Chapter 4:Structure of Crystalline Solids

Example 4.1 Page 67

In [1]:
#initiation of variable
from math import sqrt
R = 1.0 # let radius of an atom is unity

#calculation
#For FCC a=2*R*sqrt(2)
a=2.0*R*sqrt(2.0) #Edge Length
V=a**3 #Volume determination

#result
print" Volume of an FCC unit cell is %fR^3 " %V

 Volume of an FCC unit cell is 22.627417R^3

Example 4.2 Page 68

In [2]:
#initiation of variable
from math import sqrt, pi
R = 1.0 # let radius of an atom to be unity 
n=4.0 # no. of atoms for FCC are 4

#calculation
a=2.0*R*sqrt(2.0) # edge length for FCC
Vc=a**3 #Volume of cube
Vs=n*4*pi*R**3/3 #Volume of sphere
APF=Vs/Vc #Atomic packing Fraction

#result
print" Atomic packing factor is %.2f" %APF

 Atomic packing factor is 0.74

Example 4.3 Page 68

In [3]:
#initiation of variable
from math import sqrt, pi
R=1.28e-08#Atomic radius in cm
A_Cu=63.5 #Atomic wt of copper
n=4.0   #For FCC
Na=6.023e23 #Avogadro no.

#calculation
a=2*R*sqrt(2.0)
Vc=a**3
rho=n*A_Cu/(Vc*Na)

#result
print" Density of copper is %.2f g/cm^3." %rho

 Density of copper is 8.89 g/cm^3.

Example 4.4 Page 77

In [5]:
#initiation of variable
from math import sqrt, pi, cos
theta = 30.0 # angle between a line joining centres of anion with their median in degree

#calculation
r_ratio = (1- cos(theta*pi/180))/ cos(theta*pi/180)

#result
print" Minimum cation to anion radius ratio is %.3f" %r_ratio

 Minimum cation to anion radius ratio is 0.155

Example 4.5 Page 82

In [8]:
#initiation of variable
from math import sqrt, pi, cos
r_fe2 = 0.077 # radius of Iron ion in nm
r_o2 = 0.14 # radius of oxygen ion in nm
r_ratio = r_fe2 /r_o2 # cation - anion radius ratio for FeO

#calculation
if r_ratio > 0.414 and r_ratio < 0.732 :
    print"  As ratio lies between 0.414 and 0.732 and coordination number 6, so it is of AX crystal structure."


print "coordination number 6 is taken from table"

  As ratio lies between 0.414 and 0.732 and coordination number 6, so it is of AX crystal structure.
coordination number 6 is taken from table

Example 4.6 Page 83

In [9]:
#initiation of variable
from math import sqrt, pi, cos
n = 4.0 # number of crystals per unit cell
A_c = 22.99 # molar mass of cation i.e. sodium
A_a = 35.45 # molar mass of cation i.e. chlorine
r_c = 1.02e-8 # radius of sodium atom in cm
r_a = 1.81e-8 # radius of sodium atom in cm
N_a = 6.023e23 # Avogadro constant

#result
a = 2*(r_c+r_a)# edge length of Nacl molecule
V_c = a**3 # Volume of unit cell
rho = (n*(A_c+A_a))/(V_c*N_a)

#result
print" Theoretical density of crystal is %0.2f g/cm^3 ." %rho
print" This result compares very favourable with  experimental value of 2.16 g/cm^3 ."

 Theoretical density of crystal is 2.14 g/cm^3 .
 This result compares very favourable with  experimental value of 2.16 g/cm^3 .

Example 4.7 Page 94

In [3]:
#initiation of variable
from math import sqrt, pi, cos
n = 2 # number of repeated units within unit cell
A_c = 12.01  # molar mass of carbon
A_h = 1.008 # molar mass of hydrogen
a = 0.741 # edge length in x axis in nm
b = 0.494 # edge length in y axis in nm
c = 0.255 # edge length in z axis in nm
N_a = 6.023e23 # Avogadro constant
rho_s = 0.925 # density of branched polyethylene in g/cm^3
rho_a = 0.870 # density of totally amorphous polyethylene in g/cm^3

# Part A
#calculation
A = 2*A_c+4*A_h # Molar mass of polyethylene
V_c = a*b*c*(1.0e-7)**3 # Volume of unit cell
rho_c = (n*A)/(V_c*N_a)

print" Density of totally crystalline polyethylene is %0.3f g/cm^3 ." %rho_c

#part B
per_cry = (rho_c*(rho_s-rho_a))*100/(rho_s*(rho_c-rho_a))

#result
print" Percentage crystallinity is %0.1f  percent" %per_cry

 Density of totally crystalline polyethylene is 0.998 g/cm^3 .
 Percentage crystallinity is 46.4  percent

Example 4.8 Page 102

In [6]:
#initiation of variable
from math import sqrt, pi, cos, asin
a = 0.2866 # lattice parameter in nm
h = 2.0 # component of set of plane
k = 2.0# component of set of plane
l = 0.0# component of set of plane
n = 1.0 # order of detraction
Lambda = 0.1790 # wavelength of light in nm

#part A
#calculation
d_hkl=a/(sqrt(h**2+k**2+l**2))
theta=asin(n*Lambda/(2*d_hkl))

#result
print" Interplanar spacing is %.4f nm." %d_hkl


# Part B
print " Diffraction angle is %0.2f degree." %(2*theta*180/pi)
print "Answer in book is 124.26 degree. It is so because of consideration of different number of significant figures in calculation."

 Interplanar spacing is 0.1013 nm.
 Diffraction angle is 124.08 degree.
Answer in book is 124.26 degree. It is so because of consideration of different number of significant figures in calculation.