way we will find that h h k l is perpendicular. the vector h h k

l is perpendicular to the vectors ab, bc and ca. therefore and since all of these three

are forming a plane, if it is perpendicular to any two in fact even even if it’s perpendicular

to just two, it is certainly perpendicular to two and it is also perpendicular to all

three if it is perpendicular to them it is therefore perpendicular to the plane defined

by a b and c okay. therefore h h k l is perpendicular

to plane defined by abc ab, bc and ca and therefore which is basically the hh which

is basically the h k l plane okay. so that is how those vectors are even defined ab and

bc and ca were defined based on the intercept at 1 by h, 1 by k and 1 by l in those respective

axis. so therefore h h k l is perpendicular to h k l. so any, so again we are relating

something in reciprocal space a vector in reciprocal space to a plane in real space

okay. and again all these relations are coming about

simply because our original definitions related a reciprocal lattice vector to real lattice

vectors. so that relationship was already there within the framework of this relationship

we are finding other other relationships that hold. so we find that any h h k l plane which

is that therefore defined as h b 1 + k b 2 + l b 3, where b 1, b 2 and b 3 are the unit

vectors and reciprocal lattice reciprocal space. those h h k l vectors will automatically

be perpendicular to the h k l planes in the real space okay so this is already we have

seen this. so we have just we have just shown this okay.

so the other thing we would like to see is what is the value of modulus of h h k l and

how does this relate to the spacing of between the h k l planes. we will in fact see that

this is equal to 1 by d h k okay. so this is what we are just about to we are going

to look at. we will come back to this figure let us say that a unit vector

along h h k l okay so we will let us first define a unit vector along h h k l that is

simply h h k l by modulus of h k l of h h k l right that is the definition of a unit

vector along h h k l. we will just call this say m, m cap okay.

now when you say a h k l plane is defined the way we have just drawn it okay. so when

we say this we mean that the that there is a similar plane like it at the origin then

there is one at this location then there is one similarly spaced next to it similarly

space next to it and so on that is the way we define a family of a ah ah set of planes.

when you say a h k l plane it is not just a single plane i mean it is that set of planes

that are parallel there. we take the one process to the origin and

then take the intercept of it and that is how we come up with h k l, so there are planes

correspondingly apart. so therefore the spacing between the h k l planes is simply the distance

between the origin and this h k l plane right the closest distance ah or ah rather the perpendicular

the spacing between this origin and that location which would then be the closest distance between

the this plane and the origin is therefore the d h k l right.

so now ah we have basically seen that the ah at at that point the ah line drawn from

the origin which goes closest to this plane will then they will then go perpendicular

to that plane all right so that is how you will get the ah location. so therefore ah

if you look at the distance ah between the ah plane and the origin ah we basically see

that we can write that by saying h h k l or n cap

dot a 1 by h ok. if you take the ah ah dot product of a 1 by

h and you take its ah ah what shall i say the the dot product of oa with this ah vector

in the direction of this h h k l then you get the component of oa along that direction

and that would then be the spacing that you are interested in ok. so this is what you

will get, so this is simply a 1 by h ah dot product with ah you will have h b 1 + k b

2 + l b 3 divided by the modulus of h h k l okay.

so this is d h k l right, so d h k l is now defined this way d between those planes is

such that ah the if you take the the component of any of those ah intercepts along the perpendicular

ah to that plane okay. so that perpendicular is now going through the origin. so that perpendicular

vector that is there if you take the component of this vector along the direction of the

perpendicular you therefore get the distance in this direction okay.

so that vector unit vector is simply giving us the direction there but if you take the

component of this ah intercept oa along that direction or any of this ob along the direction

or oc along the direction if you take the component along the direction that would then

represent that spacing between the origin and that point which is then the closest spacing

between the origin and that point and that would therefore and at that point that vector

will be perpendicular to that plane which is what how we have defined it okay.

so if you take the component of the intercept along the perpendicular that is the closest

approach that the plane makes to the origin and that is therefore that d spacing of that

plane because the next plane is sitting at the origin. so the d h k l is the component

of this vector a 1 by h along this unit vector here. the unit vector is simply defined by

this and this is and since it is a dot product we can interchange the i does not matter which

order we do it so we will get this. so now if we carry out this dot product ah

what do we see again you have an a 1 here you have a b 1, b 2 and b 3 here and clearly

a 1 dot b 2 is 0 and a 1 dot b 3 is 0, the way we have already seen. so only a 1 dot

b 1 is going to count for anything else so okay. so a 1 dot b 1 is going to be 1 because

of the way it is defined and the h and h is are going to cancel. therefore d h k l is

equal to ah a 1, so we will just write it here, a 1 by h dot h b 1, b 1 or sorry a 1

dot h by h b 1 dot h b 1 divided by modulus of h h k l.

so this and this will cancel a 1 dot b 1 equals 1 therefore this is equal to 1 by h h k l

here. so we see ah which is the proof that we wanted to be set out to prove. so we see that h h k l is perpendicular to

and modulus of h h k l is equal to 1 by d h k l okay. so ah what we have seen is ah

we started off this class by saying that we need a notation or which is this reciprocal

lattice notation because it will help us understand the interaction between ah the wave vectors

corresponding to the electrons and the crystal structure which is which basically conveys

to us the periodic structure that is present between all the components that are present

all the atomic components that are present within the ah ah lattice.

so to speak the material that we have so in that context we found ah we indicated that

ah that basically we might need to develop something in this reciprocal lattice notation

and it is that notation that will help us capture this interaction between the wave

vectors and the periodic crystal structure. so in that context we define reciprocal lattice

to be consisting of b 1, b 2 and b 3 with specific relationships to the real lattice

vectors a 1, a 2 and a 3. on the strength of that relationship we found

that on the strength of the definition we found that already b 1, b 2 and b 3 had specific

relations to a 1, a 2 and a 3 which then translated to a general reciprocal lattice vector h h

k l having specific relationships to the plane h k l in real space the relationships are

that the vector h h k l is perpendicular to the plane h k l. reciprocal lattice vector

perpendicular to a plane in real lattice. and the modulus of the reciprocal lattice

vector is one by the spacing between those h k l planes okay. so this is the framework

of our reciprocal space we already seen some important relationships here. in the next

class we will see how diffraction which is the interaction of waves ah with the ah periodic

crystal structure how the diffraction phenomena can be represented in the reciprocal lattice

notation.