Chapter 1:Bonding in Solids

Example 1.3 , Page no:15

In [1]:
#given
r=2; #in angstrom(distance)
e=1.6E-19; # in C (charge of electron)
E_o= 8.85E-12;# absolute premittivity

#calculate
r=2*1*10**-10; # since r is in angstrom
V=-e**2/(4*3.14*E_o*r); # calculate potential
V1=V/e; # changing to eV

#result
print "\nThe potential energy is  V = ",'%.4E'%V,"J";
print "In electron-Volt V = ",round(V1,3),"eV"; 
print "Note: the answer in the book is wrong due to calculation mistake";

The potential energy is  V =  -1.1515E-18 J
In electron-Volt V =  -7.197 eV
Note: the answer in the book is wrong due to calculation mistake

Example 1.4 , Page no:15

In [2]:
import math
from __future__ import division
#given
r0=0.236; #in nanometer(interionic distance)
e=1.6E-19; # in C (charge of electron)
E_o= 8.85E-12;# absolute premittivity
N=8; # Born constant
IE=5.14;# in eV (ionisation energy of sodium)
EA=3.65;# in eV (electron affinity of Chlorine)
pi=3.14; # value of pi used in the solution

#calculate
r0=r0*1E-9; # since r is in nanometer
PE=(e**2/(4*pi*E_o*r0))*(1-1/N); # calculate potential energy
PE=PE/e; #changing unit from J to eV
NE=IE-EA;# calculation of Net energy
BE=PE-NE;# calculation of Bond Energy

#result
print"The potential energy is PE= ",round(PE,2),"eV";
print"The net energy is NE= ",round(NE,2),"eV";
print"The bond energy is BE= ",round(BE,2),"eV";
# Note: (1)-In order to make the answer prcatically feasible and avoid the unusual answer, I have used r_0=0.236 nm instead of 236 nm. because using this value will give very much irrelevant answer.
# Note: (2) There is slight variation in the answer due to round off.

The potential energy is PE=  5.34 eV
The net energy is NE=  1.49 eV
The bond energy is BE=  3.85 eV

Example 1.5 , Page no:16

In [3]:
#given
r_0=.41; #in mm(lattice constant)
e=1.6E-19; #in C (charge of electron)
E_o= 8.85E-12; #absolute premittivity
n=0.5; #repulsive exponent value
alpha=1.76; #Madelung constant
pi=3.14; # value of pi used in the solution

#calculate
r=.41*1E-3; #since r is in mm
Beta=72*pi*E_o*r**4/(alpha*e**2*(n-1)); #calculation compressibility

#result
print"The compressibility  is	Beta=",'%.4E'%Beta;

The compressibility  is	Beta= -2.5097E+15

Example 1.6 , Page no:16

In [4]:
#given
r_0=3.56; #in Angstrom
e=1.6E-19; #in C (charge of electron)
IE=3.89; #in eV (ionisation energy of Cs)
EA=-3.61; #in eV (electron affinity of Cl)
n=10.5; #Born constant
E_o= 8.85E-12; #absolute premittivity
alpha=1.763; #Madelung constant
pi=3.14; #value of pi used in the solution

#calculate
r_0=r_0*1E-10; #since r is in nanometer
U=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy
U=U/e; #changing unit from J to eV
ACE=U+EA+IE; #calculation of atomic cohesive energy

#result
print"The ionic cohesive energy is ",round(U),"eV";
print"The atomic cohesive energy is",round(ACE),"eV";

The ionic cohesive energy is  -6.0 eV
The atomic cohesive energy is -6.0 eV

Example 1.7 , Page no:17

In [5]:
#given
r_0=2.81; #in Angstrom
e=1.6E-19; #in C (charge of electron)
n=9; #Born constant
E_o= 8.85E-12; #absolute premittivity
alpha=1.748; #Madelung constant
pi=3.14; #value of pi used in the solution

#calculate
r_0=r_0*1E-10; #since r is in nanometer
V=-alpha*(e**2/(4*pi*E_o*r_0))*(1-1/n); #calculate potential energy
V=V/e; #changing unit from J to eV
V_1=V/2; #Since only half of the energy contribute per ion to the cohecive energy therfore

#result
print"The potential energy is V=",round(V,2),"eV";
print"The energy contributing per ions to the cohesive energy is ",round(V_1,2),"eV";

The potential energy is V= -7.96 eV
The energy contributing per ions to the cohesive energy is  -3.98 eV