Chapter 2:Crystal Structure

Example 2.1 , Page no:40

In [1]:
#given
N=6.02*10**26; #in /Kg-molecule (Avogadro's number)
n=4; #number of molecules per unit cell ofr NaCl
M=58.5; #in Kg/Kg-molecule (molecular weight of NaCl)
p=2189; #in Kg/m^3 (density)

#calculate
a=pow(((n*M)/(N*p)),0.3333333);
a1=a*10**10; #changing unit to Angstrom

#result
print"The lattice constant is a=",'%.2E'%a,"m";
print"\t\t\ta=",round(a1,2),"Angstrom";

The lattice constant is a= 5.62E-10 m
			a= 5.62 Angstrom

Example 2.2 , Page no:41

In [2]:
import math
from __future__ import division

#given
N=6.02*10**23; #in /gram-atom (Avogadro's number)
n=4; #number of atom per unit cell for fcc structure
M=63.5; #in gram/gram-atom (atomic weight of Cu)
p=8.96; #in g/cm^3 (density)

#calculate
a=pow((n*M/(N*p)),0.3333333);
a1=a*1E8; #changing unit from cm to Angstrom
d=a1/math.sqrt(2); #distance infcc lattice

#result
print"lattice constant is a=",'%.2E'%a,"cm";
print"\t\t    a=",round(a1,2),"Angstrom";
print"distance between two nearest Cu atoms is d=",round(d,2),"Angstrom";

lattice constant is a= 3.61E-08 cm
		    a= 3.61 Angstrom
distance between two nearest Cu atoms is d= 2.55 Angstrom

Example 2.3 , Page no:41

In [3]:
import math
from __future__ import division

#given
N=6.02E26; #in /Kg-atom (Avogadro's number)
n=2; #number of molecules per unit cell for bcc lattice
M=55.85; #in Kg/Kg-atom (atomic weight of Iron)
p=7860; #in Kg/m^3 (density)

#calculate
a=pow((n*M/(N*p)),0.33333);
a1=a*1E10; #changing unit to Angstrom

#result
print"The lattice constant is a=",'%.2E'%a,"m";
print"\t\t\ta=",round(a1,2),"Angstrom";

The lattice constant is a= 2.87E-10 m
			a= 2.87 Angstrom

Example 2.4 , Page no:42

In [4]:
import math
from __future__ import division

#given
N=6.02*10**26; #in /Kg-atom (Avogadro's number)
n=2; #number of molecules per unit cell for bcc lattice
M=6.94; #in Kg/Kg-atom (atomic weight of Iron)
p=530; #in Kg/m^3 (density)

#calculate
a=pow((n*M/(N*p)),0.33333);
a1=a*1E10; #changing unit to Angstrom

#result
print"The lattice constant is a=",'%.2E'%a,"m";
print"\t\t\ta=",round(a1,2),"Angstrom";

The lattice constant is a= 3.52E-10 m
			a= 3.52 Angstrom

Example 2.5 , Page no:42

In [5]:
import math
from __future__ import division

#given
N=6.02*10**23; #in /gram-molecule (Avogadro's number)
M=58.5; #in gram/gram-molecule (atomic weight of NaCl)
p=2.17; #in g/cm^3 (density)

#calculate
#since V=M/p
#(1/d)^-3=2N/V=2Np/M
#therefore d= (M/2Np)^-3
d=pow((M/(2*N*p)),0.33333333);
d1=d*1*10**8; #changing unit from cm to Angstrom

#result
print"The distance between two adjacent atoms of NaCl is d=",'%.2E'%d,"m";
print"\t\t\t\t\t\t   d=",round(d1,2),"Angstrom";

The distance between two adjacent atoms of NaCl is d= 2.82E-08 m
						   d= 2.82 Angstrom

Example 2.6 , Page no:43

In [6]:
import math
from __future__ import division

#given
r_Na=0.98; #in Angstrom (radius of sodium ion)
r_Cl=1.81; #in Angstrom (radius of chloride ion)
M_Na=22.99; #in amu (atomic mass of sodium)
M_Cl=35.45; #in amu (atomic mass of chlorine)

#calculate
a=2*(r_Na+r_Cl); #lattice parameter
#PF=volume of ions present in the unit cell/volume of unit cell
PF=((4*(4/3)*3.14)*r_Na**3+(4*(4/3)*3.14)*r_Cl**3)/a**3;
#Density=mass of unit cell/volume of unit cell
p=4*(M_Na+M_Cl)*1.66E-27/(a*1E-10)**3;
p1=p*1E-3; #changing unit to gm/cm^-3

#result
print"Lattice constant  is a=",round(a,3),"Angstrom";
print"Packing fraction is =",round(PF,3); 
print"Density is p=",round(p),"Kg/m^3";
print"Density is p=",round(p1,2),"g/cm^3";

Lattice constant  is a= 5.58 Angstrom
Packing fraction is = 0.662
Density is p= 2233.0 Kg/m^3
Density is p= 2.23 g/cm^3