4: Crystallography

Example number 4.2, Page number 70

In [5]:
#import modules
import math
from __future__ import division

#Variable declaration
d=2180;      #density of NaCl(kg/m^3)
M=23+35.5;      #Molecular weight of NaCl(gm)
Na=6.02*10**26;         #Avgraodo no(per kg mole)
n=4;        #for f.c.c

#calculation
a=(n*M/(Na*d))**(1/3);       #lattice constant(m)
d=a/2;               #distance(m)
d=d*10**10;        #distance(angstrom)
d=math.ceil(d*10**3)/10**3;   #rounding off to 3 decimals

#Result
print "distance between two adajcent atoms is",d,"angstrom"

distance between two adajcent atoms is 2.815 angstrom

Example number 4.3, Page number 70

In [17]:
#import modules
import math
from __future__ import division

#Variable declaration
d=2.163;      #density(gm/cm^3)
M=58.45;      #molecular weight(gm)
Na=6.02*10**23;      #Avgraodo no.(molecules/gm mole)

#calcualtion
n=Na/M;      #no. of molecules(per gram)
n=n*d;       #no. of molecules(per cm^3) 
n=2*n;       #no. of atom(per cm^3)
n=n**(1/3);        #no. of atoms in a row 1cm long
d=1/n;           #spacing between atoms(cm)

#Result
print "spacing between atoms is",round(d/1e-8,2),"angstrom"
print "answer in the book varies due to rounding off errors"

spacing between atoms is 2.82 angstrom
answer in the book varies due to rounding off errors

Example number 4.4, Page number 74

In [29]:
#import modules
import math
from __future__ import division

#Variable declaration
r=1.278;     #radius(A.U)
n=4;        #structure is f.c.c
M=63.54;      #atomic weight(gm)
Na=6.02*10**23;       #Avgraodo no.(per gm mole)

#calculation 
a=4*r/(math.sqrt(2));        #lattice constant(A.U)
V=a**3;          #volume(cm^3)
rho=n*M/(Na*V);    #density(gm/cm^3)
rho=rho*(10**8)**3      #density(gm/m^3)

#Result
print "Density is",round(rho,2),"g/m^3"
print "answer in the textbook varies due to rounding off errors"

Density is 8.94 g/m^3
answer in the textbook varies due to rounding off errors

Example number 4.10, Page number 85

In [34]:
#import modules
import math
from __future__ import division

#Vaiable declaration
r=1.746;    #atomic radius(AU)

#calulation
a=4*r/math.sqrt(2);         #lattice constant(AU)
#for (200)
h=2;k=0;l=0;
d=a/math.sqrt(h**2+k**2+l**2);          #interplanar spacing(AU)
#for (220)
h=2;k=2;l=0;            
d1=a/math.sqrt(h**2+k**2+l**2);         #interplanar spacing(AU)
#for (111)
h=1;k=1;l=1;
d2=a/math.sqrt(h**2+k**2+l**2);         #interplanar spacing(AU)
d=math.ceil(d*10**4)/10**4;   #rounding off to 4 decimals
d1=math.ceil(d1*10**3)/10**3;   #rounding off to 3 decimals
d2=math.ceil(d2*10**3)/10**3;   #rounding off to 3 decimals

#Result
print "spacing for (200) is",d,"A.U"
print "answer in the book varies in 3rd decimal due to rounding off errors"
print "spacing for (220) is",d1,"A.U"
print "spacing for (111) is",d2,"A.U"

spacing for (200) is 2.4693 A.U
answer in the book varies in 3rd decimal due to rounding off errors
spacing for (220) is 1.746 A.U
spacing for (111) is 2.852 A.U

Example number 4.11, Page number 86

In [1]:
#import modules
import math
from __future__ import division

#calculation
#for (i)
l=1;m=0;n=0;
p=0;q=1;r=0;
d=math.acos(((l*p)+(m*q)+(n*r))/(math.sqrt(l**2+m**2+n**2)*math.sqrt(p**2+q**2+r**2)));      #angle between 2 normals(radian)
theta=math.degrees(d);         #angle between 2 normals(degrees) 

#for (ii)
l=1;m=2;n=1;
p=1;q=1;r=1;
d1=math.acos(((l*p)+(m*q)+(n*r))/(math.sqrt(l**2+m**2+n**2)*math.sqrt(p**2+q**2+r**2)));       #angle between 2 normals(radian)
theta1=math.degrees(d1);              #angle between 2 normals(degrees)  
deg=int(theta1);             #angle(degrees)
t=60*(theta1-deg);
m=int(t);                   #angle(minutes)
 
#Result
print "angle between the normal to pair of miller incdices (100) and (010) is",theta,"degrees"
print "angle between the normal to pair of miller incdices (121) and (111) is",deg,"degrees",m,"minutes"

angle between the normal to pair of miller incdices (100) and (010) is 90.0 degrees
angle between the normal to pair of miller incdices (121) and (111) is 19 degrees 28 minutes

Example number 4.13, Page number 87

In [8]:
#import modules
import math
from __future__ import division

#variable declaration
a=3.61*10**-7;      #lattice constant(mm)

#Calcualtion
#for plane (100)
SA=a*a;      #surface area(mm^2)
tamc=2;     #total atoms included
ans=tamc/SA;        #number of atoms per mm^2
#for (ii) plane (110)
A=a*(math.sqrt(2)*a);      #area of the plane(mm^2)
tamc=2;     #total atoms included according to sketch
ans1=tamc/A;       #number of atoms per mm^2
#for (iii) plane (111)
A=0.866*a*a;      #area of the plane(mm^2)
tamc=2;         #total atoms included according to sketch
ans2=tamc/A;      #number of atoms per mm^2

#Result
print "atoms per mm^2 for (100) is",round(ans/1e+13,3),"*10**13 atoms/mm^2"
print "atoms per mm^2 for (110) is",round(ans1/1e+13,3),"*10**13 atoms/mm^2"
print "atoms per mm^2 for (111) is",round(ans2/1e+13,3),"*10**13 atoms/mm^2"
print "answer for plane (111) given in the book is wrong"

atoms per mm^2 for (100) is 1.535 *10**13 atoms/mm^2
atoms per mm^2 for (110) is 1.085 *10**13 atoms/mm^2
atoms per mm^2 for (111) is 1.772 *10**13 atoms/mm^2
answer for plane (111) given in the book is wrong