Concepts of Solid State (Problem Solving) - The Solid State Video Lecture - Class 12

hello my name is Pierre NGO though I am
a bit a graduate from IIT Roorkee
and I did a nice place for itj exam
today I want to discuss two important
concepts of solid steel many times I
found that students ignore this chapter
during the preparation of itj exam but
if we analyze PVCs iit-jee papers we can
find it every year at least one question
is asked from this chapter in fact in
2008 a passage of 20 marks was based
upon a solid steel here basically I want
to discuss different types of questions
that can be framed with the help of the
concepts of solid steel few modern
concepts of this chapter are like close
packing of spheres packing fraction
space lean LCP concepts of voids and
radius ratio I am a concepts of ionic
solids and defense Here I am going to
discuss different types of questions
that can be framed with the help of
concepts of solid state consider this
question this question says marbles of
radius 5 limiter are to be placed in a
square area of site 40 millimeter
such that the centres are within this
area what will be the arrangements of
marbles on a plane so that maximum
number of marbles can be placed inside
this area in fact similar type of
question was there in itj means 2003
exam now to deal with these types of
questions we need to know close packing
of the sphere the concept of
close packing of the sphere
if you try to place spheres closely or a
plane the two possible arrangements are
there one is like this
this is called a square close packing
another arrangement is
like this
this is called hydroponic rows back if
we compare the two we have one a sphere
is in contact with for his fear one in
this case one sphere is in contact with
six is fear the packing in this case is
better than square close packing so from
the packing point of view we can say
this is more economical now if we extend
this structure three dimensionally to
close packings have been observed one is
God as a baby type of packing or a
tentatively discard as xep hexagonal
roast packing and the other is ABC ABC
packing alternatively it is called as
CCP cubic closed packing in both the
cases the packing is approximately 74
percent now if you look at this question
in this question oh we need to put
maximum number of marbles within the
given area so obviously we have to take
this second arrangement to get maximum
number of marbles in a particular area
since the side length of this square is
for Emily we go so this side will go
like this
5 + 10 15 10 25 10:35 and 540 so by this
way for cable here also we need to get
14 meter to get the actual length we'll
drop this instruction let this as P Q R
this n we need to get this length H this
is 10 millimeter 10 millimeter 10 meter
so this is an equilateral triangle so
each angle will be 60 degree here e Q R
and we need to get this length x so here
we can use silence 60 is equal to x by
10 so X will be 10 sine 60 is root 3 by
2 which will be 5 root 3 meter
we'll extend this
x-x-x
X so if we try to find 4x equal to 4
into 5 root 3 so 20 into 1 point 7 3 2
this will be equal to thirty four point
six four millimeter if you add 5 meter
mole to this the total length will
become thirty four point six four five
which is approximately forty millimeters
so by this ready or draw this side no
so we need to find effective number of
items within this square so try to count
1 2 3 4 5 6 7 8 9 10 11 12 13 14 14
these fears are completely inside this
area if you look at this is fear this is
fear this is still this one this one
this one and this one 1 2 3 4 5 6 7
these seven spheres contribute only half
to this particular area so if it to
number of spheres will be 7 by 2 now if
we look at this is fear and this one
these two spheres contribute only 1 by 4
to this particular area so contribution
will be 1 by 4 so I can write 2 by 4 by
this way I get 4 e+ and here I get 4 so
18 so I can say 80 list fears can be
placed within this area so we can say to
get effective number of marbles of
maximum
Marlys basically we need to pack Spears
closely on a plane so this is the
correct way many times exams like itg
they ask questions from packing fraction
spatially in HCP in 2008 idj exam there
was a passage of 12 marks based upon
this HCP they serve an enclosed Pam now
to get the packing fraction in any unit
cell we need to draw that unit cell here
first of all we need to draw a CP unison
the sketches like this there are 12
order items here
- facing the ragams and three items are
inside this body if we join these two
atoms the line passes to st. rod of the
strangle in the fast lane
these Artemis are touching each other so
we can say if this length is e we can
write a is equal to 2 R where R is the
radius of this eigen let this as C hide
in C now first of all what is packing
fraction packing fraction is basically
fraction of total volume occupied by
sphere packing fraction can also be
written as effective number of atoms
into volume or one atom upon volume or
unit step we can say atoms are details
sphere so volume of one atom can be
written as 4 by 3 PI R cube if it will
number of atoms like this are same
volume of unit cell this is the
hexagonal close-packed structure volume
of this unit cell will be a area of the
base into height so area of base into
high now how will we find the area of
the base for this we will divide this
regular hexagon into see
equilateral triangles each side II so
this can be written as Z into 4 by 3 PI
R cube upon area of the base will be 6
into area of 1 equal added random so I
do the equilateral triangle will be root
3 by 4 and this side is so it will be L
square into height here is C so this is
the final packing fraction formula for
HCP here we will be asked to calculate
the relation between a and C we can be
asked to find the volume of the unit
cell so different types of questions can
be framed if using the concept of HCT so
when this is these things in detail
first of all I will discuss here if they
till number pattern's how to find a 50
number of atoms in this case to get
effective number of atoms we need to
know contribution of these atoms to this
particular unit cell consider this
corner I the corner atom is like this
since this angle is 120 so we can say 3
unit cells can be placed on one plane so
this corner is sounded by 3 unit cells
on Berkeley and 3 will be above this
plane that is why over on this corner
atom is surrounded by 6 unit cells so
contribution to a particle
still will be 1 by 6 now total number of
cars here are 1 2 3 4 5 6 7 8 9 10 11 12
12 corner atoms contribution is 1 by 6
so if it to number of items from the
corner will be this Plus these 2 are the
face centered atoms as we know the
contribution of face in the atom to a
particular unit cell is 1 by 2 so if you
kill number of atoms here 2 into 1 5
these 3 atoms are within this body that
is why contribution will be hundred
percent 3 2 1 so by this way if you do
the more fathoms 6 so 6 is the effect
enough atoms now next we need to get the
relation between en si
as you know this is the second layer is
the earlier
so obviously this item will touch these
three items if this either touches these
three items
so this length will be 2 R 2 R and 2 R
and since these three items are on one
plane so these three items also touching
each other so it will be cooler to our
tour as we know a is equal to R if I say
this as P Q R M and we draw here so it
will be like this
each link has a a a and they have a zoom
this is P Q R M now this is the
equilateral if you driving this with the
midpoint of Q n let this as 10 so this
is this length will be a by 2 per penny
so this is is a value we can get the
length rn rn will be root a square minus
a square by 4 so it will be root 3 by 2
a root 3 by 2 it now if you draw a
perpendicular from point P like this
let this point as oh so basically this
line passes through the centroid of this
triangle so we can say this
perpendicular will cut this are in line
as 2 is to 1 now since this is twos to
one we can get the lengths are oh it
will be 2 by 3 of this total 2 by 3 of
our it so it will be basically a by
ruthie substitute here we get this value
now if we take out this triangle
p.o are this as perpendicular who are
you know as a by root 3 P o P oh is
basically half of this total length so
it will be C by 2 and PR PR this is a so
we can use Pythagoras theorem here so I
can write a square is equal to C square
by 4 plus a square by 3 so I can get the
relation with the CND in this fan so I
can write here C square by 4 is equal to
a square minus a square by 3 it will be
R 2 a square by 3
so this C square can written as 8 a
square by 3 to the value of C will be 2
root 2 by root 3 e so here you can see
this is a relation between C and E now
to get the packing fraction we'll
substitute this relation in that formula
which we have written earlier
so the packing fraction will be as you
know infected number of items is 6 into
volume of one item was 4 by 3 PI R cube
upon 6 into volume of units a was 6 into
root 3 by 4 a square into C C value is
this 2 root 2 by root 3 e so if we
become
6 into 4 by 3 PI as we know 2 R is equal
to 8 so I the value of R in terms of a
will be 80 by 2 so I can write a cube by
8 upon 6 into root 3 by 4 into 2 root 2
by root 3 Kiki this taking equal will
get cancelled 6 cancel so effectively it
will be owed into PI into 4 root 3 upon
3 root 3 to 2 root 2 now if we calculate
this the value comes of the 0.74 so
packing fraction in lct is 0.74 and a
percentage packing will be 74 percent
and keep in mind if you are asked to
calculate the volume of the unit cell so
this will be the volume of the unit cell
SCE unit cell now I take another
question the question says is an
elevated crystallizes in fcc lattice
having each length of 100 Pico meter
calculate the maximum diameter of item
which can be placed in interstitial site
without distorting the structure
actually the question is of iit-jee 2005
exam now to deal with these types of
questions we need to know the concept of
void and radiation
consider accessibility FCC units a can
be drawn in this panel here atoms are at
corners of the cube as well as fingers
of this unison this FCC unit cell has
two types of oil one is tetrahedral void
and the other is octahedral void
nobody say dragon boy if you please
four atoms in this manner like this
they'll be void or gap between these
four atoms is called a tetrahedral void
now why you call this as a try and avoid
if you join the Centers of these four
atoms they get a tetrahedron that is why
we call this void as tetrahedral void
now what is octahedral void easy place
six atoms in this fan
the void here between these six atoms is
called octahedral void now why do we
call this as octagonal what if you join
the C in terms of these six atoms they
get an octahedron that is why a board is
called as of the middle void now to deal
with these sets of questions we need to
know the size of the item which can be
placed inside this tetrahedral void or
octahedral void first we will discuss a
tried void
now is he going to place
I am inside this a little boy we can
draw in this manner
keep in mind these items are touching
each other so lady radius of this Atanas
are this is the word rear we have placed
another item this distance can be
written as to let the radius of this
occur which is placed inside the void as
small R as we know this angle is of
tetrahedron it will be 109 degree 28
minutes so half of this angle will be
half of this angle will be 109 degree 28
minutes by 2 is equal to it will be 54
degrees and then 44 minutes this length
this length will be small R plus capital
R now this triangle this triangle we can
use here sine 50 480 44 minutes so I can
write as sine 54 degree 44 minutes is
equal to this length as our R upon small
R plus Epsilon and the value of sine 54
degree 44 minutes is zero point eight
one six now from here if you want to
find small R upon capital R the distance
approximately zero point two two fights
so
will help us helped us in finding the
size of the atom which can be placed
inside a tetrahedral void now if trying
to find out the size of the atoms can be
placed inside the octahedral point now
in case of octahedral void we have to
draw this
and the inside the world we cook another
item if you take take a cross-sectional
view we can draw in this manner like
this let the radius in this case as R
let the size of the atom which is placed
inside this October let the radius as
smaller so if if I draw this I can say
this angle is 45 degree this is
basically a square so I can use here cos
45 degree which can be written as R upon
R plus R from here
if I try to find R plus R on our it will
be 1 it will be root 2 root that is 1
point 4 1 4 so from here I get R upon R
as 0 point 4 1 4 so this will help us in
finding the size of the atom which can
be placed inside your title void now if
we look at the question the question
says you need to find the maximum
diameter of the atom which can be placed
inside the void now if we compare the
tetrahedral void and octahedral void by
looking
the radius ratio in case of tetrahedral
void the radius ratio comes 0.225 and is
octagonal what the radius ratio comes
0.4 wonderful so obviously to get the
maximum diameter we need to take
octahedral void and that is why a radius
issue will be this so in FCC unit cell
in FCC unit cell you want to put the
item in Maxim diameter inside the void
we need to consider octahedral void and
that is why we can write small R upon
capital R as zero point four one for
second these atoms are touching each
other
Neph CC unit cell basically atoms this
space a cracker will touch these corner
arm so I can say this as R to R so for R
can be written as a root to this length
will be a since it is given as 400 Pico
meter so I can write this as 400 route
to and from this we get the value of R
and I can substitute this value here so
I can write this as small R is equal to
capital R into 0.01 hole and this
capital R is nothing but 100 root 2 into
zero point 4 1 4 people since we need to
get the diameter so that is I have to
take to a so diameter of the autonomy
which can be placed in the octahedral
void will be 2 R it will be 2 into root
2 into 0 point 4 1 4 into 100 if you
solve this we get approximately 100
17 people another question that are they
here is the crystal XY which has rock
salt structure has molecular rate 6.0 to
3e further the question says if the
minimum distance between cation aniline
is into the power 1 by 3 nanometer and
you observe dense is 24 gamma did you
find the type of defect now to deal with
these sets of questions we need to know
the concepts of ionic solids and defect
or different ionic solids that we need
to study our NaCl Tycho structure z-star
infrastructure CSC with an obstruction
CF to write structure na2o and tip line
etc Here I am going to discuss any see
entity structure basically we need to
know features of these structures
two important features of nacl Titus
factions are number one DNI ca- for
example and eyes are arranged in a CCP
man
and cations of cations are there in
octahedral void
now on the base of these two features we
can conclude certain points like since
anions are in CCP manner so effective
number of anions will be will be poor if
it win over France 24 and since the
ratio is vanish to one that is by petal
number of cations is also four on this
basis we can say one unit cell of any CL
has for any seer units a very important
conclusion one unit say of ADC has four
NaCl units now by looking at the second
feature since it is cations are there
inside the octet word which means the
coordination number of cation is six and
sensations one is to one that is my
coordination number and I will also be
six now if you draw in a scale or
structure
and ions means cl- at corners of the
queue as well as at facing and the
Qataris are inside the octahedral voids
as you know octahedral voids are there
at bodhichitta and easy
so this will be the structure of NACA
solid as you see this cation and here
the to a lie on this distance is e since
the cation that is these two a tile so
we can say like this it will be R - R -
and here - R + so I can write 2 R + + -
r - is equal to a means r plus plus r
minus is equal to t by 3 so i can say
the minimum distance between katana and
i in NaCl type structure is e by 2 where
is the side length
now I discuss different types of defects
that are found in our a solid different
difference that we need to know here are
like shock a defect
find anything
met Alexis defect
method definition CD
these difference we need to study in
detail here I wanted to tell you that in
shortly different we can say caravans
missing in frankly different of cation
moves into interstitial site shortly
defect are found in those I saw its
where cations and anions in science well
in Franklin defect the size of pecan is
much smaller than the size of a dime
in short a defect density decreases
while Franco different density ribbons
see in matrix is different as you see
here the density increases because
better is in excess now intensity
increases then observe the density will
be more than calculate density while in
this shortly defective observed density
will be less Randy calculated density
now look at the question the question
since you need to find the typo defect
he observe this is given which means we
have to calculate the density on the
basis of given data
here as you see the XY structure is
basically rocks or destruction rocks or
destruction is NaCl a group structure in
initiative
spectre-1 uni sill has four units of XY
four units of XY
now how to find density density is
nothing but mass by one mass of the
units and volume of the unit cell
all right now how do you find mass of
the unix it since one unit cell has four
XY molecules so mass of burn unit cell
maybe a four into mass of XY molecule
upon volume gives you reason let the
length of the unit say as L the volume
will be L cube as we know of in initiate
their destruction the minimum distance
between cation and anion is L by 2 and
here it is given as a to the power 1 by
so I can say L by 2 is
to e to the power 1 by 3 which means L
is equal to 2 into a to the power 1 by
now substitute all the values 4 into to
find the mass of 1 XY 1 if you are right
molar mass of XY molecule which is given
as six point zero two three eight
divided by six point zero 2 3 10 to the
power 23 into volume is L cube so I will
put here is 2 e to the power 1 by 3 this
will become e e now if I can equally but
here keep in mind this a to the power
unless it is given in nanometer so L
will be nanometer and since we need to
find this in kilogram per meter cube so
I will convert this nano meter into
meters I need to put 10 to the power
minus 9 Q and since mass in graph and V
1 in kg that is why we need to put into
the car - thing finally it will become 1
by 2
into 10 to the power minus 23 will
become 26 upon 10 to the power minus 27
so it becomes 10 by 2 which is 5 5
kilogram per meter cube as you see he
calculate it in cities 5 you happily
with you while you observe their cities
20 kilogram per meter so we can say the
calculated density is less than the
observed density and that is why the
defect will be metric system