# Chapter 3 : Crystal Geometry Structure and Defects¶

## Example 3.10 Page No : 91¶

In [3]:

from numpy import degrees, arccos

# Variables
#Miller indices of plane
h_1 = 1.;
k_1 = 1.;
l_1 = 1.;
h_2 = 1.;
k_2 = 2.;
l_2 = 1.;

# Calculation
angle = degrees(arccos((h_1*h_2+k_1*k_2+l_1*l_2)/(((h_1**2+k_1**2+l_1**2)**(1./2))*((h_2**2+k_2**2+l_2**2)**(1./2)))))

# Results
print 'angle Between normals to the planes (111) and (121)(in degrees) = %.2f'%angle

angle Between normals to the planes (111) and (121)(in degrees) = 19.47


## Example 3.11 Page No : 91¶

In [5]:


import math

# Variables
r_Na = 0.98;			#Radius of Na+(in A)
r_Cl = 1.81;			#Radius of Cl-(in A)
a = 2*(r_Na+r_Cl);			#Lattice parameter (in A)

# Calculation
pi = 22./7;
V_i = 4*(4./3)*pi*((r_Na**3)+(r_Cl**3));			#Volume of ions present in unit cell
V_u = a**3;			#Volume of unit cell
Apf = V_i/V_u;			#Atomic packing fraction
Ef_p = (Apf)*100;			#Packing efficiency(in %)
AM_sodium = 22.99;			#Atomic mass of sodium(in amu)
AM_chlorine = 35.45;			#Atomic mass of chlorine(in amu)
M_1 = 4*(AM_sodium+AM_chlorine)*1.66*10**(-27);			#Mass of the unit cell
a_1 = a*10**(-10);			#Lattice parameter (in meter)
V_u1 = (a_1)**3;
Density = M_1/V_u1;

# Results
print 'Packing efficiency of sodium chloride in = %.1f %%'%Ef_p
print 'density of sodium chloride in = %.0f Kg/m3'%Density

Packing efficiency of sodium chloride in = 66.3 %
density of sodium chloride in = 2233 Kg/m3


## Example 3.12 Page No : 91¶

In [2]:


import math

# Variables
Density = 2.7;			#(in g/cm**3)
n = 4;
m = 26.98;			#atomic weight of Al

# Calculation and Results
a = ((n*m/(Density*N_a))**(1./3));			#Lattice parameter(in Cm)
A = a*10**(8);			                    #Lattice parameter(in A)
print 'radius in = %.3f A'%A
r = A/(2*1.414);			#radius for fcp structure
print 'Diameter in %.2f A'%(2*r)

radius in = 4.049 A
Diameter in 2.86 A


## Example 3.13 Page No : 92¶

In [9]:

import math

# Variables
r = 1.245;			#radius of nickel (in A)
a = 4*r/(2)**(1./2);			#Lattice consmath.tant(in A)
#Miller indices of plane 200
h_1 = 2;
k_1 = 0;
l_1 = 0;
#Miller indices of plane 111
h_2 = 1;
k_2 = 1;
l_2 = 1;

# Calculation
d_200 = a/((h_1**2)+(k_1**2)+(l_1**2))**(1./2);
d_111 = a/((h_2**2)+(k_2**2)+(l_2**2))**(1./2);

# Results
print 'interplaner distance of (200) plane of nickel crystal in = %.2f A'%d_200
print 'interplaner distance of (111) plane of nickel crystal in = %.2f A'%d_111

interplaner distance of (200) plane of nickel crystal in = 1.76 A
interplaner distance of (111) plane of nickel crystal in = 2.03 A


## Example 3.14 Page No : 92¶

In [3]:

import math

# Variables
a = 3.03*10**(-7);			#lattice consmath.tant(in mm)

# Calculation
N_100 = 1/(a**2);			#Number of atoms in the (100) plane of a simple cubic structure
N_110 = 0.707/(a**2);			#Number of atoms in the (110) plane of a simple cubic structure
N_111 = 0.58/(a**2);			#Number of atoms in the (111) plane of a simple cubic structure

# Results
print 'Number of atoms in the (100) plane of a simple cubic structure(in per mm**2) = %.2e'%N_100
print 'Number of atoms in the (110) plane of a simple cubic structure(in per mm**2) = %.1e'%N_110
print 'Number of atoms in the (111) plane of a simple cubic structure(in per mm**2) = %.1e'%N_111

Number of atoms in the (100) plane of a simple cubic structure(in per mm**2) = 1.09e+13
Number of atoms in the (110) plane of a simple cubic structure(in per mm**2) = 7.7e+12
Number of atoms in the (111) plane of a simple cubic structure(in per mm**2) = 6.3e+12


## Example 3.15 Page No : 92¶

In [13]:

import math

# Variables
r = 1.245*10**(-7);			#Radius of the Ni atom(in mm)

# Calculation
NA_100 = 1+(1./4)*4;			#Numbers of atom in (100) plane
a = 4*r/(2)**(1./2);			#Lattice consmath.tant(in mm)
Area = a**2;
P_density = NA_100/Area;

# Results
print 'the planer density of Ni (in atoms per mm**2) = %.1e'%P_density

the planer density of Ni (in atoms per mm**2) = 1.6e+13


## Example 3.16 Page No : 93¶

In [14]:

import math

# Variables
N_a1 = 4*(1./4)+1;			#Number of atoms contained in (100) plane
a_1 = 2*2**(1./2)*r;			#edge of unit  cell in case of (100) plane
PD_100 = N_a1/(a_1**2);			#Planar density of plane (100)

# Calculation
N_a2 = 4*(1./4)+2*(1./2);			#Number of atoms contained in (110) plane
a_21 = 4*r;			#top edge of the plane (110)
a_22 = 2*2**(1./2)*r;			#vertical edge of the plane (110)
PD_110 = N_a2/(a_21*a_22);			#Planar density of plane (110)
N_a3 = 3*(1./6)+3./2;			#Number of atom contained in (111) plane
Ar_111 = 4*(3**(1./2))*r**2;			#area of (111) plane
PD_111 = N_a3/Ar_111;			#Planar density of plane (111)

# Results
print 'Planar density of plane 100(in atoms/mm**2) = %.1e'%PD_100
print 'Planar density of plane 110(in atoms/mm**2) = %.1e'%PD_110
print 'Planar density of plane 111(in atoms/mm**2) = %.1e'%PD_111

Planar density of plane 100(in atoms/mm**2) = 8.2e+12
Planar density of plane 110(in atoms/mm**2) = 5.8e+12
Planar density of plane 111(in atoms/mm**2) = 9.4e+12


## Example 3.17 Page No : 94¶

In [5]:

import math

# Variables
N_a1 = (1./2)+1+(1./2);			#Number of diameters of atom along (110) direction
a = 3.61*10**(-7);			#lattice consmath.tant of copper in mm

# Calculation
L_d1 = 2**(1./2)*a;			#length of the face diagonal in case of (110) direction
p_110 = N_a1/L_d1;			#linear atomic density along (110) of copper crystal lattice(in atoms/mm)
N_a2 = (1./2)+(1./2);			#Number of diameters of atom along (111) direction
L_d2 = 3**(1./2)*a;			#length of the face diagonal in case of (111) direction
p_111 = N_a2/L_d2;			#linear atomic density along (110) of copper crystal lattice(in atoms/mm)

# Results
print 'linear atomic density along (110) of copper crystal lattice in = %.2e atoms/mm'%p_110
print 'linear atomic density along (111) of copper crystal lattice in = %.2e atoms/mm'%p_111

linear atomic density along (110) of copper crystal lattice in = 3.92e+06 atoms/mm
linear atomic density along (111) of copper crystal lattice in = 1.60e+06 atoms/mm


## Example 3.18 Page No : 95¶

In [18]:


import math

# Variables
A = 55.8;			#atomic weight of Fe
n = 2;			#number of atoms per unit cell

# Calculation
p = 7.87*10**3;			#density of Fe(in kg/m**3)
a = ((A*n/(N*p))**(1./3))*10**10;			#Value of lattice consmath.tant

# Results
print 'Value of lattice constant in = %.3f A'%a

Value of lattice constant in = 2.867 A


## Example 3.19 Page No : 95¶

In [19]:

import math

# Variables
a = 2.9*10**(-10);			#lattice parameter(in m)
A = 55.8;			#atomic weight of Fe

# Calculation
p = 7.87*10**3;			#density of Fe(in kg/m**3
n = (a**3)*N*p/A;			#Numbers of atoms per unit cell

# Results
print 'Numbers of atoms per unit cell = ',floor(n)

Numbers of atoms per unit cell =  2.0


## Example 3.20 Page No : 109¶

In [20]:

import math

# Variables
a = 2.87*10**(-10);			#lattice parameter for bcc iron
b = a*(3**(1./2))/2;			#Magnitude of burgers vector

# Calculation
u = 80*10**9;			#shear modulus
E = (1./2)*u*b**2;			#line energy of disslocation

# Results
print 'line energy of disslocation in = %.2e J/m'%E

line energy of disslocation in = 2.47e-09 J/m


## Example 3.22 Page No : 109¶

In [6]:

import math

# Variables
T = 1000.;			#absolute temperature
R = 8.314;			#consmath.tant

# Calculation
H_f = 100*1000;			#enthalpy of formation of vacancies(in J/mol)
n = N*math.exp(-(H_f)/(R*T));			#number of vacancies created during heating(in per mol)
V = 5.5*10**(-6);			#volume of 1 mole of the crystal in m**3
n_1 = n/V;			#number of vacancies created during heating(in per m**3)

# Results
print 'number of vacancies created during heating in = %.2e m**-3'%n_1

number of vacancies created during heating in = 6.54e+23 m**-3


## Example 3.23 Page No : 109¶

In [7]:

import math

# Variables
#bond energy per atom of copper = bond energy per bond*numbers of bond per atom*(1/2)
A = 56.4*1000;			#
n_1 = 12.;			#numbers of bond per atom
n_2 = 3.;			#bonds broken at the surface

# Calculation and Results
E = A*n_1/(2*N);			#Energy of total bonds
E_b = E*(n_2/n_1);			#Energy of broken bonds on surface
n_a = 1.77*10**19;			#no. of atoms on {111} planes in copper(in m**-2)
E_c = n_a*E_b;			#Surface energy (enthalpy) of copper
print 'Surface energy (enthalpy) of copper in = %.2f J/m**2'%E_c

Surface energy (enthalpy) of copper in = 2.49 J/m**2


## Example 3.24 Page No : 110¶

In [23]:
import math

# Variables
H_f = 68.*1000;			#enthalpy of formation of vacancies(in J/mol)
T_1 = 0;			#temp (in K)
T_2 = 300.;			#temp (in K)
R = 8.314;			#consmath.tant

# Calculation

n = math.exp(-H_f/(R*T_2));			#equilibrium concentration of vacancies in aluminium at 300 K

# Results
print 'equilibrium concentration of vacancies in aluminium at 300 K = %.2e'%n

equilibrium concentration of vacancies in aluminium at 300 K = 1.44e-12


## Example 3.25 Page No : 113¶

In [24]:

import math

# Variables
Wavelength = 1.54*10**(-10);			#in meter
Angle = 20.3;			#in degree
n = 1;			#First order

# Calculation
d = Wavelength*n/(2*math.sin(math.radians(Angle)));			#the interplanar spacing(in Meter)

# Results
print 'the interplanar spacing between atomic plane in = %.2f A'%(d/(10**-10))

the interplanar spacing between atomic plane in = 2.22 A


## Example 3.26 Page No : 113¶

In [10]:

import math

# Variables
wavelength = 0.58;			#in Angstrom
angle = 9.5;			#in degree
n = 1;			#First order

# Calculation
#d_200 = wavelength*n/(2*math.sin(math.radians(angle)));			#interplanar spacing(in Angstrom)
d_200 = n/math.sqrt(2**2+0**2+0**2)
#Miller indices of plane
h = 2;
k = 0;
l = 0;
a = 0.58/(math.sin(math.radians(angle))*2*d_200);			#Size of unit cell(in Angstrom)
# Results
print 'Size of unit cell in %.2f Angstrom'%a


Size of unit cell in 3.51 Angstrom


## Example 3.27 Page No : 114¶

In [11]:

import math

# Variables
#Miller indices of plane
h = 1.;
k = 1.;
l = 1.;
wavelength = 0.54;			#in angstrom
a = 3.57;			#size of a cube
n = 1;

# Calculation
d_111 = a/(h**2+k**2+l**2)**(1./2);			#interplanar spacing(in Angstrom)
sinangle = (n*wavelength)/(2*d_111)
angle = math.degrees(math.asin(sinangle))

# Results
print 'Bragg angle(in degree) = %.3f'%angle

# rounding off error

Bragg angle(in degree) = 7.527


## Example 3.28 Page No : 114¶

In [12]:

import math

# Variables
d = 1.181;			  #A
wavelength = 1.540;	  #in angstrom
angle = 90;			  #in degrees

# Calculation
n = 2*d*math.sin(math.radians(angle))/(wavelength);			#the bragg reflection index

# Results
print 'Bragg reflection index for BCC crystal = %.2f'%n

Bragg reflection index for BCC crystal = 1.53


## Example 3.29 Page No : 114¶

In [16]:

import math

# Variables
n_1 = 1;			#1st order reflection index
angle_1 = 10;			#1st order reflection angle
n_3 = 3;			#3rd order reflection index

# Calculation
sinangle_3 = n_3 * math.sin(math.radians(angle_1)/n_1);			#
angle_3 = math.degrees(math.asin(sinangle_3))

# Results
print '3rd order reflection angle = %.2f'%angle_3

3rd order reflection angle = 31.40


## Example 3.30 Page No : 115¶

In [15]:

import math

# Variables
angle = 20.3;			#in degree
wavelength = 1.54;			#in angstrom
n = 1;
a = 3.16;			#lattice parameter in angstrom

# Calculation
M_indices = a**2/(d**2);

# Results
print 'interplanar spacing of reflection plane %.2f A'%d
print 'miller indices of the reflection plane',floor(M_indices)
print "((110),(101),(011))"

interplanar spacing of reflection plane 2.22 A
miller indices of the reflection plane 2.0
((110),(101),(011))


## Example 3.31 Page No : 115¶

In [18]:


import math

# Variables
#Miller indices of plane
n = 1;
h = 1;
k = 1;
l = 1;
angle = 30;			#in degree
wavelength = 2;			#in angstrom

# Calculation

interatomic spacing(in angstrom) = 3.46