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+{
+"cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 1: Crystallography"
+ ]
+ },
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_10: calculate_free_electron_concentration.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_10,pg 1-62\n",
+"\n",
+"A=63.546 //atomic weight of Cu\n",
+"\n",
+"N=6.023*10^26 //Avogadro's number\n",
+"\n",
+"p=8930 //Density\n",
+"\n",
+"n=1.23 //no.of electron per atom\n",
+"\n",
+"//density=mass/volume\n",
+"\n",
+"//therfore 1/volume=density/mass\n",
+"\n",
+"//since electron concentration is needed, let us find out no of atoms/volume(x)\n",
+"\n",
+"x=N*p/A\n",
+"\n",
+"//now one atom contribute n=1.23 electron\n",
+"\n",
+"//therefore x atoms contribute y no of free electron\n",
+"\n",
+"y=x*n\n",
+"\n",
+"printf('free electron concentration=')\n",
+"\n",
+"disp(y)\n",
+"\n",
+"printf('electron/m^3')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_11: calculate_Y_and_Z_intercept.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_11,pg 1-62\n",
+"\n",
+"//primitive vectors\n",
+"\n",
+"a=1.5 //in amstrong unit\n",
+"\n",
+"b=2 //in amstrong unit\n",
+"\n",
+"c=4 //in amstrong unit\n",
+"\n",
+"//miller indices of the plane\n",
+"\n",
+"h=3\n",
+"\n",
+"k=2\n",
+"\n",
+"l=6\n",
+"\n",
+"//therefore intercepts are a/h,b/k,c/l\n",
+"\n",
+"x=a/h\n",
+"\n",
+"y=b/k\n",
+"\n",
+"z=c/l\n",
+"\n",
+"//this gives intercepts along x axis as x amstrong but it is given that plane cut x axis at 1.2 amstrong .\n",
+"\n",
+"t=1.5/x\n",
+"\n",
+"//this shows that the plane under consideration is another plane which is parallel to it(to keep miller indices same)\n",
+"\n",
+"n=t*y //Y intercept\n",
+"\n",
+"p=t*z //Z intercept\n",
+"\n",
+"printf(' 1) Y intercept=')\n",
+"\n",
+"disp(n)\n",
+"\n",
+"printf('amstrong')\n",
+"\n",
+"printf(' 2)Z intercept=')\n",
+"\n",
+"disp(p)\n",
+"\n",
+"printf('amstrong')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_12: calculate_Number_of_atom_per_unit_cell.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_12,pg 1-63\n",
+"\n",
+"ro=7.87 //density of metal\n",
+"\n",
+"A=55.85 //atomic wt of metal\n",
+"\n",
+"N=6.023*10^23 //Avogadro's number\n",
+"\n",
+"a=2.9*10^-8 //lattice constant of metal\n",
+"\n",
+"n=(N*(a^3)*ro)/A\n",
+"\n",
+"printf('Number of atom per unit cell of a metal=')\n",
+"\n",
+"disp(int32(n))"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_13: calculate_Lattice_constant.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_13,pg 1-63\n",
+"\n",
+"n=2 //BCC structure\n",
+"\n",
+"ro=9.6*10^2 //density of sodium crystal\n",
+"\n",
+"A=23 //atomic weight of sodium crystal\n",
+"\n",
+"N=6.023*10^26 //Avogadro's number\n",
+"\n",
+"a=((n*A)/(N*ro))^(1/3)\n",
+"\n",
+"printf('Lattice constant=')\n",
+"\n",
+"disp(a)\n",
+"\n",
+"printf('m')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_15: calculate_Number_of_atom_per_unit_cell_and_atomic_radius.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_15,pg 1-64\n",
+"\n",
+"ro=2.7*10^3 //density of metal\n",
+"\n",
+"A=27 //atomic wt of metal\n",
+"\n",
+"N=6.023*10^26 //Avogadro's number\n",
+"\n",
+"a=4.05*10^-10 //lattice constant of metal\n",
+"\n",
+"n=(N*(a^3)*ro)/A\n",
+"\n",
+"printf('1) Number of atom per unit cell of a metal=')\n",
+"\n",
+"disp(int32(n))\n",
+"\n",
+"r=sqrt(2)*a/4 //radius of metal\n",
+"\n",
+"printf('2) atomic radius of a metal=')\n",
+"\n",
+"disp(r)\n",
+"\n",
+"printf('m')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_16: calculate_Lattice_constant_and_APF.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_16,pg 1-64\n",
+"\n",
+"n=2 //BCC structure\n",
+"\n",
+"ro=5.98*10^3 //density of chromium\n",
+"\n",
+"A=50 //atomic wt of chromium\n",
+"\n",
+"N=6.023*10^26 //Avogadro's number\n",
+"\n",
+"a=((n*A)/(N*ro))^(1/3)\n",
+"\n",
+"printf(' 1) Lattice constant=')\n",
+"\n",
+"disp(a)\n",
+"\n",
+"printf('m')\n",
+"\n",
+"//for BCC\n",
+"\n",
+"r=sqrt(3)*a/4 //radius of chromium\n",
+"\n",
+"APF=(n*(4/3)*%pi*(r^3))/(a^3)\n",
+"\n",
+"printf(' 2) A.P.F. for chromium=')\n",
+"\n",
+"disp(APF)\n",
+""
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_17: calculate_Lattice_constant.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_17,pg 1-65\n",
+"\n",
+"n=4 //FCC structure\n",
+"\n",
+"ro=6250 //density\n",
+"\n",
+"M=60.2 //molecular weight\n",
+"\n",
+"N=6.023*10^26 //Avogadro's number\n",
+"\n",
+"a=((n*M)/(N*ro))^(1/3)\n",
+"\n",
+"printf('Lattice constant=')\n",
+"\n",
+"disp(a)\n",
+"\n",
+"printf('m')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_19: calculate_wavlength.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_19,pg 1-66\n",
+"\n",
+"a=2.82*10^-9 //lattice constant\n",
+"\n",
+"n=2 //FCC crystal\n",
+"\n",
+"t=17.167 //glancing angle in degree\n",
+"\n",
+"q=%pi/180*t //glancing angle in radians\n",
+"\n",
+"//assuming reflection in (1,0,0) plane\n",
+"\n",
+"h=1\n",
+"\n",
+"k=0\n",
+"\n",
+"l=0\n",
+"\n",
+"d=a/sqrt(h^2+k^2+l^2)\n",
+"\n",
+"//using Bragg's law , 2*d*sin(q)=n*la\n",
+"\n",
+"la=2*d*sin(q)/n\n",
+"\n",
+"printf('wavlength of X-ray=')\n",
+"\n",
+"disp(la)\n",
+"\n",
+"printf('m')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_1: calculate_lattice_constant.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_1,pg 1-58\n",
+"\n",
+"n=4 //FCC structure\n",
+"\n",
+"ro=2180 //density of NaCl\n",
+"\n",
+"M=23+35.5 //molecular weight of NaCl\n",
+"\n",
+"N=6.023*10^26 //Avogadro's number\n",
+"\n",
+"a=((n*M)/(N*ro))^(1/3)\n",
+"\n",
+"printf('Lattice constant=')\n",
+"\n",
+"disp(a)\n",
+"\n",
+"printf('m')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_20: calculate_Lattice_constant_and_atomic_radius.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_20,pg 1-66\n",
+"\n",
+"n=8 //Diamond structure\n",
+"\n",
+"ro=2.33*10^3 //density of diamond\n",
+"\n",
+"M=28.9 //atomic weight of diamond\n",
+"\n",
+"N=6.023*10^26 //Avogadro's number\n",
+"\n",
+"a=((n*M)/(N*ro))^(1/3)\n",
+"\n",
+"printf(' 1) Lattice constant=')\n",
+"\n",
+"disp(a)\n",
+"\n",
+"printf('m')\n",
+"\n",
+"r=sqrt(3)*a/8 //radius of diamond structure\n",
+"\n",
+"printf(' 2) atomic radius of a metal=')\n",
+"\n",
+"disp(r)\n",
+"\n",
+"printf('m')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_21: calculate_mass_of_one_atom.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_21,pg 1-66\n",
+"\n",
+"n=2 //BCC structure\n",
+"\n",
+"ro=8.57*10^3 //density of chromium\n",
+"\n",
+"d=2.86*10^-10 //nearest atoms distance\n",
+"\n",
+"//d=sqrt(3)/2*a\n",
+"\n",
+"a=2*d/sqrt(3)\n",
+"\n",
+"//now use formulae a^3*ro=n*A/N\n",
+"\n",
+"//therefore a^3*ro/n=mass of unit cell/(no of atoms pre unit cell)=mass of one atom\n",
+"\n",
+"m=a^3*ro/n\n",
+"\n",
+"printf('mass of one atom=')\n",
+"\n",
+"disp(m)\n",
+"\n",
+"printf('kg')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_2: calculate_Lattice_constant_and_diameter.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_2,pg 1-58\n",
+"\n",
+"n=4 //FCC structure\n",
+"\n",
+"ro=8.9 //density of Cu atom\n",
+"\n",
+"A=63.55 //atomic weight of Cu atom\n",
+"\n",
+"N=6.023*10^23 //Avogadro's number\n",
+"\n",
+"a=((n*A)/(N*ro))^(1/3)\n",
+"\n",
+"printf(' 1) Lattice constant=')\n",
+"\n",
+"disp(a)\n",
+"\n",
+"printf('cm')\n",
+"\n",
+"r=sqrt(2)*a/4 //radius of Cu atom\n",
+"\n",
+"d=2*r //diameter of Cu atom\n",
+"\n",
+"printf(' 2) Diameter of Cu atom=')\n",
+"\n",
+"disp(d)\n",
+"\n",
+"printf('cm')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_3: calculate_Density_of_diamond.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_3,pg 1-59\n",
+"\n",
+"n=8 //diamond structure\n",
+"\n",
+"A=12.01 //atomic wt\n",
+"\n",
+"N=6.023*10^23 //Avogadro's number\n",
+"\n",
+"a=3.75*10^-8 //lattice constant of diamond\n",
+"\n",
+"ro=(n*A)/(N*(a^3))\n",
+"\n",
+"printf('Density of diamond=')\n",
+"\n",
+"disp(ro)\n",
+"\n",
+"printf('gm/cc')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_4: calculate_miller_indices.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_4,pg 1-59\n",
+"\n",
+"//intercept of planeare in proportion 3a:4b:infinity (plane parallel to z axis)\n",
+"\n",
+"//as a,b and c are basic vectors the proportin of intercepts 3:4:infinity\n",
+"\n",
+"//therefore reciprocal\n",
+"\n",
+"r1=1/3\n",
+"\n",
+"r2=1/4\n",
+"\n",
+"r3=0\n",
+"\n",
+"//taking LCM\n",
+"\n",
+"v=int32([3,4])\n",
+"\n",
+"l=double(lcm(v))\n",
+"\n",
+"m1=(l*r1)\n",
+"\n",
+"m2=(l*r2)\n",
+"\n",
+"m3=(l*r3)\n",
+"\n",
+"printf('miler indices=')\n",
+"\n",
+"disp(m3,m2,m1)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_5: calculate_miller_indices.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_5,pg 1-59\n",
+"\n",
+"//intercept of planeare in proportion 3a:-2b:3/2c\n",
+"\n",
+"//as a,b and c are basic vectors the proportin of intercepts 3:-2:3/2\n",
+"\n",
+"//therefore reciprocal\n",
+"\n",
+"r1=1/3\n",
+"\n",
+"r2=-1/2\n",
+"\n",
+"r3=2/3\n",
+"\n",
+"//taking LCM\n",
+"\n",
+"v=int32([3,2,3/2])\n",
+"\n",
+"l=double(lcm(v))\n",
+"\n",
+"m1=(l*r1)\n",
+"\n",
+"m2=(l*r2)\n",
+"\n",
+"m3=(l*r3)\n",
+"\n",
+"printf('miler indices=')\n",
+"\n",
+"disp(m3,m2,m1)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_6: calculate_ratio_of_intercepts.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_6,pg 1-59\n",
+"\n",
+"//if a plane cut at length m,n,p on the three crystal axes,then\n",
+"\n",
+"//m:n:p=xa:yb:zc\n",
+"\n",
+"//when primitive vectors of unit cell and numbers x,y,z,are related to miller indices (h,k,l)of the plane by relation\n",
+"\n",
+"//1/x:1/y:1/z=h:k:l\n",
+"\n",
+"//since a=b=c (crystal is simple cubic)\n",
+"\n",
+"//and (h,k,l)=(1,2,3)\n",
+"\n",
+"//therefore reciprocal\n",
+"\n",
+"r1=1/1\n",
+"\n",
+"r2=1/2\n",
+"\n",
+"r3=1/3\n",
+"\n",
+"//taking LCM\n",
+"\n",
+"v=int32([1,2,3])\n",
+"\n",
+"l=double(lcm(v))\n",
+"\n",
+"m=(l*r1)\n",
+"\n",
+"n=(l*r2)\n",
+"\n",
+"p=(l*r3)\n",
+"\n",
+"printf('ratio of intercepts=')\n",
+"\n",
+"disp(m)\n",
+"\n",
+"disp(n)\n",
+"\n",
+"disp(p)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_7: calculate_y_and_z_intercepts.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_7,pg 1-60\n",
+"\n",
+"//primitive vectors\n",
+"\n",
+"a=1.2 //in amstrong unit\n",
+"\n",
+"b=1.8 //in amstrong unit\n",
+"\n",
+"c=2 //in amstrong unit\n",
+"\n",
+"//miller indices of the plane\n",
+"\n",
+"h=2\n",
+"\n",
+"k=3\n",
+"\n",
+"l=1\n",
+"\n",
+"//therefore intercepts are a/h,b/k,c/l\n",
+"\n",
+"x=a/h\n",
+"\n",
+"y=b/k\n",
+"\n",
+"z=c/l\n",
+"\n",
+"//this gives intercepts along x axis as x amstrong but it is given tthat plane cut x axis at 1.2 amstrong .\n",
+"\n",
+"t=1.2/x\n",
+"\n",
+"//this shows that the plane under consideration is another plane which is parallel to it(to keep miller indices same)\n",
+"\n",
+"n=t*y //Y intercept\n",
+"\n",
+"p=t*z //Z intercept\n",
+"\n",
+"printf(' 1) Y intercept=')\n",
+"\n",
+"disp(n)\n",
+"\n",
+"printf('amstrong')\n",
+"\n",
+"printf(' 2)Z intercept=')\n",
+"\n",
+"disp(p)\n",
+"\n",
+"printf('amstrong')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_8: calculate_radius.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_8,pg 1-61\n",
+"\n",
+"//the interplanar spacing of plane\n",
+"\n",
+"h=1\n",
+"\n",
+"k=1\n",
+"\n",
+"l=0\n",
+"\n",
+"d=2 //interpanar spacing in amstrong unit\n",
+"\n",
+"//we know that d=a/sqrt(h^2+k^2+l^2) therefore\n",
+"\n",
+"a=d*sqrt(h^2+k^2+l^2)\n",
+"\n",
+"//for FCC structure\n",
+"\n",
+"r=sqrt(2)*a/4\n",
+"\n",
+"printf('radius r=')\n",
+"\n",
+"disp(r)\n",
+"\n",
+"printf('amstrong')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.14_9: calculate_density_and_diameter.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_14_9,pg 1-61\n",
+"\n",
+"n=4 //for FCC structure\n",
+"\n",
+"//the interplanar spacing of plane\n",
+"\n",
+"h=1\n",
+"\n",
+"k=1\n",
+"\n",
+"l=1\n",
+"\n",
+"d=2.08*10^-10 //distance\n",
+"\n",
+"A=63.54 //atomic weight of Cu\n",
+"\n",
+"N=6.023*10^26 //amstrong no\n",
+"\n",
+"//we know that d=a/sqrt(h^2+k^2+l^2) therefore\n",
+"\n",
+"a=d*sqrt(h^2+k^2+l^2)\n",
+"\n",
+"//also (a^3*q)=n*A/N\n",
+"\n",
+"q=n*A/(N*a^3)\n",
+"\n",
+"printf(' 1)density=')\n",
+"\n",
+"disp(q)\n",
+"\n",
+"printf('kg/m^3')\n",
+"\n",
+"//for FCC structure\n",
+"\n",
+"r=sqrt(2)*a/4\n",
+"\n",
+"d=r*2\n",
+"\n",
+"printf(' 2)radius r=')\n",
+"\n",
+"disp(r)\n",
+"\n",
+"printf('m')\n",
+"\n",
+"printf(' 3)diameter d=')\n",
+"\n",
+"disp(d)\n",
+"\n",
+"printf('m')\n",
+""
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_10: calculate_wavelength.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_10,pg 1-73\n",
+"\n",
+"//for line -A\n",
+"\n",
+"n1=1 //1st order maximum\n",
+"\n",
+"q1=30 //glancing angle in degree\n",
+"\n",
+"//using Bragg's law for line A n1*l1=2*d1*sin(q1)\n",
+"\n",
+"//d1=n1*l1/(2*sin(q1))\n",
+"\n",
+"//for line B\n",
+"\n",
+"l2=0.97 //wavelength in amstrong unit\n",
+"\n",
+"n2=3 //1st order maximum\n",
+"\n",
+"q2=60 //glancing angle in degree\n",
+"\n",
+"//using Bragg's law for line B n2*l2=2*d2*sin(q2)\n",
+"\n",
+"//since for both lines A and B we use same plane of same crystal,therefore\n",
+"\n",
+"//d1=d2\n",
+"\n",
+"//therefore equution became n2*l2=2*n1*l1/(2*sin(q1))*sin(q2)\n",
+"\n",
+"//by arranging terms we get\n",
+"\n",
+"\n",
+"l1=n2*l2*2*sind(q1)/(2*n1*sind(q2))\n",
+"\n",
+"printf('wavelength of the line A=')\n",
+"\n",
+"disp(l1)\n",
+"\n",
+"printf('amstrong')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_11: calculate_glancing_angle.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_11,pg 1-74\n",
+"\n",
+"n=1 //first order minimum\n",
+"\n",
+"d=5.5*10^-11 //atomic spacing\n",
+"\n",
+"e=1.6*10^-19 //charge on one electron\n",
+"\n",
+"Ee=10*10^3 //energy in eV\n",
+"\n",
+"E=e*Ee //energy in Joule\n",
+"\n",
+"m=9.1*10^-31 //mass of elelctron\n",
+"\n",
+"h=6.63*10^-34 //plank's constant\n",
+"\n",
+"l=h/sqrt(2*m*E) //wavelength\n",
+"\n",
+"//using Bragg's law\n",
+"\n",
+"Q=asind((n*l)/(2*d)) //glancing angle\n",
+"\n",
+"printf('glancing angle=')\n",
+"\n",
+"disp(Q)\n",
+"\n",
+"printf('degree')\n",
+""
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_12: calculate_glancing_angle.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_12,pg 1-74\n",
+"\n",
+"a=2.814*10^-10 //lattice constant\n",
+"\n",
+"//for rock salt\n",
+"\n",
+"d=a/2 //interplaner spacing\n",
+"\n",
+"n=1 //first order maximum\n",
+"\n",
+"l=1.541*10^-10 //wavelength of rock salt crystal\n",
+"\n",
+"//using Bragg's law\n",
+"\n",
+"m=asin((n*l)/(2*d)) //glancing angle\n",
+"\n",
+"Q=m*180/%pi\n",
+"\n",
+"printf('glancing angle=')\n",
+"\n",
+"disp(Q)\n",
+"\n",
+"printf('degree')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_1: calculate_glancing_angle_and_highest_order.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_1,pg 1-68\n",
+"\n",
+"d=4.255*10^-10 //interplaner spacing\n",
+"\n",
+"l=1.549*10^-10 //wavelength of x ray\n",
+"\n",
+"//part 1: for smallest glancing angle(n=1)\n",
+"\n",
+"n1=1\n",
+"\n",
+"//using Bragg's law n*l=2*d*sin(q)\n",
+"\n",
+"q=asind(n1*l/(2*d))\n",
+"\n",
+"printf(' 1)glancing angle=')\n",
+"\n",
+"disp(q)\n",
+"\n",
+"printf('degree')\n",
+"\n",
+"//part 2: for highst order\n",
+"\n",
+"//for highest order sin(q) not exceed one i.e maximum value is one\n",
+"\n",
+"//using Bragg's law n*l=2*d*sin(q)\n",
+"\n",
+"n2=2*d/l //since sin(q)is one\n",
+"\n",
+"printf(' 2)highest order possible =')\n",
+"\n",
+"disp(floor(n2))"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_2: calculate_glancing_angle.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_2,pg 1-69\n",
+"\n",
+"a=2.125*10^-10 //lattice constant\n",
+"\n",
+"d=a/2 //interplaner spacing\n",
+"\n",
+"n=2 //second order maximum\n",
+"\n",
+"l=0.592*10^-10 //wavelength of rock salt crystal\n",
+"\n",
+"//using Bragg's law\n",
+"\n",
+"m=asin((n*l)/(2*d)) //glancing angle\n",
+"\n",
+"Q=m*180/%pi\n",
+"\n",
+"printf('glancing angle=')\n",
+"\n",
+"disp(Q)\n",
+"\n",
+"printf('degree')\n",
+""
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_3: calculate_second_order_reflection_angle.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_3,pg 1-69\n",
+"\n",
+"n1=1 //for 1st order\n",
+"\n",
+"n2=2 //for 2nd order\n",
+"\n",
+"t=3.4 //angle where 1st order reflection done\n",
+"\n",
+"t1=t*%pi/180 //convert degree to radian\n",
+"\n",
+"m=sin(t1)\n",
+"\n",
+"//but from Bragg's law\n",
+"\n",
+"//n*l=2*d*sin(t)\n",
+"\n",
+"//for for constant distance(d) and wavelength(l) \n",
+"\n",
+"//order(n) is directly proportionl to sine of angle i.e (sin(t))\n",
+"\n",
+"//n1/n2=sin(t1)/sin(t2)\n",
+"\n",
+"//assume sin(t2)=a\n",
+"\n",
+"a=n2/n1*m\n",
+"\n",
+"t2=asind(a) //taking sin inverese in degree\n",
+"\n",
+"printf('second order reflection take place at an angle=')\n",
+"\n",
+"disp(t2)\n",
+"\n",
+"printf('degree')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_4: calculate_shortest_wavelength_and_glancing_angle.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_4,pg 1-70\n",
+"\n",
+"V=50*10^3 //operating voltage of x-ray\n",
+"\n",
+"M=74.6 //molecular weight\n",
+"\n",
+"p=1.99*10^3 //density\n",
+"\n",
+"n=4 //no of atoms per unit cell(for FCC structure)\n",
+"\n",
+"h=6.63*10^-34 //plank's constant\n",
+"\n",
+"c=3*10^8 //velocity \n",
+"\n",
+"e=1.6*10^-19 //charge on electron\n",
+"\n",
+"N=6.023*10^26 //Avogadro's number\n",
+"\n",
+"//step 1:clculating shortest wavelength\n",
+"\n",
+"l=h*c/(e*V)\n",
+"\n",
+"printf(' 1)shortest wavelength=')\n",
+"\n",
+"disp(l)\n",
+"\n",
+"printf('m')\n",
+"\n",
+"//step:2 calculating distance(d)\n",
+"\n",
+"//now a^3*p=n*M/N therefore,\n",
+"\n",
+"a=(n*M/(N*p))^(1/3)\n",
+"\n",
+"//since KCl is ionic crystal herefore,\n",
+"\n",
+"d=a/2\n",
+"\n",
+"//step 3: calculaing glancing angle\n",
+"\n",
+"//using Bragg's law\n",
+"\n",
+"//n*l=2*d*sin(t)\n",
+"\n",
+"//assume sin(t)=a, wavelength is minimum i.e l and n=1\n",
+"\n",
+"n=1\n",
+"\n",
+"a=n*l/(2*d)\n",
+"\n",
+"t=asind(a) //taking sin inverese in degree\n",
+"\n",
+"printf(' 2) glancing angle=')\n",
+"\n",
+"disp(t)\n",
+"\n",
+"printf('degree')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_5: find_possible_solution_of_planes.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_5,pg 1-70\n",
+"\n",
+"n=1 //first order maximum\n",
+"\n",
+"l=0.82*10^-10 //wavelength of X ray\n",
+"\n",
+"qd=7 //glancing angle in degree\n",
+"\n",
+"qm=51/60 //glancing angle in minute\n",
+"\n",
+"qs=48/3600 //glancing angle in second\n",
+"\n",
+"q=qd+qm+qs //total glancin angle in degree\n",
+"\n",
+"//using Bragg's law n*l=2*d*sin(q)\n",
+"\n",
+"d=n*l/(2*sind(q))\n",
+"\n",
+"a=3*10^-10 //lattice constant\n",
+"\n",
+"//we know that d=a/root(h^2+k^2+l^2)\n",
+"\n",
+"//assume root(h^2+k^2+l^2) =m\n",
+"\n",
+"//arranging terms we get\n",
+"\n",
+"m=a/d\n",
+"\n",
+"printf('square root(h^2+k^2+l^2)=') \n",
+"\n",
+"disp(int32(m))\n",
+"\n",
+"printf('hence possible solutions are (100),(010),(001)')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_6: calculate_cubic_lattice_structure.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_6,pg 1-71\n",
+"\n",
+"n=1 //first order maximum\n",
+"\n",
+"l=%i //wavelength of X ray\n",
+"\n",
+"//part 1:for(100)\n",
+"\n",
+"//using Bragg's law n*l=2*d*sin(q)\n",
+"\n",
+"q1=5.4 //glancing angle in degree\n",
+"\n",
+"dl1=n*l/(2*sind(q1))\n",
+"\n",
+"//part 2:for(110)\n",
+"\n",
+"//using Bragg's law n*l=2*d*sin(q)\n",
+"\n",
+"q2=7.6 //glancing angle in degree\n",
+"\n",
+"dl2=n*l/(2*sind(q2))\n",
+"\n",
+"//part 3:for(111)\n",
+"\n",
+"//using Bragg's law n*l=2*d*sin(q)\n",
+"\n",
+"q3=9.4 //glancing angle in degree\n",
+"\n",
+"dl3=n*l/(2*sind(q3))\n",
+"\n",
+"//for taking ratio divide all dl by dl1\n",
+"\n",
+"d1=dl1/dl1\n",
+"\n",
+"d2=dl2/dl1\n",
+"\n",
+"d3=dl3/dl1\n",
+"\n",
+"printf('cubic lattice structure is=')\n",
+"\n",
+"disp(d3,d2,d1)\n",
+""
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_7: calculate_lattice_constant.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_7,pg 1-71\n",
+"\n",
+"n=1 //first order maximum\n",
+"\n",
+"l=1.54*10^-10 //wavelength of rock salt crystal\n",
+"\n",
+"q=21.7 //glancing angle in degree\n",
+"\n",
+"//using Bragg's law n*l=2*d*sin(q)\n",
+"\n",
+"d=n*l/(2*sind(q))\n",
+"\n",
+"printf('lattice constant of crystal=')\n",
+"\n",
+"disp(d)\n",
+"\n",
+"printf('meter')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_8: calculate_glancing_angle.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_8,pg 1-72\n",
+"\n",
+"a=2.814*10^-10 //lattice constant\n",
+"\n",
+"//the interplanar spacing of plane\n",
+"\n",
+"h=1\n",
+"\n",
+"k=0\n",
+"\n",
+"l=0\n",
+"\n",
+"d=a/sqrt(h^2+k^2+l^2)\n",
+"\n",
+"n=2 //first order maximum\n",
+"\n",
+"l=0.714*10^-10 //wavelength of X-ray crystal\n",
+"\n",
+"//using Bragg's law\n",
+"\n",
+"m=asin((n*l)/(2*d)) //glancing angle\n",
+"\n",
+"Q=m*180/%pi\n",
+"\n",
+"printf('glancing angle=')\n",
+"\n",
+"disp(Q)\n",
+"\n",
+"printf('degree')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.15_9: calculate_wavelength_and_glancing_angle_and_highest_order.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_15_9,pg 1-72\n",
+"\n",
+"d=2.82*10^-10 //interplaner spacing\n",
+"\n",
+"t=10 //glancing angle\n",
+"\n",
+"//for part 1\n",
+"\n",
+"n=1 //first order maximum\n",
+"\n",
+"//using Bragg's law n*l=2*d*sin(t)\n",
+"\n",
+"l=2*d*sind(t)/n\n",
+"\n",
+"printf(' 1)wavelength=')\n",
+"\n",
+"disp(l)\n",
+"\n",
+"printf('meter')\n",
+"\n",
+"//for part 2\n",
+"\n",
+"n1=2\n",
+"\n",
+"//using Bragg's law n*l=2*d*sin(q)\n",
+"\n",
+"q=asind(n1*l/(2*d))\n",
+"\n",
+"printf(' 2)glancing angle=')\n",
+"\n",
+"disp(q)\n",
+"\n",
+"printf('degree')\n",
+"\n",
+"//for part 3\n",
+"\n",
+"//for highest order sin(q) not exceed one i.e maximum value is one\n",
+"\n",
+"//using Bragg's law n*l=2*d*sin(q)\n",
+"\n",
+"n2=2*d/l //since sin(q)is one\n",
+"\n",
+"printf(' 3)highest order possible =')\n",
+"\n",
+"disp(floor(n2))"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.16_1: calculate_ratio_of_vacancies.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_16_1,pg 1-75\n",
+"\n",
+"Ev=1.08 //average energy required to creaet a vacancy\n",
+"\n",
+"k=1.38*10^-23 //boltzman constant in J/K\n",
+"\n",
+"e=1.6*10^-19 //charge on 1 electron\n",
+"\n",
+"K=k/e //boltzman constant in eV/K\n",
+"\n",
+"//for a low concentration of vacancies a relation is\n",
+"\n",
+"//n=Nexp(-Ev/KT)\n",
+"\n",
+"//since total no atom is 1 hence N=1\n",
+"\n",
+"//at 1000k\n",
+"\n",
+"T1=1000 //temperature\n",
+"\n",
+"n1=exp(-Ev/(K*T1))\n",
+"\n",
+"//at 500k\n",
+"\n",
+"T2=500 //temperature\n",
+"\n",
+"n2=exp(-Ev/(K*T2))\n",
+"\n",
+"v=(n1)/(n2) //ratio of vacancies\n",
+"\n",
+"printf('ratio of vacancies=')\n",
+"\n",
+"disp(v)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.16_2: calculate_ratio_of_vacancies_to_no_of_atom.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_16_2,pg 1-75\n",
+"\n",
+"Ev=1.95 //average energy required to creaet a vacancy\n",
+"\n",
+"k=1.38*10^-23 //boltzman constant in J/K\n",
+"\n",
+"e=1.6*10^-19 //charge on 1 electron\n",
+"\n",
+"K=k/e //boltzman constant in eV/K\n",
+"\n",
+"T=500 //temperature\n",
+"\n",
+"//for a low concentration of vacancies a relation is\n",
+"\n",
+"//n=Nexp(-Ev/KT)\n",
+"\n",
+"m=exp(-Ev/(K*T)) //ratio of no of vacancies to no of atoms n/N\n",
+"\n",
+"printf('ratio of no of vacancies to no of atoms=')\n",
+"\n",
+"disp(m)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.16_3: calculate_ratio_of_vacancies.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_16_3,pg 1-76\n",
+"\n",
+"Ev=1.8 //average energy required to creaet a vacancy\n",
+"\n",
+"k=1.38*10^-23 //boltzman constant in J/K\n",
+"\n",
+"e=1.6*10^-19 //charge on 1 electron\n",
+"\n",
+"K=k/e //boltzman constant in eV/K\n",
+"\n",
+"//for a low concentration of vacancies a relation is\n",
+"\n",
+"//n=Nexp(-Ev/KT)\n",
+"\n",
+"//ratio of vacancy is n/N assume be r=exp(-Ev/KT)\n",
+"\n",
+"//since total no atom is 1 hence N=1\n",
+"\n",
+"//at 1000k\n",
+"\n",
+"t1=-119 //temperature in degree\n",
+"\n",
+"T1=t1+273 //temperature in kelvine\n",
+"\n",
+"r1=exp(-Ev/(K*T1))\n",
+"\n",
+"printf('1)ratio of vacancies at -119 degree=')\n",
+"\n",
+"disp(r1)\n",
+"\n",
+"//at 500k\n",
+"\n",
+"t2=80 //temperature in degree\n",
+"\n",
+"T2=t2+273 //temperature in kelvine\n",
+"\n",
+"r2=exp(-Ev/(K*T2))\n",
+"\n",
+"v=(r1)/(r2) //ratio of vacancies\n",
+"\n",
+"printf('2)ratio of vacancies at 80 degree=')\n",
+"\n",
+"disp(r2)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.16_4: calculate_no_of_frankel_defects.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_16_4,pg 1-76\n",
+"\n",
+"Ev=1.5 //energy of formaton of frankel defect\n",
+"\n",
+"k=1.38*10^-23 //boltzman constant in J/K\n",
+"\n",
+"e=1.6*10^-19 //charge on 1 electron\n",
+"\n",
+"K=k/e //boltzman constant in eV/K\n",
+"\n",
+"T=700 //temperature\n",
+"\n",
+"N=6.023*10^26 //avogadro's no\n",
+"\n",
+"//for a low concentration of vacancies a relation is\n",
+"\n",
+"//n=Nexp(-Ev/KT)\n",
+"\n",
+"m=exp(-Ev/(2*K*T)) //ratio of no of vacancies to no of atoms n/N\n",
+"\n",
+"qs=5.56 //specific density\n",
+"\n",
+"q=5.56*10^3 //real density ke/m^3\n",
+"\n",
+"M=0.143 //molecular weight in kg/m^3\n",
+"\n",
+"ma=M/N //mass of one molecule\n",
+"\n",
+"v=ma/q //vol of one molecule\n",
+"\n",
+"//v volume containe 1 molecule\n",
+"\n",
+"//therefore 1 m^3 containe x molecule\n",
+"\n",
+"x=1/v\n",
+"\n",
+"d=m*x //defect per m^3\n",
+"\n",
+"dm=d*10^-9 //defect per mm^3\n",
+"\n",
+"printf('number of frankel defects per mm^3=')\n",
+"\n",
+"disp(dm)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.3_1: calculate_Unit_cell_dimension.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_1,pg 1-14\n",
+"\n",
+"A=26.98 //atomic weight of Al\n",
+"\n",
+"N=6.023*10^26 //Avogadro's number\n",
+"\n",
+"p=2700 //Density\n",
+"\n",
+"n=4 //FCC structure\n",
+"\n",
+"a=(n*A/(N*p))^(1/3)\n",
+"\n",
+"printf('Unit cell dimension of Al=')\n",
+"\n",
+"disp(a)\n",
+"\n",
+"printf('m')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.3_2: calculate_density.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_2,pg 1-15\n",
+"\n",
+"As=28.1 //atomic weight of Si\n",
+"\n",
+"Ag=69.7 //atomic weight of Ga\n",
+"\n",
+"Aa=74.9 //atomic weight of As\n",
+"\n",
+"as=5.43*10^-8 //lattice constant of Si\n",
+"\n",
+"aga=5.65*10^-8 //lattice constant of GaAs\n",
+"\n",
+"ns=8 //no of atoms/unit cell in Si\n",
+"\n",
+"nga=4 //no of atoms/unit cell in GaAs\n",
+"\n",
+"N=6.023*10^23 //Avogadro's number\n",
+"\n",
+"//p=(n*A)/(N*a^3) this is formula for density\n",
+"\n",
+"//for Si\n",
+"\n",
+"ps=(ns*As)/(N*as^3)\n",
+"\n",
+"printf(' 1) Density of Si=')\n",
+"\n",
+"disp(ps)\n",
+"\n",
+"printf('gm/cm^3')\n",
+"\n",
+"//for GaAs\n",
+"\n",
+"Aga=Ag+Aa //molecular wt of GaAs\n",
+"\n",
+"pga=(nga*Aga)/(N*aga^3)\n",
+"\n",
+"printf(' 2) Density of GaAs=')\n",
+"\n",
+"disp(pga)\n",
+"\n",
+"printf('gm/cm^3')\n",
+"\n",
+"\n",
+"\n",
+""
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.3_3: calculate_density.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_3,pg 1-16\n",
+"\n",
+"A=63.5 //atomic weight of Cu\n",
+"\n",
+"N=6.023*10^23 //Avogadro's number\n",
+"\n",
+"n=4 //FCC structure\n",
+"\n",
+"r=1.28*10^-8 //atomic radius of Cu\n",
+"\n",
+"//for FCC\n",
+"\n",
+"a=4*r/(sqrt(2)) //lattice constant\n",
+"\n",
+"p=(n*A)/(N*a^3)\n",
+"\n",
+"printf('Density of Cu=')\n",
+"\n",
+"disp(p)\n",
+"\n",
+"printf('gm/cm^3')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.3_4: calculate_APF.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_4,pg 1-17\n",
+"\n",
+"A=50 //atomic weight of chromium\n",
+"\n",
+"N=6.023*10^23 //Avogadro's number\n",
+"\n",
+"p=5.96 //Density\n",
+"\n",
+"n=2 //BCC structure\n",
+"\n",
+"//step 1 : claculation for lattice constant (a)\n",
+"\n",
+"a=(n*A/(N*p))^(1/3)\n",
+"\n",
+"//step 2 : radius of an atom in BCC\n",
+"\n",
+"r=sqrt(3)*a/4\n",
+"\n",
+"//step 3 : Atomic packing factor (APF)\n",
+"\n",
+"APF=n*((4/3)*%pi*r^3)/a^3\n",
+"\n",
+"printf('Atomic packing factor (APF)=')\n",
+"\n",
+"disp(APF)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.3_5: calculate_no_of_unit_cell.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_5,pg 1-17\n",
+"\n",
+"A=120 //atomic weight of chromium\n",
+"\n",
+"N=6.023*10^23 //Avogadro's number\n",
+"\n",
+"p=5.2 //Density\n",
+"\n",
+"n=2 //BCC structure\n",
+"\n",
+"m=20 //mass\n",
+"\n",
+"//step 1 : claculation for volume of unit cell(a^3)\n",
+"\n",
+"a=(n*A/(N*p))\n",
+"\n",
+"//step 2 : volume of 20 gm of the element\n",
+"\n",
+"v=m/p\n",
+"\n",
+"//step 3 :no of unit cell\n",
+"\n",
+"x=v/a\n",
+"\n",
+"printf('no of unit cell=')\n",
+"\n",
+"disp(x)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.3_6: calculate_no_of_atoms_per_meter_cube.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_6,pg 1-18\n",
+"A=132.91 //atomic weight of chromium\n",
+"N=6.023*10^26 //Avogadro's number\n",
+"p=1900 //Density\n",
+"a=6.14*10^-10 //lattice constant\n",
+"//step 1 : type of structure\n",
+"n=(p*N*a^3)/A\n",
+"printf('n =')\n",
+"disp(round(n))\n",
+"printf('BCC structure')\n",
+"//step 2: no of atoms/m^3\n",
+"x=n/a^3\n",
+"printf(' no of atoms/m^3=')\n",
+"disp(x)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.3_7: calculate_no_of_unit_cell.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_3_6,pg 1-18\n",
+"\n",
+"a=0.4049*10^-9 //lattice constant\n",
+"\n",
+"t=0.006*10^-2 //thickness of Al foil\n",
+"\n",
+"A=50*10^-4 //Area of foil\n",
+"\n",
+"V1=a^3 //volume of unit cell\n",
+"\n",
+"V=A*t //volume of the foil\n",
+"\n",
+"N=V/V1 //no of unit cell in the foil\n",
+"\n",
+"printf('no of unit cell in the foil=')\n",
+"\n",
+"disp(N)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.5_1: calcukate_critical_radius_ratio_of_ligancy_three.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_5_1,pg 1-29\n",
+"\n",
+"//refer diagram from textbook\n",
+"\n",
+"//on joining centre of 3 anions,an equilateral triangle is formed and on joining centres of any anion and cation a right angle triangle ABC os formed\n",
+"\n",
+"//where AC=rc+ra\n",
+"\n",
+"//and BC=ra\n",
+"\n",
+"//m(angle (ACB))=30 degree\n",
+"\n",
+"//therefore cos (30)=ra/(rc+ra)\n",
+"\n",
+"//assume rc/ra=r\n",
+"\n",
+"r=(1-cosd(30))/cosd(30) //by arrangimg terms we get value of r\n",
+"\n",
+"printf('critical radius ratio of ligancy 3=')\n",
+"\n",
+"disp(r)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.5_2: calculate_critical_radius_ratio_for_ligancy_six.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_5_2,pg 1-30\n",
+"\n",
+"//refer diagram from textbook\n",
+"\n",
+"//in the said arrangement a cation is squeezed into 4 anions in a plane and 5th anion is in upper layer and 6th in bottom layer \n",
+"\n",
+"//join cation anion centres E and B and complete the triangle EBF\n",
+"\n",
+"//in triangle EBF m(angle F)=90 and EF=BF\n",
+"\n",
+"//m(angle B)=m(angle E)=45\n",
+"\n",
+"//and EB=rc+ra and BF=ra\n",
+"\n",
+"//cos(45)=ra/(rc+ra)\n",
+"\n",
+"//assume rc/ra=r\n",
+"\n",
+"r=(1-cosd(45))/cosd(45) //by arrangimg terms we get value of r\n",
+"\n",
+"printf('critical radius ratio for ligancy 6 =')\n",
+"\n",
+"disp(r)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.5_3: calculate_critical_radius_ratio_for_octohedral.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_5_3,pg 1-30\n",
+"\n",
+"//refer diagram from textbook\n",
+"\n",
+"//since plane is square hence it is same as ligancy 6\n",
+"\n",
+"//in the said arrangement a cation is squeezed into 4 anions in a plane and 5th anion is in upper layer and 6th in bottom layer \n",
+"\n",
+"//join cation anion centres E and B and complete the triangle EBF\n",
+"\n",
+"//in triangle EBF m(angle F)=90 and EF=BF\n",
+"\n",
+"//m(angle B)=m(angle E)=45\n",
+"\n",
+"//and EB=rc+ra and BF=ra\n",
+"\n",
+"//cos(45)=ra/(rc+ra)\n",
+"\n",
+"//assume rc/ra=r\n",
+"\n",
+"r=(1-cosd(45))/cosd(45) //by arrangimg terms we get value of r\n",
+"\n",
+"printf('critical radius ratio for ligancy 8 =')\n",
+"\n",
+"disp(r)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.5_4: calculate_critical_radius_ratio_for_ligancy_4.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_5_4,pg 1-31\n",
+"\n",
+"//a tetrahedron CAEH can be considered with C as the apex of the tetrahedron.\n",
+"\n",
+"//the edges AE,AH and EH of the tetrahedron will then be the face of the cube faces ABEF,ADHF,EFHG resp.\n",
+"\n",
+"//from fig\n",
+"\n",
+"//AO=ra+rc and AJ=ra\n",
+"\n",
+"//AE=root(2)*a and AG=root(3)*a\n",
+"\n",
+"//AO/AJ=AG/AE=(ra+rc)/ra=root(3)*a/root(2)*a\n",
+"\n",
+"//assume rc/ra=r\n",
+"\n",
+"r=(sqrt(3)-sqrt(2))/sqrt(2)\n",
+"\n",
+"printf('critical radius ratio for ligancy 4 =')\n",
+"\n",
+"disp(r)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.5_5: calculate_critical_radius_ratio_for_ligancy_8.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_5_5,pg 1-32\n",
+"\n",
+"//ligancy 8 represents cubic arrangment .8 anions are at the corners and touch along cube edgs.Along the body diagonal the central cation and the corner anion are in contact.\n",
+"\n",
+"//cube edge=2*ra\n",
+"\n",
+"//refer diagram from textbook\n",
+"\n",
+"//and body diagonal=root(3)*cube edge=root(3)[2*(rc+ra)]\n",
+"\n",
+"//assume rc/ra=r\n",
+"\n",
+"r=sqrt(3)-1\n",
+"\n",
+"printf('critical radius ratio of ligancy 8=')\n",
+"\n",
+"disp(r)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.5_6: calculate_critical_radius_ratio_for_ligancy.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_5_6,pg 1-32\n",
+"\n",
+"//for an ionic crystal exibiting HCP structure the arrangment of ions refere from textbook\n",
+"\n",
+"//at centre we have a cation with radius rc=OA\n",
+"\n",
+"//it is an touch with 6 anions with radius ra=AB\n",
+"\n",
+"//OB=OC=ra+rc\n",
+"\n",
+"//intrangle ODB ,m(angle (OBC))=60 degree ,m(angle (ODB))=90 degree\n",
+"\n",
+"//therefore cos(60)=BD/OB=AB/(OA+OB)=ra/(rc+ra)\n",
+"\n",
+"//assume rc/ra=r\n",
+"\n",
+"r=(1-cosd(60))/cosd(60) //by arrangimg terms we get value of r\n",
+"\n",
+"printf('critical radius ratio 0f HCP structure=')\n",
+"\n",
+"disp(r)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.6_2: calculate_miller_indices_of_plane.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_6_2,pg 1-35\n",
+"\n",
+"//intercept of planeare in proportion a,b/3,2*c\n",
+"\n",
+"//as a,b and c are basic vectors the proportin of intercepts 1:1/3:2\n",
+"\n",
+"//therefore reciprocal\n",
+"\n",
+"r1=1\n",
+"\n",
+"r2=3\n",
+"\n",
+"r3=1/2\n",
+"\n",
+"//taking LCM\n",
+"\n",
+"v=int32([2,1])\n",
+"\n",
+"l=double(lcm(v))\n",
+"\n",
+"m1=(l*r1)\n",
+"\n",
+"m2=(l*r2)\n",
+"\n",
+"m3=(l*r3)\n",
+"\n",
+"printf('miler indices=')\n",
+"\n",
+"disp(m3,m2,m1)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1.6_4: calculate_interplanar_spacing.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"//Chapter-1,Example1_6_4,pg 1-38\n",
+"\n",
+"r=1.414 //atomic radius in amstrong unit\n",
+"\n",
+"//for FCC structure\n",
+"\n",
+"a=4*r/sqrt(2)\n",
+"\n",
+"//part 1: plane(2,0,0)\n",
+"\n",
+"//the interplanar spacing of plane\n",
+"\n",
+"h1=2\n",
+"\n",
+"k1=0\n",
+"\n",
+"l1=0\n",
+"\n",
+"//we know that d=a/sqrt(h^2+k^2+l^2)\n",
+"\n",
+"d1=a/sqrt(h1^2+k1^2+l1^2)\n",
+"\n",
+"printf(' 1)interplanar spacing for (2,0,0) plane=')\n",
+"\n",
+"disp(d1)\n",
+"\n",
+"printf('amstrong')\n",
+"\n",
+"//part 2: plane(1,1,1)\n",
+"\n",
+"//the interplanar spacing of plane\n",
+"\n",
+"h2=1\n",
+"\n",
+"k2=1\n",
+"\n",
+"l2=1\n",
+"\n",
+"//we know that d=a/sqrt(h^2+k^2+l^2)\n",
+"\n",
+"d2=a/sqrt(h2^2+k2^2+l2^2)\n",
+"\n",
+"printf(' 2)interplanar spacing for(1,1,1) plane=')\n",
+"\n",
+"disp(d2)\n",
+"\n",
+"printf('amstrong')"
+ ]
+ }
+],
+"metadata": {
+ "kernelspec": {
+ "display_name": "Scilab",
+ "language": "scilab",
+ "name": "scilab"
+ },
+ "language_info": {
+ "file_extension": ".sce",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "scilab",
+ "version": "0.7.1"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
generated by cgit (git 2.34.1) at 2025-07-05 01:15:30 +0000