path: root/Engineering_Physics_by_V_Yadav/1-Quantum_Mechanics.ipynb

{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 1: Quantum Mechanics"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.10: de_Broglie_wavelength_associated_with_moving_proton.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.10:  Page-1.8 (2009)\n",
"clc; clear;\n",
"m = 1.67e-027;      // Mass of the proton, kg\n",
"c = 3e+08;      // Speed of light, m/s\n",
"v = 1/20*c;        // Velocity of the proton, m/s\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"lambda = h/(m*v);  // de Broglie wavelength of the neutron, m\n",
"printf('\nThe de Broglie wavelength associated with moving proton = %5.3e m', lambda);\n",
"\n",
"// Result \n",
"// The de Broglie wavelength associated with moving proton = 2.645e-14 m\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.11: Wavelength_of_matter_wave_associated_with_moving_proton.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.11: Page-1.8 (2009)\n",
"clc; clear;\n",
"m = 1.67e-027;      // Mass of the proton, kg\n",
"v = 2e+08;        // Velocity of the proton, m/s\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"lambda = h/(m*v);  // de Broglie wavelength of the neutron, m\n",
"printf('\nThe wavelength of matter wave associated with moving proton = %5.3e m', lambda);\n",
"\n",
"// Result \n",
"// The wavelength of matter wave associated with moving proton = 1.984e-15 m \n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.12: de_Broglie_wavelength_of_an_electron_accelerated_through_a_given_potential.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.12: Page-1.17 (2009)\n",
"clc; clear;\n",
"m = 9.1e-031;      // Mass of the electron, kg\n",
"q = 1.6e-019;       // Charge on an electron, C\n",
"V = 50;     // Accelearting potential, V\n",
"E = q*V;    // Energy gained by the electron, J\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"lambda = h/sqrt(2*m*E);  // de Broglie wavelength of the electron, m\n",
"printf('\nThe de Broglie wavelength of the electron accelearted through a given potential = %5.3e m', lambda);\n",
"\n",
"// Result \n",
"// The de Broglie wavelength of the electron accelearted through a given potential = 1.736e-10 m \n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.13: Interplanar_spacing_of_the_crystal.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.13: Page-1.17 (2009)\n",
"clc; clear;\n",
"theta = 45;     // Diffraction angle, degrees\n",
"h = 6.626e-034;     // Planck's constant\n",
"m = 1.67e-027;      // Mass of a neutron, kg\n",
"n = 1;      // Order of diffraction\n",
"k = 1.38e-023;  // Boltzmann constant, J/mol/K\n",
"T = 27+273;     // Absolute room temperature, K\n",
"E = 3/2*k*T;    // Energy of the neutron, J\n",
"lambda = h/sqrt(2*m*E); // de-Broglie wavelength of neutrons, m\n",
"// From Bragg's law, 2*d*sin(theta) = n*lambda, solving for d\n",
"d = n*lambda/(2*sind(theta));\n",
"printf('\nThe interplanar spacing of the crystal = %4.2f angstrom', d/1e-010);\n",
"\n",
"// Result \n",
"// The interplanar spacing of the crystal = 1.03 angstrom "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.14: Interplanar_spacing_using_Bragg_law.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.14: Page-1.18 (2009)\n",
"clc; clear;\n",
"theta = 70;     // Glancing angle at which reflection occurs, degrees\n",
"h = 6.626e-034;     // Planck's constant\n",
"m = 9.1e-031;      // Mass of a electron, kg\n",
"e = 1.6e-019;       // Electronic charge, C\n",
"V = 1000;   // Accelerating potential, V\n",
"n = 1;      // Order of diffraction\n",
"E = e*V;    // Energy of the electron, J\n",
"lambda = h/sqrt(2*m*E); // de-Broglie wavelength of electron, m\n",
"// From Bragg's law, 2*d*sin(theta) = n*lambda, solving for d\n",
"d = n*lambda/(2*sind(theta));   // Interplanar spacing, m\n",
"printf('\nThe interplanar spacing of the crystal = %6.4e m', d);\n",
"\n",
"// Result \n",
"// The interplanar spacing of the crystal = 2.0660e-11 m \n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.15: de_Broglie_wavelength_of_electron_accelerated_at_V_volts.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.15: Page-1.18 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;     // Planck's constant\n",
"m = 9.1e-031;      // Mass of a electron, kg\n",
"e = 1.6e-019;       // Electronic charge, C\n",
"V = 1;   // For simplicity the accelerating potential is assumed to be unity, V\n",
"E = e*V;    // Energy of the electron, J\n",
"lambda = h/sqrt(2*m*E); // de-Broglie wavelength of electron, m\n",
"printf('\nde-Broglie wavelength of electron accelerated at V volts = %5.2f/sqrt(V) angstrom', lambda/1e-010);\n",
"\n",
"// Result \n",
"// de-Broglie wavelength of electron accelerated at V volts = 12.23/sqrt(V) angstrom "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.16: de_Broglie_wavelength_of_electron_accelerated_from_rest.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.16: Page-1.18 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;     // Planck's constant\n",
"m = 9.1e-031;      // Mass of a electron, kg\n",
"e = 1.6e-019;       // Electronic charge, C\n",
"V = 100;   // Accelerating potential for electron, V\n",
"E = e*V;    // Energy of the electron, J\n",
"lambda = h/sqrt(2*m*E); // de-Broglie wavelength of electron, m\n",
"printf('\nde-Broglie wavelength of electron accelerated at %d volts = %6.4e m', V, lambda);\n",
"\n",
"// Result \n",
"// de-Broglie wavelength of electron accelerated at 100 volts = 1.2231e-10 m "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.17: The_wavelength_associated_with_moving_mass.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.17:  Page-1.19 (2009)\n",
"clc; clear;\n",
"m = 10e-03;     // Mass of the body, kg\n",
"v = 110;    // Velocity of the mass, m/s\n",
"h = 6.6e-034;     // Planck's constant\n",
"lambda = h/(m*v); // de-Broglie wavelength of electron, m\n",
"printf('\nThe wavelength associated with mass moving with velocity %d m/s = %1.0e m', v, lambda);\n",
"\n",
"// Result \n",
"// The wavelength associated with mass moving with velocity 110 m/s = 6e-34 m "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.18: Wavelength_of_an_electron_from_its_kinetic_energy.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.18:  Page-1.19 (2009)\n",
"clc; clear;\n",
"m = 9.1e-031;     // Mass of the electron, kg\n",
"Ek = 1.27e-017;     // Kinetic energy of electron, J\n",
"h = 6.6e-034;     // Planck's constant\n",
"lambda = h/sqrt(2*m*Ek); // de-Broglie wavelength of electron, m\n",
"printf('\nThe wavelength associated with moving electron = %4.2f angstrom', lambda/1e-010);\n",
"\n",
"// Result \n",
"// The wavelength associated with moving electron = 1.37 angstrom "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.19: Kinetic_energy_of_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.19: Page-1.19 (2009)\n",
"clc; clear;\n",
"m = 9.1e-031;     // Mass of the electron, kg\n",
"h = 6.6e-034;     // Planck's constant\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"lambda = 9.1e-012; // de-Broglie wavelength of electron, m\n",
"// We have lambda = h/(m*v), solving for v\n",
"v = h/(m*lambda);   // Velocity of the electron, m/s\n",
"K = 1/2*m*v^2;     // Kinetic energy of electron, J\n",
"printf('\nThe kinetic energy of electron having wavelength %3.1e m = %4.2e eV', lambda, K/e);\n",
"\n",
"// Result \n",
"// The kinetic energy of electron having wavelength 9.1e-12 m = 1.81e+04 eV \n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.1: Energy_of_the_particle_from_de_Broglie_wavelength.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.1:Page-1.5 (2009)\n",
"clc; clear;\n",
"lambda = 2.1e-010;  // de Broglie wavelength of the particle, m\n",
"m = 1.67e-027;      // Mass of the particle, kg\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"// From de Broglie relation, lambda = h/sqrt(2*m*E), solving for E\n",
"E = h^2/(2*m*lambda^2*e);     // Energy of the particle, eV\n",
"printf('\nThe energy of the particle from de Broglie wavelength = %5.3e eV', E);\n",
"\n",
"// Result \n",
"// The energy of the particle from de Broglie wavelength = 1.863e-002 eV "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.20: Speed_of_proton_for_an_equivalent_wavelength_of_that_of_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.20: : Page-1.19 (2009)\n",
"clc; clear;\n",
"m_e = 9.1e-031;     // Mass of the electron, kg\n",
"m_p = 1.67e-027;    // Mass of the proton, kg\n",
"v_e = 1;    // For simplicity assume velocity of electron to be unity, m/s\n",
"// From de-Broglie relation, \n",
"// lambda_p = lambda_e = h(m*v_p), solving for v_p\n",
"v_p = m_e*v_e/m_p;  // Velocity of the proton, m/s\n",
"// As lambda_e = h/sqrt(2*m_e*K_e) and lambda_p = h/sqrt(2*m_p*K_p), solving for K_e/K_p\n",
"K_ratio = m_p/m_e;      // Ratio of kinetic energies of electron and proton\n",
"\n",
"printf('\nThe speed of proton for an equivalent wavelength of that of electron = %3.1e ve', v_p);\n",
"printf('\nRatio of kinetic energies of electron and proton = %3.1e, therefore Ke > Kp', K_ratio);\n",
"\n",
"// Result \n",
"// The speed of proton for an equivalent wavelength of that of electron = 5.4e-04 ve\n",
"// Ratio of kinetic energies of electron and proton = 1.8e+03, therefore Ke > Kp "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.21: de_Broglie_wavelength_of_the_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.21: Page-1.20 (2009)\n",
"clc; clear;\n",
"V = 50; // Potential difference, V\n",
"m = 9.1e-031;     // Mass of the electron, kg\n",
"e = 1.6e-019;     // Electronic charge, C\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"lambda = h/sqrt(2*m*e*V);    // From de-Broglie relation,\n",
"printf('\nde-Broglie wavelength of the electron = %4.2f angstrom', lambda/1e-010);\n",
"\n",
"// Result \n",
"// de-Broglie wavelength of the electron = 1.73 angstrom"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.23: Minimum_accuracy_to_locate_the_position_of_an_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.23:: Page-1.31 (2009)\n",
"clc; clear;\n",
"v = 740;    // Speed of the electron, m/s\n",
"m = 9.1e-031;     // Mass of the electron, kg\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"p = m*v;    // Momentum of the electron, kg-m/s\n",
"frac_v = 0.05/100;  // Correctness in the speed \n",
"delta_p = p*frac_v;    // Uncertainty in momentum, kg-m/s\n",
"delta_x = h/(4*%pi)*1/delta_p;  // Uncertainty in position, m\n",
"\n",
"printf('\nThe minimum accuracy to locate the position of an electron = %4.2e m',delta_x);\n",
"\n",
"// Result \n",
"// The minimum accuracy to locate the position of an electron = 1.56e-04 m  "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.24: Uncertainty_in_energy_of_an_emitted_photon.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.24: : Page-1.31 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"h_cross = h/(2*%pi);    // Reduced Planck's constant, Js\n",
"delta_t = 1e-010;   // Uncertainty in time, s\n",
"// From Energy-time uncertainty, \n",
"// delta_E*delta_t = h_cross/2, solving for delta_E\n",
"delta_E = h_cross/(2*delta_t);  // Uncertainty in energy of an emitted photon, J \n",
"\n",
"printf('\nThe uncertainty in energy of an emitted photon = %5.3e eV', delta_E/1.6e-019);\n",
"\n",
"// Result \n",
"// The uncertainty in energy of an emitted photon = 3.283e-06 eV"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.25: Minimum_uncertainty_in_velocity_of_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.25: : Page-1.31 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"delta_x_max = 1e-007;   // Uncertainty in length, m\n",
"m = 9.1e-031;   // Mass of an electron, kg\n",
"// From Position-momentum uncertainty, \n",
"// delta_p_min = m*delta_v_min = h/delta_x_max, solving for delta_v_min\n",
"delta_v_min = h/(delta_x_max*m);    // Minimum uncertainty in velocity of electron, m/s\n",
"\n",
"printf('\nThe minimum uncertainty in velocity of electron = %4.2e m/s', delta_v_min);\n",
"\n",
"// Result \n",
"// The minimum uncertainty in velocity of electron = 7.25e+03 m/s "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.26: Minimum_uncertainty_in_momentum_and_minimum_kinetic_energy_of_proton.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.26:  Page-1.32 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"delta_x_max = 8.5e-014;   // Uncertainty in length, m\n",
"m = 1.67e-027;  // Mass of proton, kg\n",
"// From Position-momentum uncertainty, \n",
"// delta_p_min*delta_x_max = h, solving for delta_p_min\n",
"delta_p_min = h/delta_x_max;    // Minimum uncertainty in momentum of electron, kg-m/s\n",
"p_min = delta_p_min;    // Minimum momentum of the proton, kg.m/s\n",
"delta_E = p_min^2/(2*m); \n",
"   \n",
"printf('\nThe minimum uncertainty in momemtum of proton = %4.2e kg-m/s', p_min);\n",
"printf('\nThe kinetic energy of proton = %6.3e eV', delta_E/1.6e-019);\n",
"\n",
"// Result \n",
"// The minimum uncertainty in momemtum of proton = 7.76e-21 kg-m/s\n",
"// The kinetic energy of proton = 1.128e+05 eV "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.27: Uncertainty_in_momentum_of_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.27:: Page-1.32 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"E = 0.15*1e+03*e;   // Energy of the electron, J\n",
"m = 9.1e-031;  // Mass of electron, kg\n",
"delta_x = 0.5e-010; // Position uncertainty of electron, m\n",
"p = (2*m*E)^(1/2);  // Momentum of the electron, kg-m/s\n",
"// delta_x*delta_p = h/(4*%pi), solving for delta_p\n",
"delta_p = h/(4*%pi*delta_x);    // Uncertainty in momentum of electron, kg-m/s\n",
"frac_p = delta_p/p*100;     // Percentage uncertainty in momentum of electron, kg-m/s\n",
"\n",
"printf('\nThe percentage uncertainty in momentum of electron = %2d percent', frac_p);\n",
"\n",
"// Result \n",
"// The percentage uncertainty in momentum of electron = 15 percent   "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.28: Uncertainty_in_position_of_the_particle.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.28:: Page-1.33 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"delta_v = 7.54e-015;    // Uncertainty in velocity of the particle, m/s\n",
"m = 0.25e-06;  // Mass of particle, kg\n",
"// delta_x*delta_p = h/(4*%pi), solving for delta_x\n",
"delta_x = h/(4*%pi*m*delta_v); // Position uncertainty of particle, m\n",
"\n",
"printf('\nThe position uncertainty of particle = %4.2e m', delta_x);\n",
"\n",
"// Result \n",
"// The position uncertainty of particle = 2.79e-14 m   "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.29: Uncertainty_in_position_of_the_moving_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.29:: Page-1.33 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"v = 450;        // Velocity of the electron, m/s\n",
"delta_v = v*0.05/100;    // Uncertainty in velocity of the particle, m/s\n",
"m = 9.1e-031;  // Mass of electron, kg\n",
"// delta_x*delta_p = h/(4*%pi), solving for delta_x\n",
"delta_x = h/(4*%pi*m*delta_v); // Position uncertainty of particle, m\n",
"\n",
"printf('\nThe position uncertainty of moving electron = %4.2e m', delta_x);\n",
"\n",
"// Result \n",
"// The position uncertainty of moving electron = 2.57e-04 m "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.2: de_Broglie_wavelength_of_the_particle.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.2: Page-1.5 (2009)\n",
"clc; clear;\n",
"m = 1.67e-027;      // Mass of the particle, kg\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"E = 1e+011*e;     // Energy of the particle, J\n",
"lambda = h/sqrt(2*m*E);     // de Broglie wavelength of the particle, m\n",
"printf('\nThe de Broglie wavelength of the particle = %4.2e m', lambda);\n",
"\n",
"// Result \n",
"// The de Broglie wavelength of the particle = 9.06e-017 m "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.30: Smallest_possible_uncertainty_in_position_of_the_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.30:: Page-1.33 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"c = 3e+08;      // Speed of light, m/s\n",
"v = 3e+07;        // Velocity of the electron, m/s\n",
"m0 = 9.1e-031;  // Rest mass of electron, kg\n",
"m = m0/sqrt(1-v^2/c^2); // Mass of moving electron, kg\n",
"delta_p_max = m*v;    // Maximum uncertainty in momentum of the particle, m/s\n",
"// delta_x_min*delta_p_max = h/(4*%pi), solving for delta_x_min\n",
"delta_x_min = h/(4*%pi*delta_p_max); // Minimum position uncertainty of particle, m\n",
"\n",
"printf('\nThe smallest possible uncertainty in position of the electron = %5.3f angstrom', delta_x_min/1e-010);\n",
"\n",
"// Result \n",
"// The smallest possible uncertainty in position of the electron = 0.019 angstrom"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.31: EX1_31.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.31:  : Page-1.44 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m = 9.1e-031;   // Electronic mass, kg\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"l = 2e-002;     // Length of the side of the cube, m\n",
"E_F = 9*e;  // Fermi energy, J\n",
"// As E_F = h^2/(8*m*l^2)*(nx^2 + ny^2 + nz^2) and nx = ny = nz for a cube, solving for nx\n",
"nx = sqrt(E_F*(8*m*l^2)/(3*h^2));   // Value of integer for a cube\n",
"E = h^2/(8*m*l^2)*3*nx^2; // Fermi energy, J\n",
"E1 = h^2/(8*m*l^2)*((nx-1)^2 + nx^2 + nx^2);    // Energy of the level just below the fermi level, J\n",
"delta_E = E - E1;   // Difference in the energy between the neighbouring levels of Na at the highest state, J\n",
"printf('\nThe energy difference between the neighbouring levels of Na at the highest state = %4.2e eV', delta_E/e);\n",
"// Result \n",
"// The energy difference between the neighbouring levels of Na at the highest state = 1.06e-07 eV "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.32: Energy_of_the_neutron_confined_in_a_nucleus.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.32:: Page-1.45 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m = 1.67e-027;   // Electronic mass, kg\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"nx = 1, ny = 1, nz = 1; // Principle quantum numbers in 3D corresponding to the longest energy state\n",
"lx = 1e-014, ly = 1e-014, lz = 1e-014;    // Dimensions of the box to which the neutron is confined, m\n",
"E = h^2/(8*m)*(nx^2/lx^2+ny^2/ly^2+nz^2/lz^2);    // Energy of the neutron confined in the nucleus, J\n",
"printf('\nThe energy of the neutron confined in a nucleus = %4.2e eV', E/e);\n",
"// Result \n",
"// The energy of the neutron confined in a nucleus = 6.11e+06 eV "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.33: Energy_of_an_electron_moving_in_one_dimensional_infinitely_high_potential_box.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.33:: Page-1.46 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m = 9.1e-031;   // Electronic mass, kg\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"n = 1;     // For simplicity assume principle quantum number to be unity\n",
"l = 2.1e-010;   // Length of one dimensional potential box, m\n",
"E = h^2*n^2/(8*m*l^2);    // Energy of the electron, J\n",
"printf('\nThe energy of the electron moving in one dimensional infinitely high potential box = %4.2f n^2 eV', E/e);\n",
"// Result \n",
"// The energy of the electron moving in one dimensional infinitely high potential box = 8.48 n^2 eV "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.34: Lowest_energy_of_an_electron_in_a_one_dimensional_force_free_region.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.34:: Page-1.46 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m = 9.1e-031;   // Electronic mass, kg\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"n = 1;     // The lowest energy state of electron\n",
"l = 3.5e-010;   // Length of one dimensional potential box, m\n",
"E = h^2*n^2/(8*m*l^2);    // Energy of the electron in the lowest state, J\n",
"printf('\nThe lowest energy of the electron in a one dimensional force free region = %1d eV', E/e);\n",
"// Result \n",
"// The lowest energy of an electron in a one dimensional force free region = 3 eV"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.35: First_three_energy_levels_of_an_electron_in_one_dimensional_box.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.35:: Page-1.46 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m = 9.1e-031;   // Electronic mass, kg\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"l = 9.5e-010;   // Length of one dimensional potential box, m\n",
"// First energy level\n",
"n = 1;     // The first energy state of electron\n",
"E1 = h^2*n^2/(8*m*l^2);    // Energy of the electron in first state, J\n",
"// Second energy level\n",
"n = 2;     // The second energy state of electron\n",
"E2 = h^2*n^2/(8*m*l^2);    // Energy of the electron in second state, J\n",
"// Third energy level\n",
"n = 3;     // The third energy state of electron\n",
"E3 = h^2*n^2/(8*m*l^2);    // Energy of the electron in third state, J\n",
"printf('\nThe energy of the electron in first state  = %4.1e J', E1);\n",
"printf('\nThe energy of the electron in second state  = %4.1e J', E2);\n",
"printf('\nThe energy of the electron in third state  = %4.1e J', E3);\n",
"// Result \n",
"// The energy of the electron in first state  = 6.6e-20 J\n",
"// The energy of the electron in second state  = 2.7e-19 J\n",
"// The energy of the electron in third state  = 6.0e-19 J "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.36: Lowest_two_permitted_energy_values_of_the_electron_in_a_1D_box.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.36:: Page-1.47 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m = 9.1e-031;   // Electronic mass, kg\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"l = 2.5e-010;   // Length of one dimensional potential box, m\n",
"// First energy level\n",
"n = 1;     // The lowest energy state of electron\n",
"E1 = h^2*n^2/(8*m*l^2);    // Energy of the electron in first state, J\n",
"// Second energy level\n",
"n = 2;     // The second energy state of electron\n",
"E2 = h^2*n^2/(8*m*l^2);    // Energy of the electron in second state, J\n",
"printf('\nThe energy of the electron in lowest state  = %5.2f eV', E1/e);\n",
"printf('\nThe energy of the electron in second state  = %5.2f eV', E2/e);\n",
"// Result \n",
"// The energy of the electron in lowest state  =  5.98 eV\n",
"// The energy of the electron in second state  = 23.93 eV "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.37: Lowest_energy_of_the_neutron_confined_to_the_nucleus.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.37:: Page-1.47 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m = 1.67e-027;   // Electronic mass, kg\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"l = 2.5e-010;   // Length of one dimensional potential box, m\n",
"delta_x = 1e-014;   // Uncertainty in position of neutron, m\n",
"// From uncertainty principle, \n",
"// delta_x*delta_p = h/(4*%pi), solving for delta_p\n",
"delta_p = h/(4*%pi*delta_x);    // Uncertainty in momentum of neutron, kg-m/s\n",
"p = delta_p;    // Momemtum of neutron in the box, kg-m/s\n",
"KE = p^2/(2*m); // Kinetic energy of neutron in the box, J\n",
"printf('\nThe lowest energy of the neutron confined to the nucleus = %4.2f MeV', KE/(e*1e+06));\n",
"// Result \n",
"// The lowest energy of the neutron confined to the nucleus = 0.05 MeV "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.38: X_ray_scattering.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.38: : Page-1.56 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m0 = 9.1e-031;   // Electronic mass, kg\n",
"c = 3e+08;      // Speed of light, m/s\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"phi = 45;     // Scattering angle of X-rays, degrees\n",
"E = 75;     // Incident energy of X-rays, keV\n",
"// As from Compton shift formula\n",
"// 1/E_prime - 1/E = 1/(m0*c^2)*(1-cosd(phi))\n",
"// Solving for E_prime\n",
"E_prime = 1/((1/(m0*c^2/(e*1e+03)))*(1-cosd(phi))+1/E); // Energy of scattered photon, keV\n",
"E_recoil = E - E_prime;     // Energy of recoil electron, keV\n",
"printf('\nThe energy of scattered X-ray = %4.1f keV', E_prime);\n",
"printf('\nThe energy of recoil electron = %3.1f keV', E_recoil);\n",
"// Result \n",
"// The energy of scattered X-ray = 71.9 keV\n",
"// The energy of recoil electron = 3.1 keV "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.39: Wavelength_of_scattered_Xray.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.39: : Page-1.57 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m0 = 9.1e-031;   // Electronic mass, kg\n",
"c = 3e+08;      // Speed of light, m/s\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"phi = 60;     // Scattering angle of X-rays, degrees\n",
"E = 75;     // Incident energy of X-rays, keV\n",
"// As from Compton shift formula\n",
"delta_L = h/(m0*c)*(1-cosd(phi));   // Change in photon wavelength, m\n",
"lambda = 0.198e-010;    // Wavelength of incident photon, m\n",
"lambda_prime = (lambda+delta_L)/1e-010;  // Wavelength of scattered X-ray, angstrom \n",
"printf('\nThe wavelength of scattered X-ray = %6.4f angstrom', lambda_prime);\n",
"// Result \n",
"// The wavelength of scattered X-ray = 0.2101 angstrom"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.3: de_Broglie_wavelength_of_an_accelerated_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.3: Page-1.5 (2009)\n",
"clc; clear;\n",
"V = 20e+03;     // Accelerating voltage of electron, V\n",
"lambda = 12.25/sqrt(V);     // de Broglie wavelength of the accelerated electron, m\n",
"printf('\nThe de Broglie wavelength of the electron = %6.4f angstrom', lambda);\n",
"\n",
"// Result \n",
"// The de Broglie wavelength of the electron = 0.0866 angstrom "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.40: Wavelength_of_scattered_radiation_with_changed_angle_of_view.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.40:: Page-1.57 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m0 = 9.1e-031;   // Electronic mass, kg\n",
"c = 3e+08;      // Speed of light, m/s\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"phi = 180;     // Scattering angle of X-rays, degrees\n",
"lambda = 1.78;    // Wavelength of incident photon, m\n",
"lambda_prime = 1.798;  // Wavelength of scattered X-ray, angstrom \n",
"// As from Compton shift formula\n",
"// lambda_prime - lambda = h/(m0*c)*(1-cosd(phi)), Change in photon wavelength, m\n",
"// Or we may write, lambda_prime - lambda = k*(1-cosd(phi))\n",
"// solving for k\n",
"k = (lambda_prime - lambda)/(1-cosd(phi));  // k = h/(m0*c) value, angstrom\n",
"// For phi = 60\n",
"phi = 60;       // New angle of scattering, degrees\n",
"lambda_prime = lambda + k*(1-cosd(phi)); // Wavelength of scattered radiation at 60 degree angle, angstrom\n",
"printf('\nThe wavelength of scattered X-ray at %d degrees view = %6.4f angstrom', phi, lambda_prime);\n",
"// Recoil energy of electron\n",
"E = h*c*(1/lambda - 1/lambda_prime)*1e+010;    // Recoil energy of electron, joule\n",
"printf('\nThe recoil energy of electron scattered through %d degrees = %4.1f eV', phi, E/e);      \n",
"// Result \n",
"// The wavelength of scattered X-ray at 60 degrees view = 1.7845 angstrom\n",
"// The recoil energy of electron scattered through 60 degrees = 17.5 eV "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.41: Compton_scattering_through_aluminium_foil.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.41:: Page-1.58 (2009)\n",
"clc; clear;\n",
"h = 6.6e-034;    // Planck's constant, Js\n",
"m0 = 9.1e-031;   // Electronic mass, kg\n",
"c = 3e+08;      // Speed of light, m/s\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"phi = 90;     // Scattering angle of X-rays, degrees\n",
"E = 510*1e+03*e;    // Energy of incident photon, J\n",
"// As E = h*c/lambda, solving for lambda\n",
"lambda = h*c/E;    // Wavelength of incident photon, m\n",
"// As from Compton shift formula\n",
"// lambda_prime - lambda = h/(m0*c)*(1-cosd(phi)), solving for lambda_prime\n",
"lambda_prime = lambda + h/(m0*c)*(1-cosd(phi));  // Wavelength of scattered X-ray, m  \n",
"printf('\nThe wavelength of scattered X-ray as viewed at %d degrees = %4.2e m', phi, lambda_prime);\n",
"// Recoil energy of electron\n",
"E = h*c*(1/lambda - 1/lambda_prime);    // Recoil energy of electron, joule\n",
"printf('\nThe recoil energy of electron scattered through %d degrees = %4.2e eV', phi, E/e);\n",
"// Direction of recoil electron\n",
"theta = atand(lambda*sind(phi)/(lambda_prime-lambda*cosd(phi)));  // Direction of recoil electron, degrees\n",
"printf('\nThe direction of emission of recoil electron = %5.2f degrees', theta);\n",
"      \n",
"// Result \n",
"// The wavelength of scattered X-ray as viewed at 90 degrees = 4.84e-12 m\n",
"// The recoil energy of electron scattered through 90 degrees = 2.55e+05 eV\n",
"// The direction of emission of recoil electron = 26.61 degrees "
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.42: Energetic_electrons_in_the_Xray_tube.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.42: : Page-1.59 (2009)\n",
"clc; clear;\n",
"m = 9.1e-031;   // Electronic mass, kg\n",
"c = 3e+08;      // Speed of light, m/s\n",
"e = 1.6e-019;   // Charge on the electron, C\n",
"V = 12.4e+03;   // Potential diffeence applied across the X-ray tube, V\n",
"i = 2e-03;  // Current through the X-ray tube, A\n",
"t = 1;  // Time for which the electrons strike the target material, s\n",
"N = i*t/e;    // Number of electrons striking the target per sec, per sec\n",
"v_max = sqrt(2*e*V/m);  // Maximum speed of the electrons, m/s\n",
"printf('\nThe number of electrons striking the target per sec = %4.2e electrons/sec', N);\n",
"printf('\nThe maximum speed of the electrons when they strike = %3.1e m/s', v_max);\n",
"      \n",
"// Result \n",
"// The number of electrons striking the target per sec = 1.25e+16 electrons/sec\n",
"// The maximum speed of the electrons when they strike = 6.6e+07 m/s"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.4: Energy_of_the_electron_from_de_Broglie_wavelength.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.4:  Page-1.6 (2009)\n",
"clc; clear;\n",
"lambda = 5.2e-03;  // de Broglie wavelength of the electron, m\n",
"m = 9.1e-031;      // Mass of the electron, kg\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"// From de Broglie relation, lambda = h/sqrt(2*m*E), solving for E\n",
"E = h^2/(2*m*lambda^2*e);     // Energy of the electron, eV\n",
"printf('\nThe energy of the electron from de Broglie wavelength = %5.3e eV', E);\n",
"\n",
"// Result \n",
"// The energy of the electron from de Broglie wavelength = 5.576e-014 eV \n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.5: Velocity_and_de_Broglie_wavelength_of_a_neutron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.5: Page-1.6 (2009)\n",
"clc; clear;\n",
"m = 1.67e-027;      // Mass of the neutron, kg\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"E = 1e+04*e;    // Energy of the neutron, J\n",
"// As E = 1/2*m*v^2, solving for v\n",
"v = sqrt(2*E/m);    // Velocity of the neutron, m/s\n",
"lambda = h/(m*v);  // de Broglie wavelength of the neutron, m\n",
"printf('\nThe velocity of the neutron = %4.2e m/s', v);\n",
"printf('\nThe de Broglie wavelength of the neutron = %4.2e m', lambda);\n",
"\n",
"// Result \n",
"// The velocity of the neutron = 1.38e+006 m/s\n",
"// The de Broglie wavelength of the neutron = 2.87e-013 m \n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.6: Wavelength_of_thermal_neutron_at_room_temperature.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.6: Page-1.6 (2009)\n",
"clc; clear;\n",
"m = 1.67e-027;      // Mass of the neutron, kg\n",
"k = 1.38e-023;      // Boltzmann constant, J/mol/K\n",
"T = 27+273;         // Room temperature, K\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"v = sqrt(3*k*T/m);      // Velocity of the neutron, m/s\n",
"lambda = h/(m*v);  // de Broglie wavelength of the neutron, m\n",
"printf('\nThe de Broglie wavelength of the thermal neutrons = %4.2f angstrom', lambda/1e-010);\n",
"\n",
"// Result \n",
"// The de Broglie wavelength of the thermal neutrons = 1.45 angstrom \n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.7: Angle_of_deviation_for_first_order_diffraction_maxima.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.7: Page-1.6 (2009)\n",
"clc; clear;\n",
"m = 9.1e-031;      // Mass of the electron, kg\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"E = 20e+03*e;     // Energy of the electron, J\n",
"// As 1/2*m*v^2 = E, solving for v\n",
"v = sqrt(2*E/m);        // Velocity of the electron, m/s\n",
"lambda = h/(m*v);  // de Broglie wavelength of the electron, m\n",
"n = 1;  // First order diffraction\n",
"d = 9.8e-011;       // Atomic spacing for thin gold foil, m\n",
"// Using Bragg's equation, 2*d*sin(theta) = n*lambda and solving for theta\n",
"theta = asind(n*lambda/(2*d));      // Angle of deviation for first order diffraction maxima, degree\n",
"printf('\nThe angle of deviation for first order diffraction maxima = %4.2f degrees', theta);\n",
"\n",
"// Result \n",
"// The angle of deviation for first order diffraction maxima = 2.54 degrees \n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.8: de_Broglie_wavelength_of_a_moving_electron.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.8:  Page-1.7 (2009)\n",
"clc; clear;\n",
"m = 9.1e-031;      // Mass of the electron, kg\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"E = 5*e;     // Energy of the electron, J\n",
"// As 1/2*m*v^2 = E, solving for v\n",
"v = sqrt(2*E/m);        // Velocity of the electron, m/s\n",
"lambda = h/(m*v);  // de Broglie wavelength of the electron, m\n",
"printf('\nThe de Broglie wavelength of the electron = %3.1f angstrom', lambda/1e-010);\n",
"\n",
"// Result \n",
"// The de Broglie wavelength of the electron = 5.5 angstrom\n",
""
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 1.9: de_Broglie_wavelength_of_a_neutron_of_given_kinetic_energy.sce"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"// Scilab Code Ex1.9: Page-1.7 (2009)\n",
"clc; clear;\n",
"m = 1.67e-027;      // Mass of the neutron, kg\n",
"e = 1.6e-019;   // Energy equivalent of 1 eV, J/eV\n",
"h = 6.626e-034;     // Planck's constant, Js\n",
"E = 1*e;     // Energy of the electron, J\n",
"lambda = h/sqrt(2*m*E);  // de Broglie wavelength of the neutron, m\n",
"printf('\nThe de Broglie wavelength of the neutron = %4.2f angstrom', lambda/1e-010);\n",
"\n",
"// Result \n",
"// The de Broglie wavelength of the neutron = 0.29 angstrom \n",
""
   ]
   }
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