Chapter5-Structure of Solid

Ex1-pg86

In [2]:
## Calculate packing efficiency and density of diamond
import math
n_c = 1/8. ## sharing of corner atom in a unit cell
N_c = 8. ## Number of corner atoms in unit cell
n_b = 1. ## sharing of body centered atom in a unit cell
N_b = 4. ## Number of body centered atoms in unit cell
n_f = 0.5## sharing of face centered atom in a unit cell
N_f = 6.## Number of face centered atoms in unit cell
a = 1. ## let lattice parameter
m = 12. ## mass of carbon
print("\n Example 5.1")
print("\n Part A:")
N = n_c*N_c+n_b*N_b+n_f*N_f ## effective number of atoms
r = a*math.sqrt(3.)/8.
p_e = N*4./3.*math.pi*r**3./a**3. ## packing efficiency

print'%s %.2f %s '%("\n Packing efficiency of diamond is ",p_e,"")
print("\n\n Part B:")
a = 3.57 ## lattice parameter in angstrom
d = m*1.66e-27*N/(a*1e-10)**3.
print'%s %.2f %s '%("\n Density of diamond is ",d," Kg/m^3")## numerical answer in book is 3500
print'%s %.2f %s '%("\n Density of diamond is ",d/1000," g/cm^3")

 Example 5.1

 Part A:

 Packing efficiency of diamond is  0.34  


 Part B:

 Density of diamond is  3502.47  Kg/m^3 

 Density of diamond is  3.50  g/cm^3

Ex3-pg92

In [4]:
## calculate the c/a ratio for an ideally close packed HCP crystal
import math
a = 1. ## let
PR = a
print(" Example 5.3")
RT = a/math.sqrt(3.)
PT = math.sqrt(PR**2.-RT**2.)
c_a = 2.*PT/PR
## Calculations are made on the crystal structure drawn in book
print'%s %.2f %s '%("\n c/a ratio for an ideally close packed HCP crystal is ",c_a,"")

 Example 5.3

 c/a ratio for an ideally close packed HCP crystal is  1.63

Ex4-pg94

In [5]:
## find the size of largest sphere that can fit into a tetrahedral void
import math
r = 1. ## let
a = 3./4. 
print("\n Example 5.4")
pt = 2.*math.sqrt(2./3.)*r
s = a*pt-r ## size of sphere
print'%s %.2f %s '%("\n Size of largest sphere that can fit into a tetrahedral void is r",s,"")

 Example 5.4

 Size of largest sphere that can fit into a tetrahedral void is r 0.22

Ex5-pg99

In [6]:
## find critical radius ratio for triangular coordination
import math
theta = 60. ## angle in degree
print("\n Example 5.5")
r_c_a = (2./3.*2.*math.sin(theta*math.pi/180.))-1. ## ratio calculation
print'%s %.2f %s '%("\n Critical radius ratio for triangular coordination is  ",r_c_a,"")

 Example 5.5

 Critical radius ratio for triangular coordination is   0.15

Ex6-pg101

In [8]:
## Calculate density of MgO
import math
r_mg = 0.78 ## radius of magnesium cation in angstrom
r_o = 1.32 ## radius of oxygen anion in angstrom
n = 4. ## effective number of unit cell
m_o = 16. ## mass of oxygen
m_mg = 24.3 ## mass of magnesium
print("\n Example 5.6")
a = 2.*(r_mg+r_o)## lattice parameter
d = (m_o+m_mg)*1.66e-27*n/(a*1e-10)**3.## density 
print'%s %.2f %s '%("\n Density of MgO is ",d," Kg/m^3") ## answer is 3610 kg/m**3
print'%s %.2f %s '%("\n Density of MgO is ",d/1000," g/cm^3")

 Example 5.6

 Density of MgO is  3611.81  Kg/m^3 

 Density of MgO is  3.61  g/cm^3