# Chapter 1: Crystallography¶

## Example 1.3.1,Page number 1-14¶

In :
import math

#given data

A=26.98                                 #atomic weight of Al
N=6.023*10**26                          #Avogadro's number
p=2700                                  #Density
n=4                                     #FCC structure

a=(n*A/(N*p))**(1./3)

print"Unit cell dimension of Al=","{0:.3e}".format(a),"m"

Unit cell dimension of Al= 4.049e-10 m


## Example 1.3.2,Page number 1-15¶

In :
import math

#given data

As=28.1                                 #atomic weight of Si
Ag=69.7                                 #atomic weight of Ga
Aa=74.9                                 #atomic weight of As
a_s=5.43*10**-8                         #lattice constant of Si
aga=5.65*10**-8                         #lattice constant of GaAs
ns=8                                    #no of atoms/unit cell in Si
nga=4                                   #no of atoms/unit cell in GaAs
N=6.023*10**23                          #Avogadro's number

#p=(n*A)/(N*a**3) this is formula for density

#for Si

ps=(ns*As)/(N*a_s**3)

print"1) Density of Si=",round(ps,4),"gm/cm^3"

#for GaAs

Aga=Ag+Aa                               #molecular wt of GaAs

pga=(nga*Aga)/(N*aga**3)

print"2) Density of GaAs=",round(pga,4),"gm/cm^3"

1) Density of Si= 2.3312 gm/cm^3
2) Density of GaAs= 5.3244 gm/cm^3


## Example 1.3.3,Page number 1-16¶

In :
import math

#given data

A=63.5                                  #atomic weight of Cu
N=6.023*10**23                          #Avogadro's number
n=4                                     #FCC structure
r=1.28*10**-8                            #atomic radius of Cu

#for FCC

a=4*r/(sqrt(2))                         #lattice constant
p=(n*A)/(N*a**3)

print"Density of Cu=",round(p,4),"gm/cm^3"

Density of Cu= 8.887 gm/cm^3


## Example 1.3.4,Page number 1-17¶

In :
import math

#given data

A=50                                    #atomic weight of chromium
N=6.023*10**23                          #Avogadro's number
p=5.96                                  #Density
n=2                                     #BCC structure

#step 1 : claculation for lattice constant (a)

a=(n*A/(N*p))**(1./3)

#step 2 : radius of an atom in BCC

r=sqrt(3)*a/4

#step 3 : Atomic packing factor (APF)

APF=n*((4./3)*math.pi*r**3)/a**3

print"Atomic packing factor (APF)=",round(APF,4)

Atomic packing factor (APF)= 0.6802


## Example 1.3.5,Page number 1-17¶

In :
import math

#given data

A=120                                   #atomic weight of chromium
N=6.023*10**23                          #Avogadro's number
p=5.2                                   #Density
n=2                                     #BCC structure
m=20                                    #mass

#step 1 : claculation for volume of unit cell(a**3)

a=(n*A/(N*p))

#step 2 : volume of 20 gm of the element

v=m/p

#step 3 :no of unit cell

x=v/a

print"no of unit cell=","{0:.3e}".format(x)

no of unit cell= 5.019e+22


## Example 1.3.6,Page number 1-18¶

In :
import math

#given data

A=132.91                                 #atomic weight of chromium
N=6.023*10**26                           #Avogadro's number
p=1900                                   #Density
a=6.14*10**-10                           #lattice constant

#step 1 : type of structure

n=(p*N*a**3)/A

print"n =",round(n)

print"BCC structure"

#step 2: no of atoms/m**3

x=n/a**3

print"no of atoms/m^3=","{0:.3e}".format(x)

n = 2.0
BCC structure
no of atoms/m^3= 8.610e+27


## Example 1.3.7,Page number 1-18¶

In :
import math

#given data

a=0.4049*10**-9                          #lattice constant
t=0.006*10**-2                           #thickness of Al foil
A=50*10**-4                              #Area of foil

V1=a**3                                  #volume of unit cell

V=A*t                                   #volume of the foil

N=V/V1                                  #no of unit cell in the foil

print"no of unit cell in the foil=","{0:.3e}".format(N)

no of unit cell in the foil= 4.519e+21


## Example 1.5.1,Page number 1-29¶

In :
import math

#given data

#refer diagram from textbook

#on joining centre of 3 anions,an equilateral triangle is formed and on joining centres of any anion and cation  a right angle triangle ABC os formed

#where AC=rc+ra

#and BC=ra

#m(angle (ACB))=30 degree

#therefore cos (30)=ra/(rc+ra)

#assume rc/ra=r

r=(1.0-math.cos(30.0*math.pi/180))/math.cos(math.pi*30/180)                 #by arrangimg terms we get value of r

print"critical radius ratio of ligancy 3=",round(r,4)

critical radius ratio of ligancy 3= 0.1547


## Example 1.5.2,Page number 1-30¶

In :
import math

#given data

#refer diagram from textbook

#in the said arrangement a cation is squeezed into 4 anions in a plane and 5th anion is in upper layer and 6th in bottom layer

#join cation anion centres E and B and complete the triangle EBF

#in triangle EBF m(angle F)=90 and EF=BF

#m(angle B)=m(angle E)=45

#and EB=rc+ra and BF=ra

#cos(45)=ra/(rc+ra)

#assume rc/ra=r

p=math.cos(45*math.pi/180)
r=(1-p)/math.cos(45*math.pi/180)                 #by arrangimg terms we get value of r

print"critical radius ratio for ligancy 6 =",round(r,4)

critical radius ratio for ligancy 6 = 0.4142


## Example 1.5.3,Page number 1-30¶

In :
import math

#given data

#refer diagram from textbook

#since plane is square hence it is same as ligancy 6

#in the said arrangement a cation is squeezed into 4 anions in a plane and 5th anion is in upper layer and 6th in bottom layer

#join cation anion centres E and B and complete the triangle EBF

#in triangle EBF m(angle F)=90 and EF=BF

#m(angle B)=m(angle E)=45

#and EB=rc+ra and BF=ra

#cos(45)=ra/(rc+ra)

#assume rc/ra=r

r=(1-math.cos(45*math.pi/180))/math.cos(45*math.pi/180)                 #by arrangimg terms we get value of r

print"critical radius ratio for ligancy 8 =",round(r,4)

critical radius ratio for ligancy 8 = 0.4142


## Example 1.5.4,Page number 1-31¶

In :
import math

#given data

#a tetrahedron CAEH can be considered with C as the apex of the tetrahedron.

#the edges AE,AH and EH of the tetrahedron will then be the face of the cube faces ABEF,ADHF,EFHG resp.

#from fig

#AO=ra+rc and AJ=ra

#AE=root(2)*a and AG=root(3)*a

#AO/AJ=AG/AE=(ra+rc)/ra=root(3)*a/root(2)*a

#assume rc/ra=r
r=(math.sqrt(3)-math.sqrt(2))/math.sqrt(2)

print"critical radius ratio for ligancy 4 = ",round(r,4)

critical radius ratio for ligancy 4 =  0.2247


## Example 1.5.5,Page number 1-32¶

In :
import math

#given data

#ligancy 8 represents cubic arrangment .8 anions are at the corners and touch along cube edgs.Along the body diagonal the central cation and the corner anion are in contact.

#cube edge=2*ra

#refer diagram from textbook

#and body diagonal=root(3)*cube edge=root(3)[2*(rc+ra)]

#assume rc/ra=r

r=math.sqrt(3)-1.0

print"critical radius ratio of ligancy 8=",round(r,4)

critical radius ratio of ligancy 8= 0.7321


## Example 1.5.6,Page number 1-32¶

In :
import math

#given data

#for an ionic crystal exibiting HCP structure the arrangment of ions refere from textbook

#at centre we have a cation with radius rc=OA

#it is an touch with 6 anions with radius ra=AB

#OB=OC=ra+rc

#intrangle ODB ,m(angle (OBC))=60 degree ,m(angle (ODB))=90 degree

#therefore cos(60)=BD/OB=AB/(OA+OB)=ra/(rc+ra)

#assume rc/ra=r

r=(1.-math.cos(60*math.pi/180))/math.cos(60*math.pi/180)                 #by arrangimg terms we get value of r

print"critical radius ratio 0f HCP structure=",round(r,4)

critical radius ratio 0f HCP structure= 1.0


## Example 1.6.2,Page number 1-35¶

In :
import math

#given data

#intercept of planeare in proportion a,b/3,2*c

#as a,b and c are basic vectors the proportin of intercepts 1:1/3:2

#therefore reciprocal

r1=1

r2=3

r3=1./2

#taking LCM of 2 and 1 is 2

l=2

m1=(l*r1)

m2=(l*r2)

m3=(l*r3)

print"miler indices=",m3,m2,m1

miler indices= 1.0 6 2


## Example 1.6.4,Page number 1-38¶

In :
import math

#given data

r=1.414                                 #atomic radius in amstrong unit

#for FCC structure

a=4*r/math.sqrt(2)

#part 1: plane(2,0,0)

#the interplanar spacing of plane

h1=2
k1=0
l1=0

#we know that d=a/sqrt(h**2+k**2+l**2)

d1=a/sqrt(h1**2+k1**2+l1**2)

print"1)interplanar spacing for (2,0,0) plane=",round(d1,4),"amstrong"

#part 2: plane(1,1,1)

#the interplanar spacing of plane

h2=1
k2=1
l2=1

#we know that d=a/sqrt(h**2+k**2+l**2)

d2=a/sqrt(h2**2+k2**2+l2**2)

print"2)interplanar spacing for(1,1,1) plane=",round(d2,4),"amstrong"

1)interplanar spacing for (2,0,0) plane= 1.9997 amstrong
2)interplanar spacing for(1,1,1) plane= 2.3091 amstrong


## Example 1.14.1,Page number 1-58¶

In :
import math

#given data

n=4                                     #FCC structure
ro=2180                                 #density of NaCl
M=23+35.5                               #molecular weight of NaCl
N=6.023*10**26                          #Avogadro's number

a=((n*M)/(N*ro))**(1.0/3)

print"Lattice constant=","{0:.3e}".format(a),"m"

Lattice constant= 5.627e-10 m


## Example 1.14.2,Page number 1-58¶

In :
import math

#given data

n=4                                     #FCC structure
ro=8.9                                  #density of Cu atom
A=63.55                                 #atomic weight of Cu atom
N=6.023*10**23                          #Avogadro's number

a=((n*A)/(N*ro))**(1./3)

print"1) Lattice constant=","{0:.3e}".format(a),"cm"

r=math.sqrt(2)*a/4                       #radius of Cu atom

d=2*r                                    #diameter of Cu atom

print"2) Diameter of Cu atom=","{0:.3e}".format(d),"cm"

1) Lattice constant= 3.620e-08 cm
2) Diameter of Cu atom= 2.559e-08 cm


## Example 1.14.3,Page number 1-59¶

In :
import math

#given data

n=8                                     #diamond structure
A=12.01                                 #atomic wt
N=6.023*10**23                          #Avogadro's number
a=3.75*10**-8                           #lattice constant of diamond

ro=(n*A)/(N*(a**3))

print"Density of diamond=",round(ro,4),"gm/cc"

Density of diamond= 3.025 gm/cc


## Example 1.14.4,Page number 1-59¶

In :
import math

#given data

#intercept of planeare in proportion 3a:4b:infinity (plane parallel to z axis)

#as a,b and c are basic vectors the proportin of intercepts 3:4:infinity

#therefore reciprocal

r1=1./3
r2=1./4
r3=0

#taking LCM of 3 and 4 i.e. 12

l=12

m1=(l*r1)

m2=(l*r2)

m3=(l*r3)

print"miler indices=",(m3,m2,m1)

miler indices= (0, 3.0, 4.0)


## Example 1.14.5,Page number 1-59¶

In :
import math

#given data

#intercept of planeare in proportion 3a:-2b:3/2c

#as a,b and c are basic vectors the proportin of intercepts 3:-2:3/2

#therefore reciprocal

r1=1./3
r2=-1./2
r3=2./3

#taking LCM of 3, 2 and 3/2 is 6

l=6

m1=(l*r1)

m2=(l*r2)

m3=(l*r3)

print"miler indices=",(m3,m2,m1)

miler indices= (4.0, -3.0, 2.0)


## Example 1.14.6,Page number 1-59¶

In :
import math

#given data

#if a plane cut at length m,n,p on the three crystal axes,then

#m:n:p=xa:yb:zc

#when primitive vectors of unit cell and numbers x,y,z,are related to miller indices (h,k,l)of the plane by relation

#1/x:1/y:1/z=h:k:l

#since a=b=c (crystal is simple cubic)

#and (h,k,l)=(1,2,3)

#therefore reciprocal

r1=1./1
r2=1./2
r3=1./3

#taking LCM of 1 ,2 and 3 is 6

l=6

m=(l*r1)

n=(l*r2)

p=(l*r3)

print"ratio of intercepts=",(m,n,p)

ratio of intercepts= (6.0, 3.0, 2.0)


## Example 1.14.7,Page number 1-60¶

In :
import math

#given data

#primitive vectors

a=1.2                                   #in amstrong unit
b=1.8                                   #in amstrong unit
c=2                                    #in amstrong unit

#miller indices of the plane

h=2
k=3
l=1

#therefore intercepts are a/h,b/k,c/l

x=a/h
y=b/k
z=c/l

#this gives intercepts along x axis as x amstrong but it is given tthat plane cut x axis at 1.2 amstrong .

t=1.2/x

#this shows that the plane under consideration is another plane which is parallel to it(to keep miller indices same)

n=t*y                                   #Y intercept

p=t*z                                   #Z intercept

print"1) Y intercept=",n,"amstrong"

print"2)Z intercept=",p,"amstrong"

1) Y intercept= 1.2 amstrong
2)Z intercept= 4.0 amstrong


## Example 1.14.8,Page number 1-61¶

In :
import math

#given data

#the interplanar spacing of plane

h=1
k=1
l=0
d=2                                     #interpanar spacing in amstrong unit

#we know that d=a/sqrt(h**2+k**2+l**2) therefore

a=d*math.sqrt(h**2+k**2+l**2)

#for FCC structure

r=math.sqrt(2)*a/4

print"radius r=",(r),"amstrong"

radius r= 1.0 amstrong


## Example 1.14.9,Page number 1-61¶

In :
import math

#given data

n=4                                     #for FCC structure

#the interplanar spacing of plane

h=1
k=1
l=1
d=2.08*10**-10                          #distance
A=63.54                                 #atomic weight of Cu
N=6.023*10**26                          #amstrong no

#we know that d=a/sqrt(h**2+k**2+l**2) therefore

a=d*math.sqrt(h**2+k**2+l**2)

#also (a**3*q)=n*A/N

q=n*A/(N*a**3)

print"1)density=",round(q,4),"kg/m^3"

#for FCC structure

r=math.sqrt(2)*a/4

d=r*2

print"2)radius r=","{0:.3e}".format(r),"m"

print"3)diameter d=","{0:.3e}".format(d),"m"

1)density= 9024.4855 kg/m^3
2)radius r= 1.274e-10 m
3)diameter d= 2.547e-10 m


## Example 1.14.10,Page number 1-62¶

In :
import math

#given data

A=63.546                                #atomic weight of Cu
N=6.023*10**26                           #Avogadro's number
p=8930                                  #Density
n=1.23                                  #no.of electron per atom

#density=mass/volume

#therfore 1/volume=density/mass

#since electron concentration is needed, let us find out no of atoms/volume(x)

x=N*p/A

#now one atom contribute n=1.23 electron

#therefore x atoms contribute  y no of free electron

y=x*n

print"free electron concentration=","{0:.3e}".format(y),"electron/m^3"

free electron concentration= 1.041e+29 electron/m^3


## Example 1.14.11,Page number 1-62¶

In :
import math

#given data

#primitive vectors

a=1.5                                   #in amstrong unit
b=2                                     #in amstrong unit
c=4.                                    #in amstrong unit

#miller indices of the plane

h=3
k=2
l=6

#therefore intercepts are a/h,b/k,c/l

x=a/h
y=b/k
z=c/l

#this gives intercepts along x axis as x amstrong but it is given that plane cut x axis at 1.2 amstrong .

t=1.5/x

#this shows that the plane under consideration is another plane which is parallel to it(to keep miller indices same)

n=t*y                                   #Y intercept

p=t*z                                   #Z intercept

print"1) Y intercept=",(n),"amstrong"

print"2)Z intercept=",(p),"amstrong"

1) Y intercept= 3.0 amstrong
2)Z intercept= 2.0 amstrong


## Example 1.14.12,Page number 1-63¶

In :
import math

#given data

ro=7.87                                 #density of metal
A=55.85                                 #atomic wt of metal
N=6.023*10**23                          #Avogadro's number
a=2.9*10**-8                            #lattice constant of metal

n=(N*(a**3)*ro)/A

print"Number of atom per unit cell of a metal=",round(n,0)

Number of atom per unit cell of a metal= 2.0


## Example 1.14.13,Page number 1-63¶

In :
import math

#given data

n=2                                     #BCC structure
ro=9.6*10**2                             #density of sodium crystal
A=23                                    #atomic weight of sodium crystal
N=6.023*10**26                           #Avogadro's number

a=((n*A)/(N*ro))**(1./3)

print"Lattice constant=","{0:.3e}".format(a),"m"

Lattice constant= 4.301e-10 m


## Example 1.14.15,Page number 1-64¶

In :
import math

#given data

ro=2.7*10**3                             #density of metal
A=27                                    #atomic wt of metal
N=6.023*10**26                           #Avogadro's number
a=4.05*10**-10                           #lattice constant of metal

n=(N*(a**3)*ro)/A

print"1) Number of atom per unit cell of a metal=",round(n,0)

r=math.sqrt(2)*a/4                           #radius of metal

print"2) atomic radius of a metal=","{0:.3e}".format(r),"m"

1) Number of atom per unit cell of a metal= 4.0
2) atomic radius of a metal= 1.432e-10 m


## Example 1.14.16,Page number 1-64¶

In :
import math

#given data

n=2                                     #BCC structure
ro=5.98*10**3                           #density of chromium
A=50                                    #atomic wt of chromium
N=6.023*10**26                          #Avogadro's number

a=((n*A)/(N*ro))**(1./3)

print"1) Lattice constant=","{0:.3e}".format(a),"m"

#for BCC

r=math.sqrt(3)*a/4                           #radius of chromium

APF=(n*(4./3)*math.pi*(r**3))/(a**3)

print"2) A.P.F. for chromium=",round(APF,4)

1) Lattice constant= 3.028e-10 m
2) A.P.F. for chromium= 0.6802


## Example 1.14.17,Page number 1-65¶

In :
import math

#given data

n=4                                     #FCC structure
ro=6250                                 #density
M=60.2                                  #molecular weight
N=6.023*10**26                           #Avogadro's number

a=((n*M)/(N*ro))**(1./3)

print"Lattice constant=","{0:.3e}".format(a),"m"

Lattice constant= 3.999e-10 m


## Example 1.14.19,Page number 1-66¶

In :
import math

#given data

a=2.82*10**-9                           #lattice constant
n=2                                     #FCC crystal
t=17.167                                #glancing angle in degree
q=math.pi/180*t                         #glancing angle in radians

#assuming reflection in (1,0,0) plane

h=1
k=0
l=0

d=a/math.sqrt(h**2+k**2+l**2)

#using Bragg's law , 2*d*sin(q)=n*la

la=2*d*sin(q)/n

print"wavlength of X-ray=","{0:.3e}".format(la),"m"

wavlength of X-ray= 8.323e-10 m


## Example 1.14.20,Page number 1-66¶

In :
import math

#given data

n=8                                     #Diamond structure
ro=2.33*10**3                           #density of diamond
M=28.9                                  #atomic  weight of diamond
N=6.023*10**26                          #Avogadro's number

a=((n*M)/(N*ro))**(1./3)

print"1) Lattice constant=","{0:.3e}".format(a),"m"

r=math.sqrt(3)*a/8                           #radius of diamond structure

print"2) atomic radius of a metal=","{0:.3e}".format(r),"m"

1) Lattice constant= 5.482e-10 m
2) atomic radius of a metal= 1.187e-10 m


## Example 1.14.21,Page number 1-66¶

In :
import math

#given data

n=2                                     #BCC structure
ro=8.57*10**3                            #density of chromium
d=2.86*10**-10                           #nearest atoms distance

#d=sqrt(3)/2*a

a=2*d/math.sqrt(3)

#now use formulae a**3*ro=n*A/N

#therefore a**3*ro/n=mass of unit cell/(no of atoms pre unit cell)=mass of one atom

m=a**3*ro/n

print"mass of one atom=","{0:.3e}".format(m),"kg"

mass of one atom= 1.543e-25 kg


## Example 1.15.1,Page number 1-68¶

In :
import math

#given data

d=4.255*10**-10                          #interplaner spacing
l=1.549*10**-10                          #wavelength of x ray

#part 1: for smallest glancing angle(n=1)

n1=1

#using Bragg's law n*l=2*d*sin(q)

q=math.degrees(math.asin(n1*l/(2*d)))

print"1)glancing angle=",round(q,4),"degree"

#part 2: for highst order

#for highest order sin(q) not exceed one i.e maximum value is one

#using Bragg's law n*l=2*d*sin(q)

n2=2*d/l                                #since sin(q)is one

print"2)highest order possible =",math.floor(n2)

1)glancing angle= 10.4875 degree
2)highest order possible = 5.0


## Example 1.15.2,Page number 1-69¶

In :
import math

#given data

a=2.125*10**-10                          #lattice constant
d=a/2                                   #interplaner spacing
n=2                                     #second order maximum
l=0.592*10**-10                          #wavelength of rock salt crystal

#using Bragg's law

q=math.degrees(math.asin((n*l)/(2*d)))                    #glancing angle

print"glancing angle=",round(q,4),"degree"

glancing angle= 33.8608 degree


## Example 1.15.3,Page number 1-69¶

In :
import math

#given data

n1=1                                    #for 1st order
n2=2                                    #for 2nd order
t=3.4                                   #angle where 1st order reflection done
t1=t*math.pi/180                        #convert degree to radian

m=math.sin(t1)

#but from Bragg's law

#n*l=2*d*sin(t)

#for for constant distance(d) and wavelength(l)

#order(n) is directly proportionl to sine of angle i.e (sin(t))

#n1/n2=sin(t1)/sin(t2)

#assume sin(t2)=a

a=n2/n1*m

t2=math.degrees(math.asin(a))                              #taking sin inverese in degree

print"second order reflection take place at an angle=",round(t2,4),"degree"

second order reflection take place at an angle= 6.812 degree


## Example 1.15.4,Page number 1-70¶

In :
import math

#given data

V=50*10**3                              #operating voltage of x-ray
M=74.6                                  #molecular weight
p=1.99*10**3                            #density
n=4                                     #no of atoms per unit cell(for FCC structure)
h=6.63*10**-34                           #plank's constant
c=3*10**8                                #velocity
e=1.6*10**-19                            #charge on electron
N=6.023*10**26                           #Avogadro's number

#step 1:clculating shortest wavelength

l=h*c/(e*V)

print"1)shortest wavelength=",(l),"m"

#step:2 calculating distance(d)

#now a**3*p=n*M/N therefore,

a=(n*M/(N*p))**(1./3)

#since KCl is ionic crystal herefore,

d=a/2

#step 3: calculaing glancing angle

#using Bragg's law

#n*l=2*d*sin(t)

#assume sin(t)=a, wavelength is minimum i.e l and n=1

n=1

a=n*l/(2*d)

t=math.degrees(math.asin(a))                             #taking sin inverese in degree

print"2) glancing angle=",round(t,4),"degree"

1)shortest wavelength= 2.48625e-11 m
2) glancing angle= 2.265 degree


## Example 1.15.5,Page number 1-70¶

In :
import math

#given data

n=1.0                                     #first order maximum
l=0.82*10**-10                          #wavelength of X ray
qd=7.0                                    #glancing angle in degree
qm=51./60                                #glancing angle in minute
qs=48./3600                              #glancing angle in second

q=qd+qm+qs                              #total glancin angle in degree

#using Bragg's law n*l=2*d*sin(q)

d=n*l/(2*math.sin(q*math.pi/180))

a=3*10**-10                              #lattice constant

#we know that d=a/root(h**2+k**2+l**2)

#assume root(h**2+k**2+l**2) =m

#arranging terms we get

m=a/d

print"square root(h**2+k**2+l**2)=",round(m,0)

print"hence possible solutions are (100),(010),(001)"

square root(h**2+k**2+l**2)= 1.0
hence possible solutions are (100),(010),(001)


## Example 1.15.6,Page number 1-71¶

In :
import math

#given data

n=1                                     #first order maximum
l=1j                                    #wavelength of X ray

#part 1:for(100)

#using Bragg's law n*l=2*d*sin(q)

q1=5.4                                   #glancing angle in degree

dl1=n*l/(2*math.sin(q1*math.pi/180))

#part 2:for(110)

#using Bragg's law n*l=2*d*sin(q)

q2=7.6                                   #glancing angle in degree

dl2=n*l/(2*math.sin(q2*math.pi/180))

#part 3:for(111)

#using Bragg's law n*l=2*d*sin(q)

q3=9.4                                   #glancing angle in degree

dl3=n*l/(2*math.sin(q3*math.pi/180))

#for taking ratio divide all dl by dl1

d1=dl1/dl1

d2=dl2/dl1

d3=dl3/dl1

print"cubic lattice structure is=",d1,d2,d3

 cubic lattice structure is= (1+0j) (0.711559669333+0j) (0.576199350225+0j)


## Example 1.15.7,Page number 1-71¶

In :
import math

#given data

n=1                                     #first order maximum
l=1.54*10**-10                           #wavelength of rock salt crystal
q=21.7                                  #glancing angle in degree

#using Bragg's law n*l=2*d*sin(q)

d=n*l/(2*math.sin(q*math.pi/180))

print"lattice constant of crystal=","{0:.3e}".format(d),"meter"

lattice constant of crystal= 2.083e-10 meter


## Example 1.15.8,Page number 1-72¶

In :
import math

#given data

a=2.814*10**-10                          #lattice constant

#the interplanar spacing of plane

h=1
k=0
l=0

d=a/math.sqrt(h**2+k**2+l**2)

n=2                                     #first order maximum

l=0.714*10**-10                         #wavelength of X-ray crystal

#using Bragg's law

q=math.degrees(math.asin((n*l)/(2*d)))                     #glancing angle

print"glancing angle=",round(q,4),"degree"

glancing angle= 14.6984 degree


## Example 1.15.9,Page number 1-72¶

In :
import math

#given data

d=2.82*10**-10                           #interplaner spacing
t=10                                    #glancing angle

#for part 1

n=1                                     #first order maximum

#using Bragg's law n*l=2*d*sin(t)

l=2*d*math.sin(math.pi*t/180)/n

print"1)wavelength=","{0:.3e}".format(l),"meter"

#for part 2

n1=2

#using Bragg's law n*l=2*d*sin(q)

q=math.degrees(math.asin(n1*l/(2*d)))

print"2)glancing angle=",round(q,4),"degree"

#for part 3

#for highest order sin(q) not exceed one i.e maximum value is one

#using Bragg's law n*l=2*d*sin(q)

n2=2*d/l                                #since sin(q)is one

print"3)highest order possible =",(floor(n2))

1)wavelength= 9.794e-11 meter
2)glancing angle= 20.322 degree
3)highest order possible = 5.0


## Example 1.15.10,Page number 1-73¶

In :
import math

#given data

#for line -A

n1=1                                    #1st order maximum
q1=30                                   #glancing angle in degree

#using Bragg's law for line A n1*l1=2*d1*sin(q1)

#d1=n1*l1/(2*sin(q1))

#for line B

l2=0.97                                 #wavelength in amstrong unit
n2=3                                    #1st order maximum
q2=60                                   #glancing angle in degree

#using Bragg's law for line B n2*l2=2*d2*sin(q2)

#since for both lines A and B we use same plane of same crystal,therefore

#d1=d2

#therefore equution became n2*l2=2*n1*l1/(2*sin(q1))*sin(q2)

#by arranging terms we get

l1=n2*l2*2*math.sin(q1*math.pi/180)/(2*n1*math.sin(q2*math.pi/180))

print"wavelength of the line A=",round(l1,4),"amstrong"

wavelength of the line A= 1.6801 amstrong


## Example 1.15.11,Page number 1-74¶

In :
import math

#given data

n=1.0                                     #first order minimum
d=5.5*10**-11                            #atomic spacing
e=1.6*10**-19                            #charge on one electron
Ee=10*10**3                              #energy in eV
E=e*Ee                                  #energy in Joule
m=9.1*10**-31                            #mass of elelctron
h=6.63*10**-34                           #plank's constant

l=h/math.sqrt(2*m*E)                         #wavelength

#using Bragg's law

q=math.degrees(math.asin((n*l)/(2*d)))                    #glancing angle

print"glancing angle=",round(q,4),"degree"

glancing angle= 6.4129 degree


## Example 1.15.12,Page number 1-74¶

In :
import math

#given data

a=2.814*10**-10                          #lattice constant

#for rock salt

d=a/2                                   #interplaner spacing

n=1                                     #first order maximum

l=1.541*10**-10                          #wavelength of rock salt crystal

#using Bragg's law

q=math.degrees(math.asin((n*l)/(2*d)))                     #glancing angl

print"glancing angle=",round(q,4),"degree"

glancing angle= 33.2038 degree


## Example 1.16.1,Page number 1-75¶

In :
import math

#given data

Ev=1.08                                 #average energy required to creaet a vacancy
k=1.38*10**-23                           #boltzman constant in J/K
e=1.6*10**-19                            #charge on 1 electron

K=k/e                                   #boltzman constant in eV/K

#for a low concentration of vacancies a relation is

#n=Nexp(-Ev/KT)

#since total no atom is 1 hence N=1

#at 1000k

T1=1000                                  #temperature

n1=math.exp(-Ev/(K*T1))

#at 500k

T2=500                                   #temperature

n2=math.exp(-Ev/(K*T2))

v=(n1)/(n2)                                #ratio of vacancies

print"ratio of vacancies=",round(v,4)

ratio of vacancies= 274234.5745


## Example 1.16.2,Page number 1-75¶

In :
import math

#given data

Ev=1.95                                 #average energy required to creaet a vacancy
k=1.38*10**-23                           #boltzman constant in J/K
e=1.6*10**-19                            #charge on 1 electron
K=k/e                                   #boltzman constant in eV/K
T=500                                   #temperature

#for a low concentration of vacancies a relation is

#n=Nexp(-Ev/KT)

m=math.exp(-Ev/(K*T))                           #ratio of no of vacancies to no of atoms n/N

print"ratio of no of vacancies to no of atoms=","{0:.3e}".format(m)

ratio of no of vacancies to no of atoms= 2.303e-20


## Example 1.16.3,Page number 1-76¶

In :
import math

#given data

Ev=1.8                                  #average energy required to creaet a vacancy
k=1.38*10**-23                           #boltzman constant in J/K
e=1.6*10**-19                            #charge on 1 electron
K=k/e                                   #boltzman constant in eV/K

#for a low concentration of vacancies a relation is

#n=Nexp(-Ev/KT)

#ratio of vacancy is n/N assume be r=exp(-Ev/KT)

#since total no atom is 1 hence N=1

#at 1000k

t1=-119                                  #temperature in degree
T1=t1+273                               #temperature in kelvine
r1=math.exp(-Ev/(K*T1))

print"1)ratio of vacancies at -119 degree=","{0:.3e}".format(r1)

#at 500k

t2=80                                    #temperature in degree

T2=t2+273                               #temperature in kelvine

r2=exp(-Ev/(K*T2))

v=(r1)/(r2)                                #ratio of vacancies

print"2)ratio of vacancies at 80 degree=","{0:.3e}".format(r2)

1)ratio of vacancies at -119 degree= 1.399e-59
2)ratio of vacancies at 80 degree= 2.110e-26


## Example 1.16.4,Page number 1-76¶

In :
import math

#given data

Ev=1.5                                  #energy of formaton of frankel defect
k=1.38*10**-23                          #boltzman constant in J/K
e=1.6*10**-19                           #charge on 1 electron
K=k/e                                   #boltzman constant in eV/K
T=700                                   #temperature
N=6.023*10**26                          #avogadro's no

#for a low concentration of vacancies a relation is

#n=Nexp(-Ev/KT)

m=math.exp(-Ev/(2*K*T))                        #ratio of no of vacancies to no of atoms n/N

qs=5.56                                 #specific density
q=5.56*10**3                            #real density ke/m**3
M=0.143                                 #molecular weight in kg/m**3
ma=M/N                                  #mass of one molecule
v=ma/q                                  #vol of one molecule

#v volume containe 1 molecule

#therefore 1 m**3 containe x molecule

x=1./v
d=m*x                                   #defect per m**3
dm=d*10**-9                              #defect per mm**3

print"number of frankel defects per mm^3=","{0:.3e}".format(dm)

number of frankel defects per mm^3= 9.432e+16

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