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diff --git a/Modern_Physics_by_R_A_Serway/14-Nuclear_Physics_Applications_.ipynb b/Modern_Physics_by_R_A_Serway/14-Nuclear_Physics_Applications_.ipynb new file mode 100644 index 0000000..355badd --- /dev/null +++ b/Modern_Physics_by_R_A_Serway/14-Nuclear_Physics_Applications_.ipynb @@ -0,0 +1,226 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14: Nuclear Physics Applications " + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.1: Energy_released_in_Fission.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex14.1: Pg 513 (2005)\n", +"clc; clear;\n", +"// Part (a)\n", +"u = 931.5; // Atomic mass unit, Mev \n", +"M_Li = 7.016003; // Mass of Lithium, kg\n", +"M_H = 1.007825; // Mass of Hydrogen, kg\n", +"M_He = 4.002603; // Mass of Helium, kg\n", +"Q = (M_Li + M_H - 2*M_He)*u; // Q-value of the reaction, MeV\n", +"// Part (b)\n", +"K_incident = 0.6; // Kinetic energy of the incident protons, MeV\n", +"K_products = Q + K_incident; // Kinetic energy of the products\n", +"printf('\nThe Q value of the reaction = %4.1f MeV', Q);\n", +"printf('\nThe kinetic energy of the products (two alpha particles) = %4.1f MeV', K_products);\n", +"\n", +"// Result\n", +"// The Q value of the reaction = 17.3 MeV\n", +"// The kinetic energy of the products (two alpha particles) = 17.9 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.2: Neutron_capture_by_Al.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex14.2: Pg 509 (2005)\n", +"clc; clear;\n", +"roh = 2.7e+06; // Density of Al, g/cm^3\n", +"A = 27; // Mass number of Al\n", +"n = (6.02e+23*roh)/A; // Number of nuclei/m^3\n", +"sigma = 2.0e-31; // Effective area of nucleas normal to motion, m^2\n", +"R_0 = 5.0e+12; // Rate of incident particles per unit area, neutrons/cm^2-s\n", +"x = 0.30e-03; // Thickness of foil, m\n", +"R = (R_0*sigma*n*x) // Number of neutrons captured by foil, neutrons/cm^2-s\n", +"printf('\nThe number of neutrons captured by foil = %3.1fe+07 neutrons/Sq.cm-s', R*1e-07);\n", +"\n", +"// Result\n", +"// The number of neutrons captured by foil = 1.8e+07 neutrons/Sq.cm-s " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.4: Energy_released_in_the_Fission_of_U235.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex14.4: Pg 513 (2005)\n", +"clc; clear;\n", +"m = 1; // Mass of Uranium taken, kg\n", +"Q = 208; // Disintegration energy per event, MeV\n", +"A = 235; // Mass number of Uranium\n", +"N = (6.02e+23*m)/A; // Number of nuclei\n", +"E = N*Q; // Disintegration energy, MeV\n", +"printf('\nThe total energy released if %1d kg of Uranium undergoes fission = %4.2fe+26 MeV', m, E*1e-23);\n", +"\n", +"// Result\n", +"// The total energy released if 1 kg of Uranium undergoes fission = 5.33e+26 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.5: A_Rough_Mechanism_for_Fission_Process.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex14.5: Pg 513 (2005)\n", +"clc; clear;\n", +"A_Ba = 141; // Mass number of Barium\n", +"A_Kr = 92; // Mass number of Barium\n", +"r_0 = 1.2e-15; // Separation constant, m\n", +"r_Ba = r_0*A_Ba^(1/3); // Nuclear radius of Barium, m \n", +"r_Kr = r_0*A_Kr^(1/3); // Nuclear radius of Krypton, m\n", +"r = r_Ba + r_Kr; // Separation between two atoms, m\n", +"Z_1 = 56; // Atomic number of Barium\n", +"Z_2 = 36; // Atomic number of Barium\n", +"k = 1.440e-09; // Coulomb constant, eV-m\n", +"U = k*Z_1*Z_2/r // Coulomb Potential energy of two charges, MeV\n", +"printf('\nThe Coulomb potential energy for two charges = %3d MeV' , U/1e+06);\n", +"printf('\nThis shows that the fission mechanism is plausible');\n", +"\n", +"// Result\n", +"// The Coulomb potential energy for two charges = 248 MeV\n", +"// This shows that the fission mechanism is plausible " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.6: The_Fusion_of_Two_Deutrons.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex14.6: Pg 519 (2005)\n", +"clc; clear;\n", +"// Part (a)\n", +"e = 1.6e-19; // Charge on electron, C\n", +"k = 8.99e-09; // Coulomb constant, N-m^2/C^2\n", +"r = 1.0e-14; // Distance between two duetrons, m\n", +"// We have U = (k*q1*q2)/r, for duetrons q1 = q2 = e, therefore we get\n", +"U = (k*e^2)/r; // Potential energy of duetrons, J\n", +"E_C = 1.1e-014; // The coulomb energy per deutron, J\n", +"k_B = 1.38e-023; // Boltzmann constant, J/mol/K\n", +"T = 2/3*E_C/k_B; // Effective temperature required for deutron to overcome the potential barrier, K\n", +"printf('\nThe potential energy of two duetrons separated by the distance of %1.0de-14 m = %4.2f MeV', r*1e+14, (U*1e+12)/e);\n", +"printf('\nThe effective temperature required for deutron to overcome the potential barrier = %3.1e K', T);\n", +"\n", +"// Result\n", +"// The potential energy of two duetrons separated by the distance of 1e-14 m = 0.14 MeV\n", +"// The effective temperature required for deutron to overcome the potential barrier = 5.3e+008 K \n", +"// Result\n", +"// The potential energy of two duetrons separated by the distance of 1e-14 m = 0.14 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.7: Half_value_thickness.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Ex14.7: Pg 530 (2005)\n", +"clc; clear;\n", +"mew = 55e-02; // Linear absoption coefficient, per m\n", +"// In equation I(x) = I_o*exp(-mew*x), replacing I(x) by I_o/2 & solving for x, we get\n", +"x = log(2)/mew; // Half value thickness, m \n", +"printf('\nThe half value thickness for lead = %4.2fe-02 cm', x);\n", +"\n", +"// Result\n", +"// The half value thickness for lead = %1.26e-02 cm" + ] + } +], +"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 +} |