# Chapter 6 : Imperfections, Defects and Dislocations in Solids¶

### Example 6.1 pageno : 125¶

In [3]:
import math

# Variables
p = 3170.;	        		#density in kg/m**3

# Calculations
mw = 7.9*10**-2;			#molecular weight of CaF2
nl = na*p/mw;			    #calcium ions/cubic m
ni = 2.*nl;			        # /cubic m
t = 1300.;	        		#in K
ef = 2.7*1.6*10**-19.;			#energy of formation of one frenkel defect
k = 1.38*10**-23;			#boltzmann constant
nf = math.sqrt(nl*ni)*math.exp(-ef/(2*k*t));			#in /cubic m

# Results
print "No. of Frenkel defect per unit volume of Calcium Fluoride (in /m**3)  =  %.2e /m**3"%nf

# note: Answers may vary because of rouding error.

No. of Frenkel defect per unit volume of Calcium Fluoride (in /m**3)  =  2.02e+23 /m**3


### Example 6.2 pageno : 127¶

In [5]:
import math

# Variables
r = 8.314;			# J/mol K
t1 = 300.;			#in K
t2 = 1000.;			#in K

# Calculations
ent = 168*10**3;        			#enthalpy of formation of vacancy in J/mol
x1 = math.exp(-ent/(r*t1));			#x1 = n/Na
x2 = math.exp(-ent/(r*t2));			#x2 = n/Na
rt = x1/x2;             			#ratio

# Results
print "Ratio of no. of vacancies  =  %.2e"%rt

Ratio of no. of vacancies  =  3.34e-21


### Example 6.3 page no : 132¶

In [5]:
# Calculations
# for 1/6[121] dislocation
a1 = 1./6 * 1              # x - component
b1 = 1./6 * 2              # y - component
c1 = 1./6 * 1              # z - component

# for 1/6[211] dislocation
a2 = 1./6 * 2              # x - component
b2 = 1./6 * 1              # y - component
c2 = 1./6 * -1              # z - component

a = a1 + a2              # x - component
b = b1 + b2              # y - component
c = c1 + c2              # z - component

# Results
print "X-component  a = %.1f"%a
print "Y-component  b = %.1f"%b
print "Z-component  c = %.1f"%c

print "Thus,  [abc] = [%.1f %.1f %.1f] = 1/2[110]"%(a,b,c)
print "Hence, 1/6[121] + 1/6[211] -> 1/2[110] proved"

X-component  a = 0.5
Y-component  b = 0.5
Z-component  c = 0.0
Thus,  [abc] = [0.5 0.5 0.0] = 1/2[110]
Hence, 1/6[121] + 1/6[211] -> 1/2[110] proved


### Example 6.4 pageno : 136¶

In [6]:
import math

# Variables
v = 0.31;       			#poisson's ratio
bv = .25*10**-9;			#burger's vector in m
ri = 1.1*10**-9;			#in m

# Calculations
r0 = 10.**5*bv;	    		#in m
sm = 45.*10**9;		    	#shear modulous in n/sqm
gb_2 = sm*bv**2;
u_ed = (gb_2/(4*3.14*(1-v)))*math.log(r0/ri);
u_sd = (gb_2/(4*3.14))*math.log(r0/ri);
r = u_ed/u_sd;			#ratio

# Results
print "Elastic Strain Energy of Edge dislocation (in J/m)  =  %.2e J/m"%u_ed
print "Elastic Strain Energy of Screw dislocation (in J/m)  =  %.2e J/m"%u_sd
print "Ratio of energies of edge dislocation over screw dislocation  =  %.2f"%r

Elastic Strain Energy of Edge dislocation (in J/m)  =  3.26e-09 J/m
Elastic Strain Energy of Screw dislocation (in J/m)  =  2.25e-09 J/m
Ratio of energies of edge dislocation over screw dislocation  =  1.45


### Example 6.5 pageno : 139¶

In [8]:
# Variables
r = 1.7*10**-10;			#atomic radius in m
n1 = 10.**-3;   			#1mm = 10**-3m

# Calculations
a = 2*r;		        	#in m
n = n1/a;
ed = 2.*10**-6; 			#edge dislocation in m
ns = ed/a;
nv = n*ns;

# Results
print "Total no. of created vacancies  =  %.2e"%nv

Total no. of created vacancies  =  1.73e+10


### Example 6.6 pageno : 142¶

In [11]:
import math

# Variables
c1 = 3.61           # lattice parameter of copper
c2 = 44             # shear modulus of copper

# Calculations
b = c1*math.sqrt(1**2 + 1**2 + 0**2)/2.
U = c2*b**2 * 10**9 * 10**-20/2.

# Results
print "The magnitude of the Burgers vector : %.3f A"%b
print "LIne energy of dislocation U : %.3e J/m"%U

The magnitude of the Burgers vector : 2.553 A
LIne energy of dislocation U : 1.434e-09 J/m


### Example 6.7 pageno : 144¶

In [1]:
import math

# Variables
a = 3.84          # lattice constant
h = 1
k = 1
l = 0

# calculations
b = a*math.sqrt(h**2  + k**2 + l**2)/2.

# Results
print "b = %.3f A"%b
print "Spacing between dislocaitons in a low angle tilt boundary in iridium"
print "When,"
print "Angle of tilts 1 = %.2f A"%h1
print "Angle of tilts 3 = %.2f A"%h3

b = 2.715 A
Spacing between dislocaitons in a low angle tilt boundary in iridium
When,
Angle of tilts 1 = 155.56 A
Angle of tilts 3 = 51.81 A