initial commit / add all books

27 files changed, 487 insertions, 0 deletions

diff --git a/1736/CH1/EX1.1/Ch01Ex1.sce b/1736/CH1/EX1.1/Ch01Ex1.sce
new file mode 100755
index 000000000..dafad3434
--- /dev/null
+++ b/1736/CH1/EX1.1/Ch01Ex1.sce
@@ -0,0 +1,11 @@
+// Scilab Code Ex1.1  Page-13 (2006)

+clc; clear;

+r = 1.278e-010;     // Atomic radius of fcc structure, m

+a = 4*r/sqrt(2);    // Lattice parameter of fcc strucure, m

+V = a^3;    // Volume of fcc unit cell, metre, cube

+printf("\nThe lattice parameter of fcc strucure = %4.2e m", a);

+printf("\nThe volume of fcc unit cell = %5.2e metre, cube", V);

+

+// Result 

+// The lattice parameter of fcc strucure = 3.61e-010 m

+// The volume of fcc unit cell = 4.72e-029 metre cube 

diff --git a/1736/CH1/EX1.10/Ch01Ex10.sce b/1736/CH1/EX1.10/Ch01Ex10.sce
new file mode 100755
index 000000000..183f896cc
--- /dev/null
+++ b/1736/CH1/EX1.10/Ch01Ex10.sce
@@ -0,0 +1,45 @@
+// Scilab Code Ex 1.10 : Page-24 (2006)

+clc; clear;

+function str = structure(r_ratio)

+    if r_ratio > 0.732 then

+        str = 'Caesium Chloride';

+    else if  r_ratio < 0.732 & r_ratio > 0.414 then

+            str = 'Rock Salt';

+        else if  r_ratio < 0.414 then

+                str = 'Rutile'

+             end

+        end

+    end

+endfunction

+

+crystal = cell(6,2);    // Declare cells of 6 rows and 2 columns

+crystal(1,1).entries = 'I';

+crystal(1,2).entries = 2.19;        // Ionic radius of I, angstrom

+crystal(2,1).entries = 'Cl';

+crystal(2,2).entries = 1.81;        // Ionic radius of Cl, angstrom

+crystal(3,1).entries = 'Na';

+crystal(3,2).entries = 0.95;        // Ionic radius of Na, angstrom

+crystal(4,1).entries = 'Cs';

+crystal(4,2).entries = 1.69;        // Ionic radius of Cs, angstrom

+crystal(5,1).entries = 'Mg';

+crystal(5,2).entries = 0.99;        // Ionic radius of Mg2+, angstrom

+crystal(6,1).entries = 'O';

+crystal(6,2).entries = 1.40;        // Ionic radius of O2-, angstrom

+

+printf("\nThe crystal structure of %s%s with radius ratio = %6.4f is %s", crystal(3,1).entries, crystal(1,1).entries, crystal(3,2).entries/crystal(1,2).entries, structure(crystal(3,2).entries/crystal(1,2).entries));

+

+printf("\nThe crystal structure of %s%s with radius ratio = %6.4f is %s", crystal(3,1).entries, crystal(2,1).entries, crystal(3,2).entries/crystal(2,2).entries, structure(crystal(3,2).entries/crystal(2,2).entries));

+

+printf("\nThe crystal structure of %s%s with radius ratio = %6.4f is %s", crystal(4,1).entries, crystal(2,1).entries, crystal(4,2).entries/crystal(2,2).entries, structure(crystal(4,2).entries/crystal(2,2).entries));

+

+printf("\nThe crystal structure of %s%s with radius ratio = %6.4f is %s", crystal(4,1).entries, crystal(1,1).entries, crystal(4,2).entries/crystal(1,2).entries, structure(crystal(4,2).entries/crystal(1,2).entries));

+

+printf("\nThe crystal structure of %s%s with radius ratio = %6.4f is %s", crystal(5,1).entries, crystal(6,1).entries, crystal(5,2).entries/crystal(6,2).entries, structure(crystal(5,2).entries/crystal(2,2).entries));

+

+// Result 

+//The crystal structure of NaI with radius ratio = 0.4338 is Rock Salt

+//The crystal structure of NaCl with radius ratio = 0.5249 is Rock Salt

+//The crystal structure of CsCl with radius ratio = 0.9337 is Caesium Chloride

+//The crystal structure of CsI with radius ratio = 0.7717 is Caesium Chloride

+//The crystal structure of MgO with radius ratio = 0.7071 is Rock Salt  

+

diff --git a/1736/CH1/EX1.11/Ch01Ex11.sce b/1736/CH1/EX1.11/Ch01Ex11.sce
new file mode 100755
index 000000000..68ded7eef
--- /dev/null
+++ b/1736/CH1/EX1.11/Ch01Ex11.sce
@@ -0,0 +1,16 @@
+// Scilab Code Ex 1.11 :Page-25 (2006)

+clc; clear;

+R = 1;  // For simplicity we assume radius of atom to be unity, m

+// For bcc Structure,

+a = 4*R/sqrt(3);    // Lattice parameter of bcc crystal, m

+// We have R+r = a/2, solving for r

+r = a/2-R   // Relation between radius of the void and radius of the atom, m 

+printf("\nThe maxiumum radius of the sphere that can fit into void between two bcc unit cells = %5.3fR", r); 

+

+// Result 

+// The maxiumum radius of the sphere that can fit into void between two bcc unit cells = 0.155R 

+

+

+

+

+

diff --git a/1736/CH1/EX1.12/Ch01Ex12.sce b/1736/CH1/EX1.12/Ch01Ex12.sce
new file mode 100755
index 000000000..43482af50
--- /dev/null
+++ b/1736/CH1/EX1.12/Ch01Ex12.sce
@@ -0,0 +1,16 @@
+// Scilab Code Ex 1.12 :Page-25 (2006)

+clc; clear;

+R = 1;  // For simplicity we assume radius of atom to be unity, m

+// For fcc Structure,

+a = 4*R/sqrt(2);    // Lattice parameter of fcc crystal, m

+// We have R+r = a/2, solving for r

+r = a/2-R   // Relation between radius of the void and radius of the atom, m 

+printf("\nThe maxiumum radius of the sphere that can fit into void between two fcc unit cells = %5.3fR", r); 

+

+// Result 

+// The maxiumum radius of the sphere that can fit into void between two fcc unit cells = 0.414R 

+

+

+

+

+

diff --git a/1736/CH1/EX1.13/Ch01Ex13.sce b/1736/CH1/EX1.13/Ch01Ex13.sce
new file mode 100755
index 000000000..afd9555b9
--- /dev/null
+++ b/1736/CH1/EX1.13/Ch01Ex13.sce
@@ -0,0 +1,23 @@
+// Scilab Code Ex 1.13 :Page-26 (2006)

+clc; clear;

+R = 1;  // For simplicity we assume radius of atom to be unity, m

+// For bcc Structure,

+a = 4*R/sqrt(3);    // Lattice parameter of bcc crystal, m

+// We have (R+r)^2 = (a/2)^2+(a/4)^2, solving for r

+r = sqrt(5)*a/4-R   // Relation between radius of the void and radius of the atom, m 

+printf("\nThe radius of largest void in the bcc lattice = %4.2fR", r); 

+

+// For fcc Structure,

+a = 4*R/sqrt(2);    // Lattice parameter of fcc crystal, m

+// We have (R+r)^2 = (a/2)^2+(a/4)^2, solving for r

+r_fcc = a/2-R   // Relation between radius of the void and radius of the atom, m 

+printf("\nThe radius of largest void in the fcc lattice is %4.2f times larger than that in the bcc lattice", r_fcc/r); 

+

+// Result 

+// The radius of largest void in the bcc lattice = 0.29R

+// The radius of largest void in the fcc lattice is 1.42 times larger than that in the bcc lattice 

+

+

+

+

+

diff --git a/1736/CH1/EX1.14/Ch01Ex14.sce b/1736/CH1/EX1.14/Ch01Ex14.sce
new file mode 100755
index 000000000..ff09bf3da
--- /dev/null
+++ b/1736/CH1/EX1.14/Ch01Ex14.sce
@@ -0,0 +1,17 @@
+// Scilab Code Ex 1.14 :Page-26 (2006)

+clc; clear;

+R = 1;  // For simplicity we assume radius of atom to be unity, m

+

+// For bcc Structure,

+a = 4*R/sqrt(3);    // Lattice parameter of bcc crystal, m

+// We have (R+r)^2 = (a/2)^2+(a/4)^2, solving for r

+r = a/2-R   // Relation between radius of the void and radius of the atom, m 

+printf("\nThe radius of void for carbon atoms in iron = %5.3fR", r); 

+

+// Result 

+//The radius of void for carbon atoms in iron = 0.155R 

+

+

+

+

+

diff --git a/1736/CH1/EX1.15/Ch01Ex15.sce b/1736/CH1/EX1.15/Ch01Ex15.sce
new file mode 100755
index 000000000..645998e94
--- /dev/null
+++ b/1736/CH1/EX1.15/Ch01Ex15.sce
@@ -0,0 +1,15 @@
+// Scilab Code Ex 1.15 :Page-27 (2006)

+clc; clear;

+R = 1;  // For simplicity we assume radius of atom to be unity, m

+// From the right triangle LMO, LM/LO = R/(R + r) = cosd(30), solving for r

+r = poly(0, 'r');

+r = roots(R/cosd(30)-R-r);

+printf("\nThe radius of triangular void = %5.3fR", r); 

+

+// Result 

+// The radius of triangular void = 0.155R 

+

+

+

+

+

diff --git a/1736/CH1/EX1.16/Ch01Ex16.sce b/1736/CH1/EX1.16/Ch01Ex16.sce
new file mode 100755
index 000000000..dad7d23d6
--- /dev/null
+++ b/1736/CH1/EX1.16/Ch01Ex16.sce
@@ -0,0 +1,15 @@
+// Scilab Code Ex 1.16 :Page-27 (2006)

+clc; clear;

+R = 1;  // For simplicity we assume radius of atom to be unity, m

+// From the right triangle LMN similar to trinagle LPO, LM/LO = R/(R + r) = LP/LO = sqrt(2/3), solving for r

+r = poly(0, 'r');

+r = roots(R/sqrt(2/3)-R-r);

+printf("\nThe radius ratio of tetragonal void = %5.3f", r/R); 

+

+// Result 

+// The radius ratio of tetragonal void = 0.225 

+

+

+

+

+

diff --git a/1736/CH1/EX1.17/Ch01Ex17.sce b/1736/CH1/EX1.17/Ch01Ex17.sce
new file mode 100755
index 000000000..44451586c
--- /dev/null
+++ b/1736/CH1/EX1.17/Ch01Ex17.sce
@@ -0,0 +1,15 @@
+// Scilab Code Ex 1.17 :Page-28 (2006)

+clc; clear;

+R = 1;  // For simplicity we assume radius of atom to be unity, m

+// From the isosceles right triangle LMN, LM/LO = (R + r)/R = sqrt(2)/1, solving for r

+r = poly(0, 'r');

+r = roots(R*sqrt(2)-R-r);

+printf("\nThe radius ratio of octahedral void = %5.3f", r/R); 

+

+// Result 

+// The radius ratio of octahedral void = 0.414 

+

+

+

+

+

diff --git a/1736/CH1/EX1.18/Ch01Ex18.sce b/1736/CH1/EX1.18/Ch01Ex18.sce
new file mode 100755
index 000000000..c377cb15b
--- /dev/null
+++ b/1736/CH1/EX1.18/Ch01Ex18.sce
@@ -0,0 +1,14 @@
+// Scilab Code Ex 1.18 Page-32 (2006)

+clc; clear;

+p = 3; q = -3; r = 3/2; // Coefficients of intercepts along three axes

+h = 1/p;        // Reciprocate the first coefficient

+k = 1/q;        // Reciprocate the second coefficient

+l = 1/r;        // Reciprocate the third coefficient

+mul_fact = double(lcm(int32([p,q,r]))); // Find l.c.m. of m,n and p

+h = h*mul_fact;    // Clear the first fraction

+k = k*mul_fact;    // Clear the second fraction

+l = l*mul_fact;    // Clear the third fraction

+printf("\nThe required miller indices are : (%d %d %d) ", h,k,l);

+

+// Result 

+// The required miller indices are : (1 -1 2)  

diff --git a/1736/CH1/EX1.19/Ch01Ex19.sce b/1736/CH1/EX1.19/Ch01Ex19.sce
new file mode 100755
index 000000000..e42e2aab8
--- /dev/null
+++ b/1736/CH1/EX1.19/Ch01Ex19.sce
@@ -0,0 +1,14 @@
+// Scilab Code Ex 1.19 Page-32 (2006)

+clc; clear;

+p = 2; q = 3; r = 4; // Coefficients of intercepts along three axes

+h = 1/p;        // Reciprocate the first coefficient

+k = 1/q;        // Reciprocate the second coefficient

+l = 1/r;        // Reciprocate the third coefficient

+mul_fact = double(lcm(int32([p,q,r]))); // Find l.c.m. of m,n and p

+h = h*mul_fact;    // Clear the first fraction

+k = k*mul_fact;    // Clear the second fraction

+l = l*mul_fact;    // Clear the third fraction

+printf("\nThe required miller indices are : (%d %d %d) ", h,k,l);

+

+// Result 

+// The required miller indices are : (6 4 3) 

diff --git a/1736/CH1/EX1.2/Ch01Ex2.sce b/1736/CH1/EX1.2/Ch01Ex2.sce
new file mode 100755
index 000000000..97ffa47b7
--- /dev/null
+++ b/1736/CH1/EX1.2/Ch01Ex2.sce
@@ -0,0 +1,23 @@
+// Scilab Code Ex1.2 Page-14 (2006)

+clc; clear;

+r = 0.143e-09;  // Radius of Nb unit cell, m

+d = 8.57e+03; // Density of Nb unit cell, kg/metre-cube

+M = 92.91e-03;        // Atomic weight of Nb, kg per mole

+N = 6.023D+23;    // Avogadro's No.

+

+// For fcc

+a = 4*r/sqrt(2);    // Lattice parameter for fcc structure of Nb, m

+n = a^3*d*N/M; // Number of lattice points per unit cell

+if (modulo(n, int(n)) < 0.001) then

+printf("\nThe number of atoms associated with the cell is %d, Nb should have fcc structure", int(n));

+end

+

+// For bcc

+a = 4*r/sqrt(3);    // Lattice parameter for bcc structure of Nb, m

+n = a^3*d*N/M; // Number of lattice points per unit cell

+if (modulo(n, int(n)) < 0.001) then

+printf("\nThe number of atoms associated with the cell is %d, Nb should have bcc structure", int(n));

+end

+

+// Result 

+// The number of atoms associated with the cell is 2, Nb should have bcc structure 

diff --git a/1736/CH1/EX1.20/Ch01Ex20.sce b/1736/CH1/EX1.20/Ch01Ex20.sce
new file mode 100755
index 000000000..f2006a73b
--- /dev/null
+++ b/1736/CH1/EX1.20/Ch01Ex20.sce
@@ -0,0 +1,14 @@
+// Scilab Code Ex 1.20 Page-32 (2006)

+clc; clear;

+p = 4; q = 4; r = %inf; // Coefficients of intercepts along three axes

+h = 1/p;        // Reciprocate the first coefficient

+k = 1/q;        // Reciprocate the second coefficient

+l = 1/r;        // Reciprocate the third coefficient

+mul_fact = double(lcm(int32([p,q]))); // Find l.c.m. of m,n and p

+h = h*mul_fact;    // Clear the first fraction

+k = k*mul_fact;    // Clear the second fraction

+l = l*mul_fact;    // Clear the third fraction

+printf("\nThe required miller indices are : (%d %d %d) ", h,k,l);

+

+// Result 

+// The required miller indices are : (1 1 0)  

diff --git a/1736/CH1/EX1.21/Ch01Ex21.sce b/1736/CH1/EX1.21/Ch01Ex21.sce
new file mode 100755
index 000000000..e17a02069
--- /dev/null
+++ b/1736/CH1/EX1.21/Ch01Ex21.sce
@@ -0,0 +1,12 @@
+// Scilab Code Ex 1.21 Page-32 (2006)

+clc; clear;

+a = 0.424; b = 2; c = 0.367;    // Intercepts on planes along three axes, m

+// Here pa = 0.424; qb = 2; rc = 0.183, solving for p, q and r, we have

+p = 0.424/a; q = 2/b; r = 0.183/c; // Coefficients of intercepts along three axes

+h = 1/p;        // Reciprocate the first coefficient

+k = 1/q;        // Reciprocate the second coefficient

+l = 1/r;        // Reciprocate the third coefficient

+printf("\nThe required miller indices are : (%d %d %d) ", h,k,l);

+

+// Result 

+// The required miller indices are : (1 1 2)  

diff --git a/1736/CH1/EX1.22/Ch01Ex22.sce b/1736/CH1/EX1.22/Ch01Ex22.sce
new file mode 100755
index 000000000..e49a91a1f
--- /dev/null
+++ b/1736/CH1/EX1.22/Ch01Ex22.sce
@@ -0,0 +1,20 @@
+// Scilab Code Ex 1.22 Page-33 (2006)
+clc; clear;
+r = 1.746e-010;     // Atomic radius of lead atom, angstrom
+a = 4*r/sqrt(2);    // Interatomic spacing, m
+h = 1; k = 0; l = 0; // Miller Indices for planes in a cubic crystal
+d_100 = a/(h^2+k^2+l^2)^(1/2);  // The interplanar spacing for cubic crystals, m
+printf("\nThe interplanar spacing between consecutive (100) planes = %4.2f angstrom", d_100/1e-010);
+
+h = 1; k = 1; l = 0; // Miller Indices for planes in a cubic crystal
+d_110 = a/(h^2+k^2+l^2)^(1/2);  // The interplanar spacing for cubic crystals, m
+printf("\nThe interplanar spacing between consecutive (110) planes = %5.3f angstrom", d_110/1e-010);
+
+h = 1; k = 1; l = 1; // Miller Indices for planes in a cubic crystal
+d_111 = a/(h^2+k^2+l^2)^(1/2);  // The interplanar spacing for cubic crystals, m
+printf("\nThe interplanar spacing between consecutive (111) planes = %4.2f angstrom", d_111/1e-010);
+
+// Result
+// The interplanar spacing between consecutive (100) planes = 4.94 angstrom
+// The interplanar spacing between consecutive (110) planes = 3.492 angstrom
+// The interplanar spacing between consecutive (111) planes = 2.85 angstrom 
diff --git a/1736/CH1/EX1.23/Ch01Ex23.sce b/1736/CH1/EX1.23/Ch01Ex23.sce
new file mode 100755
index 000000000..fc6e1121a
--- /dev/null
+++ b/1736/CH1/EX1.23/Ch01Ex23.sce
@@ -0,0 +1,13 @@
+// Scilab Code Ex 1.23 Page-34 (2006)
+clc; clear;
+e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV
+h = 6.626e-034;     // Planck's constant, Js
+c = 3.0e+08;    // Speed of light, m/s
+E_K = 13.6*29^2;    // Energy of electron in the K-shell
+E_L = 13.6*29^2/4;  // Energy of electron in the L-shell
+// As E_K - E_L = h*c/lambda, solving for lambda
+lambda = h*c/((E_K - E_L)*e);       // Wavelength of K_alpha radiation of tungsten, m
+printf("\nThe wavelength of K_alpha radiation of Cu = %5.3f angstrom", lambda/1e-010);
+
+// Result
+// The wavelength of K_alpha radiation of Cu = 1.448 angstrom 
diff --git a/1736/CH1/EX1.24/Ch01Ex24.sce b/1736/CH1/EX1.24/Ch01Ex24.sce
new file mode 100755
index 000000000..bf578beef
--- /dev/null
+++ b/1736/CH1/EX1.24/Ch01Ex24.sce
@@ -0,0 +1,13 @@
+// Scilab Code Ex 1.24 Page-35 (2006)
+clc; clear;
+e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV
+h = 6.626e-034;     // Planck's constant, Js
+c = 3.0e+08;    // Speed of light, m/s
+E_K = 13.6*74^2;    // Energy of electron in the K-shell
+E_L = 13.6*74^2/4;  // Energy of electron in the L-shell
+// As E_K - E_L = h*c/lambda, solving for lambda
+lambda = h*c/((E_K - E_L)*e);       // Wavelength of K_alpha radiation of tungsten, m
+printf("\nThe wavelength of K_alpha radiation of tungsten = %4.2e angstrom", lambda/1e-010);
+
+// Result
+// The wavelength of K_alpha radiation of tungsten = 2.22e-01 angstrom 
diff --git a/1736/CH1/EX1.25/Ch01Ex25.sce b/1736/CH1/EX1.25/Ch01Ex25.sce
new file mode 100755
index 000000000..3411f22c1
--- /dev/null
+++ b/1736/CH1/EX1.25/Ch01Ex25.sce
@@ -0,0 +1,30 @@
+// Scilab Code Ex 1.25 Page-35 (2006)
+clc; clear;
+a_Cu = 3.61;    // Lattice constant of Cu, angstrom
+a_Pd = 3.89;    // Lattice constant of Pd, angstrom
+
+// For x = 20% of Pd
+x = 0.20;   // Percentage of Pd in Cu-Pd alloy
+a_Cu_Pd = ((1-x)*a_Cu + x*a_Pd);
+printf("\nFor %2d percent of Pd in Cu-Pd alloy, a = %4.2f angstrom", x*100, a_Cu_Pd);
+
+// For x = 40% of Pd
+x = 0.40;   // Percentage of Pd in Cu-Pd alloy
+a_Cu_Pd = ((1-x)*a_Cu + x*a_Pd);
+printf("\nFor %2d percent of Pd in Cu-Pd alloy, a = %5.3f angstrom", x*100, a_Cu_Pd);
+
+// For x = 60% of Pd
+x = 0.60;   // Percentage of Pd in Cu-Pd alloy
+a_Cu_Pd = ((1-x)*a_Cu + x*a_Pd);
+printf("\nFor %2d percent of Pd in Cu-Pd alloy, a = %5.3f angstrom", x*100, a_Cu_Pd);
+
+// For x = 80% of Pd
+x = 0.80;   // Percentage of Pd in Cu-Pd alloy
+a_Cu_Pd = ((1-x)*a_Cu + x*a_Pd);
+printf("\nFor %2d percent of Pd in Cu-Pd alloy, a = %5.3f angstrom", x*100, a_Cu_Pd);
+
+// Result
+// For 20 percent of Pd in Cu-Pd alloy, a = 3.67 angstrom
+// For 40 percent of Pd in Cu-Pd alloy, a = 3.722 angstrom
+// For 60 percent of Pd in Cu-Pd alloy, a = 3.778 angstrom
+// For 80 percent of Pd in Cu-Pd alloy, a = 3.834 angstrom 
diff --git a/1736/CH1/EX1.26/Ch01Ex26.sce b/1736/CH1/EX1.26/Ch01Ex26.sce
new file mode 100755
index 000000000..c76759583
--- /dev/null
+++ b/1736/CH1/EX1.26/Ch01Ex26.sce
@@ -0,0 +1,16 @@
+// Scilab Code Ex 1.26 Page-36 (2006)
+clc; clear;
+a_Rh = 3.80;    // Lattice constant of Rh, angstrom
+a_Pt = 3.92;    // Lattice constant of Pt, angstrom
+a_Pt_Rh = 3.78;    // Lattice constant of unit cell of Pt-Rh alloy, angstrom
+V = (a_Pt*1e-08)^3; // Volume of unit cell of Pt, metre cube
+V_90 = 0.9*V;   // 90 percent of the cell volume of Pt, metre cube
+
+// For x = 20% of Rh in Pt-Rh alloy, we have
+// a_Pt_Rh = ((1-x)*a_Pt + x*a_Rh), solving for x
+x = poly(0, 'x');
+x = roots (a_Pt_Rh - a_Pt + x*a_Pt - x*a_Rh);      // Amount of required Rh in Pt to change the unit cell volume
+printf("\nThe amount of Rh required in Pt to change the unit cell volume = %4.2f percent", x);
+
+// Result
+// The amount of Rh required in Pt to change the unit cell volume = 1.17 percent 
diff --git a/1736/CH1/EX1.27/Ch01Ex27.sce b/1736/CH1/EX1.27/Ch01Ex27.sce
new file mode 100755
index 000000000..92fe04769
--- /dev/null
+++ b/1736/CH1/EX1.27/Ch01Ex27.sce
@@ -0,0 +1,16 @@
+// Scilab Code Ex 1.27 :Page-36 (2006)
+clc; clear;
+r_bcc = 0.126;     // Atomic radius of the iron atoms in the bcc structure, nm
+r_fcc = 0.129;      // Atomic radius of the iron atoms in the fcc structure, nm
+a_bcc = 4*r_bcc/sqrt(3);
+a_fcc = 4*r_fcc/sqrt(2);
+V_bcc = 2*a_bcc^3;      // Volume of bcc unit cell, nm cube
+V_fcc = a_fcc^3;      // Volume of fcc unit cell, nm cube
+delta_V = V_fcc - V_bcc;    // Change in volume from bcc to fcc structure, nm cube
+V = V_bcc;
+V_frac = delta_V/V;     // Fractional change in volume from bcc to fcc structure
+
+printf("\nThe percentage change in volume from bcc to fcc structure = %3.1f percent", V_frac*100);
+
+// Result
+// The percentage change in volume from bcc to fcc structure = -1.4 percent 
diff --git a/1736/CH1/EX1.3/Ch01Ex3.sce b/1736/CH1/EX1.3/Ch01Ex3.sce
new file mode 100755
index 000000000..502ad6be2
--- /dev/null
+++ b/1736/CH1/EX1.3/Ch01Ex3.sce
@@ -0,0 +1,12 @@
+// Scilab Code Ex1.3 : Page-17 (2006)

+clc; clear;

+V = 10.58e-29;  // Volume of the unit cell, metre cube

+a = poly(0, 'a');      // Declare a variable

+a = roots(3*sqrt(3)/2*1.58*a^3-V);        // First lattice parameter, m

+c = 1.58*a(3);     // Third lattice parameter, m

+printf("\nThe lattice parameters of hcp structure of Ti are:");

+printf("\na = %4.2f angstorm, c = %4.2f angstorm", a(3)/1e-010, c/1e-010);

+

+// Result 

+// The lattice parameters of hcp structure of Ti are:

+// a = 2.95 angstorm, c = 4.67 angstorm 

diff --git a/1736/CH1/EX1.4/Ch01Ex4.sce b/1736/CH1/EX1.4/Ch01Ex4.sce
new file mode 100755
index 000000000..e45d0ae38
--- /dev/null
+++ b/1736/CH1/EX1.4/Ch01Ex4.sce
@@ -0,0 +1,22 @@
+// Scilab Code Ex1.4 : Page-17 (2006)

+clc; clear;

+c_by_a_ratio = 1.633;   // Ideal c/a ratio

+A = cell(2,4);  // Declare a cell

+// Assign values to the elements of the cell from the table

+A(1,1).entries = 'Mg';

+A(2,1).entries = 'Cd';

+A(1,2).entries = 5.21;

+A(2,2).entries = 5.62;

+A(1,3).entries = 3.21;

+A(2,3).entries = 2.98;

+A(1,4).entries = A(1,2).entries/A(1,3).entries;

+A(2,4).entries = A(2,2).entries/A(2,3).entries;

+if (A(1,4).entries - c_by_a_ratio) < 0.01 then

+    printf("\n%s satisfies ideal c/a ratio and %s has large deviation from this value.", A(1,1).entries, A(2,1).entries);

+else if (A(1,4).entries - c_by_a_ratio) < 0.01 then

+    printf("\n%s satisfies ideal c/a ratio and %s has large deviation from this value.", A(2,1).entries, A(1,1).entries);       

+    end

+end

+

+// Result 

+// Mg satisfies ideal c/a ratio and Cd has large deviation from this value. 

diff --git a/1736/CH1/EX1.5/Ch01Ex5.sce b/1736/CH1/EX1.5/Ch01Ex5.sce
new file mode 100755
index 000000000..bcb0bd176
--- /dev/null
+++ b/1736/CH1/EX1.5/Ch01Ex5.sce
@@ -0,0 +1,15 @@
+// Scilab Code Ex 1.5 : Page-18 (2006)

+clc; clear;5

+M_Na = 23;        // Atomic weight of Na, gram per mole

+M_Cl = 35.5;        // Atomic weight of Cl, gram per mole

+d = 2.18e+06;    // Density of Nacl salt, g per metre cube

+n = 4;    // No. of atoms per unit cell for an fcc lattice of NaCl crystal

+N = 6.023D+23;    // Avogadro's No.

+// Volume of the unit cell is given by

+// a^3 = M*n/(N*d)

+// Solving for a

+a = (n*(M_Na + M_Cl)/(d*N))^(1/3);    // Lattice constant of unit cell of NaCl

+printf("\nLattice constant for the NaCl crystal = %4.2f angstorm", a/1e-010);

+

+// Result 

+// Lattice constant for the NaCl crystal = 5.63 angsotrm 

diff --git a/1736/CH1/EX1.6/Ch01Ex6.sce b/1736/CH1/EX1.6/Ch01Ex6.sce
new file mode 100755
index 000000000..884407ee1
--- /dev/null
+++ b/1736/CH1/EX1.6/Ch01Ex6.sce
@@ -0,0 +1,11 @@
+// Scilab Code Ex 1.6 : Page-18 (2006)

+clc; clear;

+r = 1.33;       // Ionic radii of K+ ion, angstrom

+R = 1.81;     // Ionic radii of Cl- ion, angstrom

+n = 4;    // No. of atoms per unit cell for an fcc lattice of NaCl crystal

+APF = (n*(4*%pi*r^3/3)+n*(4*%pi*R^3/3))/(2*r+2*R)^3;    // Atomic packing factor of fcc KCl

+printf("\nThe ionic packing factor of fcc KCl = %4.2f", APF);

+

+// Result 

+// The ionic packing factor of fcc KCl = 0.56 

+

diff --git a/1736/CH1/EX1.7/Ch01Ex7.sce b/1736/CH1/EX1.7/Ch01Ex7.sce
new file mode 100755
index 000000000..efb36c6bb
--- /dev/null
+++ b/1736/CH1/EX1.7/Ch01Ex7.sce
@@ -0,0 +1,23 @@
+// Scilab Code Ex 1.7 : Page-20 (2006)

+clc; clear;

+N = 6.023e+23;  // Avogadro's number

+M = 12.01e-03;  // Atomic weight of diamond/graphite, kg

+

+// For diamond

+a = 3.568e-010;     // Lattice parameter of diamond, m

+rho = 3.518e+03;    // Density of diamond, kg per metre cube

+n = a^3*rho*N/M;    // Number of atoms in the unit cell of diamond structure

+printf("\nThe number of atoms in the unit cell of diamond structure = %1d", n);

+

+// For graphite

+a = 2.451e-010;     // First lattice parameter of graphite, m

+c = 6.701e-010;     // Third lattice parameter of graphite, m

+rho = 2.2589e+03;    // Density of graphite, kg per metre cube

+V = 3*sqrt(3)*a^2*c/2;  // Volume of hexagonal unit cell of graphite, metre cube

+n = V*rho*N/M;    // Number of atoms in the unit cell of graphite structure

+printf("\nThe number of atoms in the unit cell of graphite structure = %2d", ceil(n));

+

+// Result 

+// The number of atoms in the unit cell of diamond structure = 8

+// The number of atoms in the unit cell of graphite structure = 12 

+

diff --git a/1736/CH1/EX1.8/Ch01Ex8.sce b/1736/CH1/EX1.8/Ch01Ex8.sce
new file mode 100755
index 000000000..da195a8fd
--- /dev/null
+++ b/1736/CH1/EX1.8/Ch01Ex8.sce
@@ -0,0 +1,24 @@
+// Scilab Code Ex 1.8 :Page-21 (2006)

+clc; clear;

+N = 6.023e+23;  // Avogadro's number

+

+// For silicon crystallized into diamond structure

+a = 5.43e-08;   // Lattice parameter of Si, cm

+M = 28.1;   // Atomic mass of Si, g/mol

+n = 8/a^3;       // Number of atoms per unit volume, atoms per cm cube

+d = n*M/N;      // Density of Si crytal, g/cm

+printf("\nThe density of crystallized Si = %4.2f gram per cm cube", d);

+

+// For GaAs crystallized into Zinc Blende structure

+a = 5.65e-08;   // Lattice parameter of GaAs, cm

+M_Ga = 69.7;    // Atomic weight of Ga, g/mol

+M_As = 74.9;    // Atomic weight of As, g/mol

+M = M_Ga + M_As;   // Atomic weight of GaAs, g/mol

+n = 4/a^3;       // Number of atoms per unit volume, atoms per cm cube

+d = n*M/N;      // Density of Si crytal, g/cm

+printf("\nThe density of crystallized GaAs = %5.3f gram per cm cube", d);

+

+// Result 

+// The density of crystallized Si = 2.33 gram per cm cube

+// The density of crystallized GaAs = 5.324 gram per cm cube 12 

+

diff --git a/1736/CH1/EX1.9/Ch01Ex9.sce b/1736/CH1/EX1.9/Ch01Ex9.sce
new file mode 100755
index 000000000..0ae360591
--- /dev/null
+++ b/1736/CH1/EX1.9/Ch01Ex9.sce
@@ -0,0 +1,22 @@
+// Scilab Code Ex 1.9 :Page-21 (2006)

+clc; clear;

+N = 6.023e+23;  // Avogadro's number

+

+r1 = 0.122e-09;     // Ionic radii of Ga, m

+r2 = 0.125e-09;     // Ionic radii of As, m

+r3 = 0.11e-09;     // Ionic radii of P, m

+

+// For GaP

+r = r1 + r3;        // Interatomic separation between Ga and P atoms, m

+a = 4*r/3^(1/2);    // Lattice parameter of GaP structure, m

+printf("\nThe lattice parameter of GaP structure = %5.3f angstrom", a/1e-10);

+

+// For GaAs

+r = r1 + r2;        // Interatomic separation between Ga and As atoms, m

+a = 4*r/3^(1/2);    // Lattice parameter of GaP structure, m

+printf("\nThe lattice parameter of GaAs structure = %4.2f angstrom", a/1e-10);

+

+// Result 

+// The lattice parameter of GaP structure = 5.358 angstrom

+// The lattice parameter of GaAs structure = 5.70 angstrom 

+