{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 4: Quantum Physics"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.10: Finding_Energy_level_and_Temperature_of_molecules.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 10\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"l       = 0.1*10^-9;        // side of cubical box\n",
"h       = 6.625*10^-34       // plancks constant in Jsec\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"Kb      = 1.38*10^-23       // Boltzmann constant \n",
"\n",
"// Calculations\n",
"// for cubical box the energy eigen value is Enx ny nz = (h^2/(8*m*l^2))*(nx^2 + ny^2 +nz^2)\n",
"// For the next energy level to the lowest energy level nx = 1 , ny = 1 and nz = 2\n",
"nx      = 1\n",
"ny      = 1\n",
"nz      = 2\n",
"E112    = (h^2/(8*m*l^2))*( nx^2 + ny^2 + nz^2);\n",
"\n",
"// we know the average energy of molecules of aperfect gas = (3/2)*(Kb*T)\n",
"T       = (2*E112)/(3*Kb);      // Temperature in kelvin\n",
"\n",
"// Output\n",
"mprintf('E112 = %3.4e Joules\n Temperature of the molecules T = %3.4e K',E112,T);\n",
"//==============================================================================\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.11: Finding_Minimum_energy_of_an_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 11\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"l       = 4*10^-9;           // width of infinitely deep potential\n",
"h       = 6.625*10^-34       // plancks constant in Jsec\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"n       = 1;                 // minimum energy\n",
"e       = 1.6*10^-19         // charge of electron in columbs\n",
"\n",
"// Calculations\n",
"E       = (h^2 * n^2)/(8*m*l^2)    // Energy of electron in an infinitely deep potential well \n",
"E1      = E/e                      // energy conversion from joules to eV\n",
"\n",
"// Output\n",
"mprintf('Minimum energy of an electron = %3.4f eV',E1);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.12: Finding_Energy_for_Exciting_a_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 12\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"l       = 0.1*10^-9;         // length of one dimensional box \n",
"h       = 6.625*10^-34       // plancks constant in Jsec\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"n       = 1;                 // for ground state\n",
"n5      = 6;                 // n value for fifth excited state\n",
"e       = 1.6*10^-19         // charge of electron in columbs\n",
"\n",
"// Calculations\n",
"Eg      = (h^2 * n^2)/(8*m*l^2 *e ) // Energy in ground state in eV \n",
"Ee      = (h^2 * n5^2)/(8*m*l^2 * e) // Energy in excited state  in eV\n",
"E       = Ee - Eg;                   // energy req to excite electrons from ground state to fift excited state\n",
"\n",
"// Output\n",
"mprintf('Energy required to excite an electron from ground state to fifth excited state = %3.2f eV',E);\n",
"//==============================================================================\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.13: Finding_Energy_of_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 13\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"l       = 0.1*10^-9;         // length of one dimensional box \n",
"h       = 6.625*10^-34       // plancks constant in Jsec\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"n       = 1;                 // for ground state\n",
"e       = 1.6*10^-19         // charge of electron in columbs\n",
"\n",
"// Calculations\n",
"E       = (h^2 * n^2)/(8*m*l^2 *e ) // Energy of electron in eV \n",
"// Output\n",
"mprintf('Energy of an electron = %3.3f eV',E);\n",
"//==============================================================================\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.14: Finding_Least_energy_of_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 14\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"l       = 0.5*10^-9;         // width of one dimensional box in m \n",
"h       = 6.625*10^-34       // plancks constant in Jsec\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"n       = 1;                 // for ground state\n",
"e       = 1.6*10^-19         // charge of electron in columbs\n",
"\n",
"// Calculations\n",
"E       = (h^2 * n^2)/(8*m*l^2 *e ) // Energy of electron in eV \n",
"// Output\n",
"mprintf('Least Energy of an electron = %3.4f eV',E);\n",
"//==============================================================================\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.1: Finding_Wavelength_of_the_Scattered_photons.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 1\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"\n",
"lamda   = 3*10^-10;         // wavelength of incident photons in m\n",
"theta   = 60;               // viewing angle in degrees\n",
"h       = 6.625*10^-34      // plancks constant\n",
"mo      = 9.11*10^-31      // mass in Kg\n",
"c       = 3*10^8;           // vel. of light \n",
"\n",
"// Calculatioms\n",
"// from Compton theory ,Compton shift is given by\n",
"// lamda' - lamda = (h/(mo*c))*(1-cosθ)\n",
"\n",
"theta_r = theta*%pi/180;     // degree to radian conversion\n",
"lamda1  = lamda+( (h/(mo*c))*(1-cos(theta_r))) // wavelength of scattered photons\n",
"\n",
"// Output\n",
"mprintf('Wavelength of Scattered photons = %3.4f Å',lamda1*10^10);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.2: Finding_Change_in_Wavelength.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 2\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"theta   = 135;               // angle in degrees\n",
"h       = 6.625*10^-34      // plancks constant\n",
"mo      = 9.1*10^-31       // mass in Kg\n",
"c       = 3*10^8;          // vel. of light in m/s\n",
"\n",
"// Calculatioms\n",
"// from Compton theory ,Compton shift is given by\n",
"// lamda' - lamda = (h/(mo*c))*(1-cosθ)\n",
"theta_r = theta*%pi/180;     // degree to radian conversion\n",
"c_lamda = ( (h/(mo*c))*(1-cos(theta_r))) // Change in wavelength in m\n",
"\n",
"// Output\n",
"mprintf('Change in Wavelength  = %3.5f Å',c_lamda*10^10);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.3: Finding_wavelength_of_Scattered_beam.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 3\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"\n",
"lamda   = 0.1*10^-9;         // wavelength of X-rays in m\n",
"theta   = 90;               // angle with incident beam in degrees\n",
"h       = 6.625*10^-34      // plancks constant\n",
"mo      = 9.11*10^-31      // mass in Kg\n",
"c       = 3*10^8;           // vel. of light \n",
"\n",
"// Calculatioms\n",
"// from Compton theory ,Compton shift is given by\n",
"// lamda' - lamda = (h/(mo*c))*(1-cosθ)\n",
"theta_r = theta*%pi/180;     // degree to radian conversion\n",
"lamda1  = lamda+( (h/(mo*c))*(1-cos(theta_r))) // wavelength of scattered beam\n",
"\n",
"// Output\n",
"mprintf('Wavelength of Scattered beam = %3.4f Å',lamda1*10^10);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.4: Finding_De_Broglie_wavelength.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 4\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"h       = 6.625*10^-34       // plancks constant\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"e       = 1.6*10^-19         // charge of electron\n",
"V       = 150;               // potential difference in volts\n",
"\n",
"// Calculations\n",
"\n",
"lamda   = h/(sqrt(2*m*e*V))  // de Broglie wavelength\n",
"\n",
"// Output\n",
"mprintf('The de-Broglie wavelength = %d Å',lamda*10^10);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.5: Finding_de_Broglie_wavelength.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 5\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"h       = 6.625*10^-34       // plancks constant\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"e       = 1.6*10^-19         // charge of electron\n",
"V       = 5000;               // potential in volts\n",
"\n",
"// Calculations\n",
"\n",
"lamda   = h/(sqrt(2*m*e*V))  // de Broglie wavelength\n",
"\n",
"// Output\n",
"mprintf('The de-Broglie wavelength of electron = %3.5f Å',lamda*10^10);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.6: Finding_de_Broglie_wavelength.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 6\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"E       = 100                // Energy of electron in eV\n",
"h       = 6.625*10^-34       // plancks constant\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"e       = 1.6*10^-19         // Charge of electron in Columbs\n",
"\n",
"// Calculations\n",
"\n",
"E1      = E*e               // Energy conversion from eV to Joule\n",
"lamda   = h/(sqrt(2*m*E1))  // de Broglie wavelength\n",
"\n",
"// Output\n",
"mprintf('The de-Broglie wavelength  = %3.3f Å',lamda*10^10);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.7: Finding_de_Broglie_Wavelength.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 7\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"m       = 1.675*10^-27;     // Mass of proton in kg\n",
"c       = 3*10^8;           // velocity of light in m/s\n",
"h       = 6.625*10^-34      // plancks constant\n",
"\n",
"// Calculations\n",
"\n",
"vp      = c/20;             // velocity of proton in m/s\n",
"lamda   = h/(m*vp)          // de-Broglie wavelength in m\n",
"\n",
"// Output\n",
"mprintf('de-Broglie wavelength = %e m',lamda);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.8: Finding_de_Broglie_Wavelength_of_neutron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Example 8\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"E       = 10000              // Energy of neutron in eV\n",
"h       = 6.625*10^-34       // plancks constant\n",
"m       = 1.675*10^-27       // mass of neutron in Kg\n",
"e       = 1.6*10^-19         \n",
"// Calculations\n",
"\n",
"E1      = E*e               // Energy conversion from eV to Joule\n",
"lamda   = h/(sqrt(2*m*E1))  // de Broglie wavelength\n",
"\n",
"// Output\n",
"mprintf('The de-Broglie wavelength of neutron = %3.3e m',lamda);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_10: Finding_Energy_of_system_having_two_electrons.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Additional Example 10\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"l       = 10^-10   ;         // length of one dimensional box in m \n",
"h       = 6.625*10^-34       // plancks constant in Jsec\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"n       = 1;                 // for ground state\n",
"e       = 1.6*10^-19         // charge of electron in columbs\n",
"\n",
"// Calculations\n",
"E       = 2*(h^2 * n^2)/(8*m*l^2 *e ) // Energy of system having two electrons\n",
"// Output\n",
"mprintf('Energy of the system having two electrons = %3.4f eV',E);\n",
"//==============================================================================\n",
"\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_11: Finding_Magnifying_power.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Additional Example 10\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"b       = 40;       // angle subtended by final images at eye in degrees\n",
"a       = 10        // angle subtended by the object at the eye kept at near point in degrees\n",
"\n",
"// Calculations\n",
"b_r     = b*%pi/180;    // degree to radian conversion\n",
"a_r     = a*%pi/180;    // degree to radian conversion\n",
"M       = tan(b_r)/tan(a_r);    // magnifying power\n",
"\n",
"// Output\n",
"mprintf('Magnifying power = %3.3f',M);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_1: Finding_no_of_photons_emitted.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Addutional Example 1\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"h       = 6.625*10^-34      // plancks constant\n",
"c       = 3*10^8;           // vel. of light\n",
"lamda   = 5893*10^-10;      // wavelength in m\n",
"P       = 100               // power of sodium vapour lamp\n",
"\n",
"// Calculations\n",
"E       = (h*c)/lamda;      // Energy in joules\n",
"N       =  P/E              // Number of photons emitted\n",
"\n",
"// Output\n",
"mprintf('Number of Photons emitted = %3.4e per second',N);\n",
"//==============================================================================\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_2: Finding_wavelength_of_incident_beam.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 AdditionalExample 2\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"\n",
"lamda1  = 0.022*10^-10;      // wavelength of scatterd X-rays in m\n",
"theta   = 45;               // scatterring angle in degrees\n",
"h       = 6.625*10^-34      // plancks constant\n",
"mo      = 9.11*10^-31      // mass in Kg\n",
"c       = 3*10^8;           // vel. of light \n",
"\n",
"// Calculatioms\n",
"// from Compton theory ,Compton shift is given by\n",
"// lamda' - lamda = (h/(mo*c))*(1-cosθ)\n",
"\n",
"theta_r = theta*%pi/180;     // degree to radian conversion\n",
"lamda  = lamda1-( (h/(mo*c))*(1-cos(theta_r))) // incident Wavelength\n",
"\n",
"// Output\n",
"mprintf('Wavelength of incident beam = %3.4f Å',lamda*10^10);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_3: Finding_Energy_of_scattered_photon.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Additional Example 3\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"Ei      = 1.02*10^6         // photon energy in eV\n",
"theta   = 90;               // scattered angle in degrees\n",
"h       = 6.625*10^-34      // plancks constant\n",
"mo      = 9.1*10^-31       // mass of electron in Kg\n",
"e       = 1.6*10^-19       // charge of electron\n",
"c       = 3*10^8;          // vel. of light in m/s\n",
"\n",
"// Calculatioms\n",
"// from Compton theory ,Compton shift is given by\n",
"// lamda' - lamda = (h/(mo*c))*(1-cosθ)\n",
"theta_r = theta*%pi/180;     // degree to radian conversion\n",
"c_lamda = ( (h/(mo*c))*(1-cos(theta_r))) // Change in wavelength in m\n",
"dv      = c/c_lamda;         // change in frequency of the scattered photon\n",
"dE      = (h*dv)/e           // change in energy of scattered photon in eV\n",
"// This change in energy is transferred as the KE of the recoil electron\n",
"Er      = dE;                // Energy of recoil electron\n",
"Es      = Ei - Er            // Energy of scattered photon\n",
"\n",
"\n",
"// Output\n",
"mprintf('Energy of the recoil electron = %3.4f MeV\n Energy of the Scattered photon = %3.4f MeV',Er*10^-6,Es*10^-6);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_4: Finding_wavelength_of_Scattered_X_rays.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Additional Example 4\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"\n",
"lamda   = 0.124*10^-10;     // wavelength of X-rays in  m\n",
"theta   = 180;               // Scattering angle in degrees\n",
"h       = 6.625*10^-34      // plancks constant\n",
"mo      = 9.11*10^-31      // mass in Kg\n",
"c       = 3*10^8;           // vel. of light \n",
"\n",
"// Calculatioms\n",
"// from Compton theory ,Compton shift is given by\n",
"// lamda' - lamda = (h/(mo*c))*(1-cosθ)\n",
"\n",
"theta_r = theta*%pi/180;     // degree to radian conversion\n",
"lamda1  = lamda+( (h/(mo*c))*(1-cos(theta_r))) // wavelength of scattered X-rays\n",
"\n",
"// Output\n",
"mprintf('Wavelength of Scattered X-rays = %3.4f Å',lamda1*10^10);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_5: Finding_de_Broglie_wavelength.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
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	   "outputs": [],
"source": [
"// Chapter 4 Additional Example 5\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"h       = 6.625*10^-34       // plancks constant\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"e       = 1.6*10^-19         // charge of electron\n",
"V       = 2000;               // potential in volts\n",
"\n",
"// Calculations\n",
"\n",
"lamda   = h/(sqrt(2*m*e*V))  // de Broglie wavelength\n",
"\n",
"// Output\n",
"mprintf('The de-Broglie wavelength of electron = %3.4f Å',lamda*10^10);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_6: Finding_de_Broglie_Wavelength.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Additional Example 6\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"h       = 6.625*10^-34       // plancks constant\n",
"m       = 1.678*10^-27       // mass of proton in Kg\n",
"e       = 1.6*10^-19         // charge of electron\n",
"Kb      = 1.38*10^-23;       // boltzmann constant\n",
"T       = 300                // Temperature in kelvin\n",
"// Calculations\n",
"\n",
"lamda   = h/(sqrt(3*m*Kb*T))  // de Broglie wavelength\n",
"\n",
"// Output\n",
"mprintf('The de-Broglie wavelength = %3.4f Å',lamda*10^10);\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_7: Finding_Energy_of_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Additional Example 7\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"// input data\n",
"h       = 6.625*10^-34       // plancks constant\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"lamda   = 3*10^-2;             // wavelength of electron wave\n",
"e       = 1.6*10^-19;          // charge of electron\n",
"// Calculations\n",
"\n",
"E       = (h^2)/(2*m*lamda^2);  // Energy in Joules\n",
"E1      = E/e;\n",
"// Output\n",
"mprintf('Energy of the electron E = %3.4e eV\n',E1);\n",
"mprintf(' Note: Calculation mistake in textbook')\n",
"//=============================================================================="
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_8: Proving_de_Broglie_is_equal_to_compton_wavelength.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Additional Example 8\n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"// input data\n",
"h       = 6.625*10^-34       // plancks constant\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"c       = 3*10^8;            // velocity of light in m/s\n",
"\n",
"// Calculations\n",
"ve      = 0.7071*c           // velocity of electron\n",
"lamda   = h/(m*ve*sqrt(1-(ve/c)^2))    // de Broglie wavelength\n",
"\n",
"// we know Compton wavelength  ,lamda' - lamda = (h/(mo*c))*(1-cosθ)\n",
"// maximum shift θ = 180\n",
"theta    = 180\n",
"theta1   = theta*%pi/180;\n",
"d_lamda  = (h/(m*c))*(1-cos(theta1))\n",
"mprintf('de Broglie wavelength = %e m\n',lamda);\n",
"mprintf(' compton wavelength    = %e m\n',d_lamda)\n",
"mprintf(' The de-Broglie wacelength is equal to the compton wavelength');\n",
"//==============================================================================\n",
" "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 4.a_9: Finding_Eigen_Values.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Chapter 4 Additional Example 9 \n",
"//==============================================================================\n",
"clc;\n",
"clear;\n",
"\n",
"// input data\n",
"l       = 10^-10;            // side of one dimensional box \n",
"h       = 6.625*10^-34       // plancks constant in Jsec\n",
"m       = 9.11*10^-31        // mass of electron in Kg\n",
"n1      = 1;                 // for 1st eigen value\n",
"n2      = 2;                 // for 2nd eigen value\n",
"n3      = 3;                 // for 3rd eigen value\n",
"n4      = 4;                 // for 4th eigen value\n",
"e       = 1.6*10^-19         // charge of electron in columbs\n",
"\n",
"// Calculations\n",
"E1      = (h^2 * n1^2)/(8*m*l^2 *e ) // first Eigen value\n",
"E2      = (h^2 * n2^2)/(8*m*l^2 *e ) // second Eigen value\n",
"E3      = (h^2 * n3^2)/(8*m*l^2 *e ) // third Eigen value\n",
"E4      = (h^2 * n4^2)/(8*m*l^2 *e ) // fourth Eigen value\n",
" \n",
"// Output\n",
"mprintf('1st Eigen value  = %3.1f eV\n 2nd Eigen value  = %3.1f eV\n 3rd Eigen value  = %3.1f eV\n 4th Eigen value  = %3.1f eV\n',E1,E2,E3,E4);\n",
"//==============================================================================\n",
"\n",
""
   ]
   }
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