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diff --git a/Modern_Physics_by_R_A_Serway/4-The_particle_nature_of_matter.ipynb b/Modern_Physics_by_R_A_Serway/4-The_particle_nature_of_matter.ipynb new file mode 100644 index 0000000..46e82ac --- /dev/null +++ b/Modern_Physics_by_R_A_Serway/4-The_particle_nature_of_matter.ipynb @@ -0,0 +1,380 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: The particle nature of matter" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.1: Electrolysis_of_barium_chloride.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex4.1: Pg 109 (2005)\n", +"clc; clear;\n", +"I = 10; // Electric current, A\n", +"t = 3600; // Time, s\n", +"q = I*t; // Electric charge liberated, C\n", +"mm_Ba = 137; // Molar mass of Barium, g\n", +"mm_Cl = 35.5; // Molar mass of Chlorine, g\n", +"valence_Ba = 2; // Valence electrons of Barium\n", +"valence_Cl = 1; // Valence electrons of Chlorine\n", +"// Using Faraday’s law of electrolysis, we have\n", +"m_Ba = (q*mm_Ba)/(96500*valence_Ba); // Mass of Barium obtained, g\n", +"m_Cl = (q*mm_Cl)/(96500*valence_Cl); // Mass of Chlorine obtained, g\n", +"printf('\nMass of Barium obtained = %4.1f g', m_Ba);\n", +"printf('\nMass of Chlorine obtained = %4.1f g', m_Cl);\n", +"\n", +"// Result\n", +"// Mass of Barium obtained = 25.6 g\n", +"// Mass of Chlorine obtained = 13.2 g" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2: Deflection_of_electron_beam_by_E_and_B_Fields.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex4.2: Pg 113 (2005)\n", +"clc; clear;\n", +" V = 200; // Electrc potential, V\n", +"theta = 0.20; // Angle, radians\n", +"l = 0.050; // Length of plates, m\n", +"d = 1.5e-02; // Distance between two plates, m\n", +"c_m_r = 1.76e+11; // Charge-to-mass ratio, C/kg\n", +"// Since e/m_e = (V*theta)/(B^2*l*d), solving for B\n", +"B = sqrt((V*theta)/(l*d*c_m_r)); // Magnetic field, T\n", +"printf('\nThe magnetic field required to produce the deflection of %4.2f rad = %3.1e T', theta, B);\n", +"\n", +"// Result\n", +"// The magnetic field required to produce the deflection of 0.20 rad = 5.5e-04 T " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3: Experimental_determination_of_e.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex4.3: Pg 117 (2005)\n", +"clc; clear;\n", +"\n", +"// Part (a)\n", +"delta_y = 0.600; // Distance of rise or fall of a droplet, cm\n", +"t_av = 21.0; // Average time of fall of droplet, s\n", +"delta_t = [46.0, 15.5, 28.1, 12.9, 45.3, 20.0]; // Rise time of the droplet in succession, s\n", +"v = delta_y/t_av; // Average speed of the falling droplet, cm/s\n", +"v_prime = zeros(6);\n", +"for i = 1:1:6\n", +" v_prime(i) = delta_y/delta_t(i); // Successive speeds of the rising drops, cm/s \n", +"end\n", +"\n", +"// Calculate charge ratios\n", +"q1byq2 = (v+v_prime(1))/(v + v_prime(2));\n", +"q2byq3 = (v+v_prime(2))/(v + v_prime(3));\n", +"q3byq4 = (v+v_prime(3))/(v + v_prime(4));\n", +"q4byq5 = (v+v_prime(4))/(v + v_prime(5));\n", +"q5byq6 = (v+v_prime(5))/(v + v_prime(6));\n", +"printf('\nq1/q2 = %5.3f', q1byq2);\n", +"printf('\nq2/q3 = %5.3f', q2byq3);\n", +"printf('\nq3/q4 = %5.3f', q3byq4);\n", +"printf('\nq4/q5 = %5.3f', q4byq5);\n", +"printf('\nq5/q6 = %5.3f', q5byq6);\n", +"printf('\nThe charge ratios are ratios of small whole numbers\n');\n", +"\n", +"// Part (b)\n", +"eta = 1.83e-05; // Viscosity of air, kg/ms\n", +"rho = 858; // Oil density, kg/m^3\n", +"g = 9.81; // Acceleration due to gravity, m/s^2\n", +"a = sqrt((9*eta*v*1e-02)/(2*rho*g)); // Radius of oil droplet, m\n", +"V = 4/3*(%pi)*a^3; // Volume of oil droplet, m^3\n", +"m = rho*V; // Mass of oil droplet, kg\n", +"printf('\nRadius of oil droplet = %4.2e m', a);\n", +"printf('\nVolume of oil droplet = %4.2e m^3', V);\n", +"printf('\nMass of oil droplet = %4.2e kg', m);\n", +"\n", +"// Part (c)\n", +"V = 4550; // Potential difference across the plates of the capacitor, volt\n", +"d = 0.0160; // Distance between the plates\n", +"E = V/d; // Electric field between plates, V/m\n", +"q = zeros(6), e = zeros(6);\n", +"for i=1:1:6\n", +" q(i) = m*g/E*((v+v_prime(i))/v); // Charge on first drop, C \n", +" printf('\nq%d = %4.2e V/m', i, q(i)); \n", +"end\n", +"e(1) = q(1)/5;\n", +"e(2) = q(2)/8;\n", +"e(3) = q(3)/6;\n", +"e(4) = q(4)/9;\n", +"e(5) = q(5)/5;\n", +"e(6) = q(6)/7;\n", +"e_tot = 0;\n", +"for i = 1:1:6\n", +" e_tot = e_tot + e(i);\n", +"end\n", +"e = e_tot/6;\n", +"printf('\nThe average charge on an electron = %5.3e C', e);\n", +"\n", +"// Result\n", +"// q1/q2 = 1.105\n", +"// q2/q3 = 0.958\n", +"// q3/q4 = 1.053\n", +"// q4/q5 = 0.899\n", +"// q5/q6 = 1.086\n", +"// The charge ratios are ratios of small whole numbers\n", +"\n", +"// Radius of oil droplet = 1.67e-06 m\n", +"// Volume of oil droplet = 1.96e-17 m^3\n", +"// Mass of oil droplet = 1.68e-14 kg \n", +"\n", +"// q1 = 8.44e-019 V/m\n", +"// q2 = 1.36e-018 V/m\n", +"// q3 = 1.01e-018 V/m\n", +"// q4 = 1.52e-018 V/m\n", +"// q5 = 8.48e-019 V/m\n", +"// q6 = 1.19e-018 V/m\n", +"// The average charge on an electron = 1.694e-019 C " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4: Collision_of_alpha_particle_with_proton.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex4.4: Pg 121 (2005)\n", +"clc; clear;\n", +"\n", +"// Part (b)\n", +"// For easy calculations, assume all variables to be unity\n", +"m_p = 1; // Mass of proton, a.m.u\n", +"m_a = 4*m_p; // Mass of alpha particle, a.m.u\n", +"Valpha = 1; // Velocity of alpha particle before collision, m/s\n", +"v_p = (2*m_a*Valpha)/(m_a + m_p); // Velocity of proton after collision, m/s\n", +"v_a = ((m_a - m_p)*(Valpha))/(m_a + m_p); // Velocity of alph particle after collision, m/s\n", +"p_change = ((v_a - Valpha)/(Valpha))*100; // Percentage change in velocity of alpha particle\n", +"printf('\nVelocity of proton after collision = %4.2fVa m/s', v_p);\n", +"printf('\nVelocity of alpha particle after collision = %4.2fVa m/s', v_a);\n", +"printf('\nPercentage change in velocity of alpha particle = %2d percent', p_change);\n", +"\n", +"// Result\n", +"// Velocity of proton after collision = 1.60 V_a m/s\n", +"// Velocity of alph particle after collision = 0.60 V_a m/s\n", +"// Percentage change in velocity of alpha particle = -40 percent" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5: Radius_of_Aluminium_Nucleus.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex4.5: Pg 124 (2005)\n", +"clc; clear;\n", +"Z = 13; // Atomic number of Aluminium\n", +"e = 1.6e-19; // Charge on electron, C\n", +"k = 8.99e+09; // Coulomb constant, Nm^2/C^2\n", +"K_a = 7.7e+06*e; // Since K_a = (k*Z_e*2*e)/d_min, solving for d_min\n", +"d_min = (k*2*Z*e^2)/K_a; // Radius of Aluminum, m\n", +"printf('\nRadius of Aluminum = %3.1e m', d_min);\n", +"\n", +"// Result\n", +"// Radius of Aluminum = 4.9e-15 m" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7: Collision_of_alpha_particle_with_proton.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex4.7: Pg 135 (2005)\n", +"clc; clear;\n", +"// Part (a)\n", +"n_i = 2; // Initial level of electron\n", +"n_f = 1; // Final level of electron\n", +"R = 1.097e+07; // Rydberg constant, per metre\n", +"c = 3e+08; // Velocity of light, m/s\n", +"h = 4.136e-15; // Planck's constant, eV\n", +"lamda = n_i^2*n_f^2/((n_i^2-n_f^2)*R); // Wavelength of emitted photon, m\n", +"f = c/lamda; // Frequency of emitted photon, Hz\n", +"E = h*f; // Energy of emitted photon, eV\n", +"printf('\nThe wavelength of emitted photon = %5.1f nm', lamda/1e-09);\n", +"printf('\nThe frequency of emitted photon = %4.2e Hz', f);\n", +"printf('\nEnergy of emitted photon = %4.1f eV', E);\n", +"\n", +"// Part (b)\n", +"mc_square = 938.8e+06; // Energy of recoil of hydrogen atom, eV\n", +"K = 0.5*(E^2/mc_square); // Recoil kinetic energy of H atom, eV\n", +"E_difference = K/E; // Energy difference\n", +"printf('\nRecoil kinetic energy of H atom = %4.2e eV', K);\n", +"printf('\nThe fraction of energy difference = %3.1e', E_difference);\n", +"\n", +"// Result\n", +"// The wavelength of emitted photon = 121.5 nm\n", +"// The frequency of emitted photon = 2.47e+15 Hz\n", +"// Energy of emitted photon = 10.2 eV\n", +"// Recoil kinetic energy of H atom = 5.55e-08 eV\n", +"// The fraction of energy difference = 5.4e-09 " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8: series_for_Hydrge.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex4.8: Pg 136 (2005)\n", +"clc; clear;\n", +"\n", +"// Part (a)\n", +"n_i = 3; // Initial level of electron\n", +"n_f = 2; // Final level of electron\n", +"R = 1.097e+07; // Rydberg constant, per metre\n", +"c = 3e+08; // Velocity of light, m/s\n", +"h = 6.626e-34; // Plank's constant, Js\n", +"lamda_max = (n_i^2*n_f^2)/((n_i^2-n_f^2)*R); // Maximum wavelength of emitted photon, m\n", +"E_photon = (h*c)/(lamda_max*1.6e-19); // Energy of emitted photon, eV\n", +"printf('\nThe maximum wavelength of emitted photon = %5.1f nm', lamda_max/1e-09);\n", +"printf('\nEnergy of emitted photon = %4.2f eV', E_photon);\n", +"\n", +"// Part (b)\n", +"n_i = %inf; // Initial level of electron\n", +"lamda_min = 1/(R*(1/n_f^2-1/n_i^2));\n", +"printf('\nThe wavelength corresponding to the series limit = %5.1f nm which is in the ultraviolet region', lamda_min/1e-09);\n", +"\n", +"// Result\n", +"// The maximum wavelength of emitted photon = 656.3 nm\n", +"// Energy of emitted photon =1.89 eV\n", +"//// The wavelength corresponding to the series limit = 364.6 nm which is in the ultraviolet region " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9: Hydrogen_in_its_first_excited_state.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex4.9: Pg 137 (2005)\n", +"clc; clear;\n", +"k_B = 8.62e-05; // Boltzmann constant, eV/K\n", +"delta_E = 10.2; // Average thermal energy, eV\n", +"// Since (3/2)*k_B*T = averge thermal energy per atom = 10.2eV, solving for T\n", +"T = 10.2/(3/2*k_B); // Temperature at which H-atoms jump to first excited state, K\n", +"printf('\nThe temperature at which H-atoms jump to first excited state = %5d K', T);\n", +"N_ratio = 0.10; // Number ratio of population of first excited state relative to the ground state\n", +"// As N_ratio = exp(-delta_E/(k_B*T)), solving for T\n", +"T = -delta_E/(k_B*log(N_ratio)); // Temperature at which H-atoms jump to first excited state, K\n", +"printf('\nThe temperature of excitation from Boltzmann distribution = %5d K', T);\n", +"\n", +"// Result\n", +"// The temperature at which H-atoms jump to first excited state = 78886 K\n", +"// The temperature of excitation from Boltzmann distribution = 51389 K " + ] + } +], +"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 +} |