Chapter 2: Crystal Structure

Example 2.1, Page number 2.16

In [2]:
from math import sqrt 

#Variable declaration
#Assuming r=1 for simpliciy in calculations
r = 1

#Calculations
a = (4*r)/sqrt(3)
#Let R be the radius of interstitial sphere that can fit into the void,therefore,
R = (a-2*r)/2     

#Result
print "The maximum radius of interstitial sphere is",round(R,3),"r"

The maximum radius of interstitial sphere is 0.155 r

Example 2.2, Page number 2.17

In [3]:
from math import sqrt

#Variable declaration
r1 = 1.258*10**-10   #atomic radius(m)
r2 = 1.292*10**-10   #atomic radius(m)

#Calculations
#In BCC
a_bcc = (4*r1)/sqrt(3)
v_bcc = a_bcc**3          #volume of unit cell(m^3)
n1 = ((1./8.)*8.)+1
V1 = v_bcc/n1             #volume occupied by 1 atom(m^3)

#In FCC
a_fcc = 2*sqrt(2)*r2
v_fcc = a_fcc**3          #volume of unit cell(m^3)
n2 = ((1./2.)*6.)+((1./8.)*8.)
V2 = v_fcc/n2             #volume occupied by 1 atom(m^3)

del_v = ((V1-V2)/V1)*100  #change in volume

#Result
print "During the conversion of iron from BCC to FCC, the decrease in volume is",round(del_v,1),"%"

During the conversion of iron from BCC to FCC, the decrease in volume is 0.5 %

Example 2.3, Page number 2.17

In [4]:
from math import sqrt

#Variable declaration
a = 0.27*10**-9   #nearest neighbour distance(m)
c = 0.494*10**-9  #height of unit cell(m)
M = 65.37         #atomic weight of zinc
N = 6.023*10**26  #Avogadro's number(k/mol)

#Calculations
V = (3*sqrt(3)*a**2*c)/2  #volume of unit cell
rho = (6*M)/(N*V)         #density of crystal

#Results
print "Volume of unit cell =",round((V/1E-29),2),"*10^29 m^3"
print "Density of zinc =",round(rho),"kg/m^3"
print "\nThe solution differs because of rounding-off of the digits in the textbook"

Volume of unit cell = 9.36 *10^29 m^3
Density of zinc = 6960.0 kg/m^3

The solution differs because of rounding-off of the digits in the textbook

Example 2.4, Page number 2.18

In [5]:
from math import sqrt

#Varaible declaration
#Let r be the radius of atom and R be the radius of sphere
#For simplicity in calculations, let us assume r =1
r = 1

#Calculations
#For FCC structure
a = (4*r)/sqrt(2)
R = (a/2)-r

print "Maximum radius of sphere =",round(R,3),"r"

Maximum radius of sphere = 0.414 r

Example 2.5, Page number 2.19

In [58]:
#Variable declaration
a = 0.356*10**-9  #cube edge(m)
M = 12.01         #atomic weight of carbon
N = 6.023*10**26  #Avogadro's number(k/mol)
na = 1.77*10**29  #no. of atoms per meter cube

#Calculations
#Diamond has 2 interpenetrating FCC lattices. Since each FCC unit cell has 4 atoms, the total no. of atoms per unit cell is 8
n = 8/a**3
m = M/N
rho = m*na

#Result
print "Number of atoms =",round((n/1E+29),3),"*10^29"
print "The density of diamond is",round(rho,1),"kg/m^3(Calculation mistake in the textbook)"

Number of atoms = 1.773 *10^29
The density of diamond is 3529.4 kg/m^3(Calculation mistake in the textbook)

Example 2.6, Page number 2.19

In [4]:
#Variable declaration
rho = 2.18  #density of NaCl(gm/cm^3)
N = 6.023*10**23  #Avogadro's number(/mol)

#Caculations
w = 23+35.5  #molecular weight of NaCl
m = w/N      #mass of NaCl molecule
nm = rho/m    #no. of molecules per unit volume(molecule/cm^3)
#Since NaCl is diatomic
n = 2*nm
#Let a be the distance between adjacent atoms in NaCl and
#    n be the no. of atoms along the edge of the cube
#length of an edge = na
#volume of unit cube = n^3*a^3
a = (1/n)**(1./3.)

#Result
print "The distance between two adjacent atoms is",round((a/1E-8),2),"A"

The distance between two adjacent atoms is 2.81 A

Example 2.7, Page number 2.20

In [3]:
from math import sqrt

#Variable declaration
w = 63.5           #atomic weight of copper
r = 1.278*10**-8   #atomic rdius(m)
N = 6.023*10**23   #Avogadro's number(/mol)

#Calculations
m = w/N    #mass of each copper atom(gm)
#Since copper has FCC structure lattice constant
a = (4*r)/sqrt(2) 
n = 4      #no. of atoms in unit cell of FCC structure
M = n*m    #mass of unit cell
rho = M/a**3  #density

#Result
print "Density of copper =",round(rho,2),"gm/cm^3"

Density of copper = 8.93 gm/cm^3