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diff --git a/Modern_Physics_by_K_S_Krane/8-Many_Electron_Atoms.ipynb b/Modern_Physics_by_K_S_Krane/8-Many_Electron_Atoms.ipynb new file mode 100644 index 0000000..70cd647 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/8-Many_Electron_Atoms.ipynb @@ -0,0 +1,177 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: Many Electron Atoms" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1: Energy_of_Ka_X_ray_of_sodium.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-8.1');\n", +"hc=1240*10^-9;Rinfi=1.097*10^7;Z=11; //for sodium atom;and other constants in MeV\n", +"delE=3*hc*Rinfi*(Z-1)^2/4 //change in energy\n", +"printf('The energy of the Ka x-ray of the sodium atom is %.3f KeV.',delE/10^3);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-8.2(a)');\n", +"EKa=21.990;EKb=25.145;EK=25.514 //all the values are in KeV\n", +"ELo=EKb-EKa;printf('The enrgy of La of X-ray is %.3fKeV.\n',ELo); //Energy of La X-ray\n", +"disp('Exa-8.2(b)');\n", +"EL=-EK+EKa;printf('Hence the binding energy of the L electon is %.3fKeV.',EL); // for electron L electron\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.3: Total_orbital_and_spin_quantum_numbers_of_carbon.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-8.3'); // theoretical\n", +"l=1; Lmax=l+l;Lmin=l-l;printf('Value of L ranges from %d to %d i.e %d %d %d\n.',Lmin,Lmax,Lmin,1,Lmax);\n", +"s=1/2; Smax=s+s;Smin=s-s;printf('Values of S are %d &%d',Smax,Smin);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.4: Total_orbital_and_spin_quantum_numbers_of_nitrogen.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-8.4');\n", +"l=1; Lmax=l+l;Lmin=l-l;printf('Considering any two electrons,Value of L2e ranges from %d to %d i.e %d %d %d.\n',Lmin,Lmax,Lmin,1,Lmax);\n", +"printf('Adding the angular momentum of the third electron to L2emax gives the maximum whole angular momentum as 2+1=3; and subtracting it from L2e=1 gives 0\n')\n", +"s=1/2; Smax=s+s;Smin=s-s;printf('Values of S2e are %d &%d.\n',Smax,Smin);\n", +"printf('Adding and subtracting the spin of third to S2e=1 and S2e=0 respectively gives the spins 3/2 and 1/2 for the 3 electron system.');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.5: Hunds_rule_to_find_ground_state_quantum_numbers_of_nitrogen.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-8.5');\n", +"disp('The nitrogen atom has a configuration of 1s2,2s2,2p3.');\n", +"disp('Let us maximize the net spin of all the 3 electrons by assigning a spin of 1/2 to each of them. Hence S=3/2.');\n", +"disp('To maximize Ml,the consistent values of L for the 3 electrons left are 1 -1 and 0.Thus L=0 & S=3/2 are the ground state quantum numbers for nitrogen.');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.6: Ground_state_L_and_S_of_oxygen.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-8.6');\n", +"disp('The Oxygen atom has a configuration of 1s2,2s2,2p4. 4 electrons in the outer most shell.');\n", +"disp('Let us maximize the net spin by assigning a spin of 1/2 to 3 of them but the fourth should have spin of -1/3. Hence S=3/2-1/2=1.');\n", +"disp('The consistent values of L for the 3 electrons are 1 -1 and 0.To maximize Ml, assign a L 0f +1 to the fourth electron.Thus L=1 & S=1 are the ground state quantum numbers for Oxygen.');" + ] + } +], +"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 +} |