# Chapter 5:Imperfection in Solids¶

## Example 5.1 Page 112¶

In [1]:
#initiation of variable
from math import exp
rho=8.4e6 #Density of Copper in g/m^3
A=63.5 #Atomic weight of Copper
Qv=0.9 #Activation energy in eV
k=8.62*10**-5 #Boltzmann Constant in eV/K
T=1000.0+273#Temperature in K

#calculation
N=Na*rho/A #No. of atomic site per cubic meter
Nv=N*exp(-Qv/(k*T))

#result
print" Equilibrium number of vacancies/m^3 is %.1e for 1273K" %Nv

 Equilibrium number of vacancies/m^3 is 2.2e+25 for 1273K


## Example 5.2 Page 117¶

In [2]:
#initiation of variable
#proving equation to be correct for sample values
C1=5.0;
C2=3.0;
A1=9.0;
A2=15.0;
C1_dash=50.0;

#calculation
C11=C1*A1/(C1*A1+C2*A2)*100;
if C11==C1_dash :
print "The equation derived i.e C1_dash=C1*A1/(C1*A1+C2*A2)*100 is correct"


The equation derived i.e C1_dash=C1*A1/(C1*A1+C2*A2)*100 is correct


## Example 5.3 Page 118¶

In [7]:
#initiation of variable
C_Al=97.0 #Aluminium wt%
C_Cu=3.0 #Copper wt%
A_Al=26.98 #Atomic wt of Aluminium
A_Cu=63.55 #Atomic wt of Copper

#calculation
CAl=C_Al*A_Cu*100/((C_Al*A_Cu)+(C_Cu*A_Al))
CCu=C_Cu*A_Al*100/((C_Cu*A_Al)+(C_Al*A_Cu))

#result
print " Atomic percent of Al is %.1f  percent" %CAl
print " Atomic percent of Cu is %.1f  percent" %CCu

 Atomic percent of Al is 98.7 in percent
Atomic percent of Cu is 1.3 in percent


## Example 5.4 Page 121¶

In [8]:
#initiation of variable
from math import exp
rho=1.955 #Density of KCl in g/cm^3
A_k= 39.10 #Atomic weight of potassium in g/mol
A_cl= 35.45 #Atomic weight of Chlorine in g/mol
Qs=2.6 #Activation energy in eV
k=8.62*10**-5 #Boltzmann Constant in eV/K
T=500+273.0 #Temperature in K

#result
A = A_k+A_cl # Molar mass of KCl in gram
N=Na*rho*1.0e6/A #No. of atomic site per cubic meter
Ns=N*exp(-Qs/(2*k*T))

#result
print" Number of Schottky defects are %.2e defects/m^3." %Ns

 Number of Schottky defects are 5.31e+19 defects/m^3.


## Example 5.5 Page 122¶

In [9]:
#initiation of variable
print "the are two possible scenarios"
print "scenario 1, removal of Na+ ion will intoduce vacancy in the NaCl crystal"
print "scenario 2, addition of Cl- ion wil fill up any additional void left"
print "since there is no way to predict which will happen at large thus can be assumed 50% each so, there is very less chances of creating a void"

the are two possible scenarios
scenario 1, removal of Na+ ion will intoduce vacancy in the NaCl crystal
scenario 2, addition of Cl- ion wil fill up any additional void left
since there is no way to predict which will happen at large thus can be assumed 50% each so, there is very less chances of creating a void


## Example 5.6 Page 138¶

In [11]:
#initiation of variable
from math import log
N=45.0 #Number of grains per square inch
M=85.0 # magnification

#part A
#calculation
n=(log(N)/log(2.0))+1 #calculation for grain size no.  N=2^(n-1)

#result
print" Grain size number is %.1f" %n

#part B
Nm=(100/M)**2*2**(n-1)

#result
print" At magnification of 85x"
print" Number of grains per inch square are %.2f" %Nm
print "answer in book is 62.6. It is because of rounding off at intermediate stages"

 Grain size number is 6.5
At magnification of 85x
Number of grains per inch square are 62.28
answer in book is 62.6. It is because of rounding off at intermediate stages