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diff --git a/Modern_Physics_by_K_S_Krane/1-Introduction.ipynb b/Modern_Physics_by_K_S_Krane/1-Introduction.ipynb new file mode 100644 index 0000000..6597615 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/1-Introduction.ipynb @@ -0,0 +1,107 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: Introduction" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.1: Mass_difference_between_a_proton_and_a_neutron.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-1.1');\n", +"Mn=1.008665;Mp=1.007276 //Given mass of an electron and a proton in terms of u\n", +"Md= Mn-Mp; //mass difference\n", +"printf('Mass difference in terms of U is %f ',Md); \n", +"Md=Md*931.50; //converting u into Mev/c^2 by multiplying by 931.5 MeV/c^2\n", +"printf('which equals %.3f Mev/c^2.',Md);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.2: Total_mass_of_proton_and_electron.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-1.2');\n", +"Mp=1.007276 ; Me=5.4858*10^-4; //mass of proton and electron in terms of U\n", +"Mt=Mp+Me; //Total mass= sum of above masses \n", +"printf('The combined mass of an electron and a proton was found out to be %f U.',Mt);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.3: Value_of_hc.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-1.3');\n", +"h=6.621*10^-34 ; c=2.9979*10^8; // h is in J/s and c is in m/s\n", +"hc=h*c*((10^9)/(1.6022*10^-19)); //1e=1.602*10^-19 J and 1 m=10^9 nm\n", +"printf('The value of hc is %f eV.nm\n',hc); \n", +"printf('Rounding off to 4 digits, we obtain %4.f eV.nm.',hc);\n", +"disp('Hence zero at the end is significant.');" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/10-Statistical_Physics.ipynb b/Modern_Physics_by_K_S_Krane/10-Statistical_Physics.ipynb new file mode 100644 index 0000000..72e73ff --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/10-Statistical_Physics.ipynb @@ -0,0 +1,165 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10: Statistical Physics" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.1: Various_Speeds_obtained_from_maxwell_speed_distribution.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-10.1'); //Theoretical Question\n", +"//**Install and use maxim tool for symbolic integration. remove the '//'(comment markings) below and run the program.\n", +"//Vm=integrate('(v^3)*(e^(-b*v^2))','x',0,%infi);\n", +"//rest of the results follow from above\n", +"printf('The average speed is found out to be (8*k*T/m)^1/2)\n'); \n", +"printf('The RMS speed is (3*k*T/m)^1/2\n');\n", +"printf('The Most probable speed is found out to be (2*k*T/m)^1/2 \n where all the symbols used are conventional constants.');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.2: Frequency_distribution_of_emitted_light.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-10.2'); //The solution is purely theoretical and involves a lot of approximations.\n", +"printf('The value of shift in frequency was found out to be delf=7.14*fo*10^-7*sqrt(T) for a star composing of hydrogen atoms at a temperature T.\n');\n", +"T=6000; //temperature for sun\n", +"delf=7.14*10^-7*sqrt(T);.....//change in frequency\n", +"printf('The value of frequency shift for sun(at 6000 deg. temperature) comprsing of hydrogen atoms is %.1e times the frequency of the light.',delf);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.3: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-10.3(a)');\n", +"kT=0.0252;E=10.2 // at room temperature, kT=0.0252 standard value and given value of E\n", +"n2=2;n1=1; g2=2*(n2^2);g1=2*(n1^2); //values for ground and excited states\n", +"t=(g2/g1)*%e^(-E/kT); //fraction of atoms\n", +"printf('The number of hydrogen atoms required is %e which weighs %e Kg\n',1/t,(1/t)*(1.67*10^-27));\n", +"disp('Ex-10.3(b)');\n", +"t=0.1/0.9;k=8.65*10^-5 //fracion of atoms in case-2 is given\n", +"T=-E/(log(t/(g2/g1))*k); //temperature\n", +"printf('The value of temperature at which 1/10 atoms are in excited state is %.3f K',T);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.4: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-10.4(a)'); //theoretical\n", +"printf('The energy of interaction with magnetic field is given by uB and the degeneracy of the states are +-1/2 which are identical.\nThe ratio is therefore pE2/pE1 which gives e^(-2*u*B/k*T)');\n", +"disp('Ex-10.4(b)');\n", +"uB=5.79*10^-4; //for a typical atom\n", +"t=1.1;k=8.65*10^-5; //ratio and constant k\n", +"T=2*uB/(log(t)*k); //temperature\n", +"printf('The value of temperature ar which the given ratio exists is %.2f K',T);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.5: Fermi_Energy_Ef_for_sodium.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-10.5');\n", +"p=0.971; A=6.023*10^23; m=23.0; // various given values and constants\n", +"c= (p*A/m)*10^6; // atoms per unit volume\n", +"hc=1240; mc2=0.511*10^6; // hc=1240 eV.nm\n", +"E= ((hc^2)/(2*mc2))*(((3/(8*%pi))*c)^(2/3)); //value of fermi energy\n", +"printf('The fermi energy for sodium is %f eV',E*10^-18);//multiply by 10^-18 to convert metres^2 term to nm^2" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/11-Properties_of_Ionic_Crystals.ipynb b/Modern_Physics_by_K_S_Krane/11-Properties_of_Ionic_Crystals.ipynb new file mode 100644 index 0000000..baccdc1 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/11-Properties_of_Ionic_Crystals.ipynb @@ -0,0 +1,117 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11: Properties of Ionic Crystals" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.1: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-11.1(a)');\n", +"c=769*10^3; Na=6.023*10^23; JeV=1.6*10^-19; //various constants and given values\n", +"Be=c/(Na*JeV); //Binding energy of an ion pair in the lattice\n", +"printf('The experimental value was found out to be %.4f eV.\n',Be);\n", +"disp('Exa-11.1(b)');\n", +"n=9;a=1.7476; R=0.281; k= 1.44; //Given values and consstants\n", +"Bc=k*a*(1-(1/n))/R; //ionic binding energy eperimentally\n", +"printf('The calculated value of the binding energy is %.4f eV.',Bc);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.2: Energy_per_neytral_atom_to_take_apart_a_crystal_of_Nacl.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-11.2');\n", +"a=3.61;// amount of energy required to remove an electron from Cl- ion\n", +"b=-5.14 //amount of energy returned when an electron is added to Na+ ion\\n", +"c=7.98 //binding energy of NaCl atom\n", +"E=a+b+c //suom of all the energies\n", +"printf('The net energy to be supplied is %.3f eV',E);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11.3: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-11.3(a)');\n", +"Na=6.023*10^23; p=8.96*10^3; M=63.5*10^-3; //Na=avagadro's number,p=density,M=molar mass\n", +"n= p*Na/M; //density of charge carriers\n", +"printf('The density of charge carriers in copper is %e atoms/m3\n',n);\n", +"s=5.88*10^7;m=9.11*10^-31;e=1.6*10^-19; //charge & mass of an electron,resistance per unit length\n", +"t= s*m/(n*e^2); //average time between collisions\n", +"printf('The average time between collisions of conducting electrons is %e sec.\n',t);\n", +"disp('Ex-11.3(b)');\n", +"Ef=7.03*1.6*10^-19; //converting given enrgy to J\n", +"Vf=sqrt(2*Ef/m); //fermi velocity\n", +"l=Vf*t; //mean free path\n", +"printf('The average mean free path is %e m = %.1f nm',l,l*10^9);" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/12-Nuclear_Structure_and_Radioactivity.ipynb b/Modern_Physics_by_K_S_Krane/12-Nuclear_Structure_and_Radioactivity.ipynb new file mode 100644 index 0000000..358a5b5 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/12-Nuclear_Structure_and_Radioactivity.ipynb @@ -0,0 +1,445 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12: Nuclear Structure and Radioactivity" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.10: Q_value_of_various_decays.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.10(a)');\n", +"m40K=39.963999; //mass of various particles\n", +"m40Ca=39.962591;\n", +"c2=931.5; //value of c^2 in MeV\n", +"Qb1=(m40K-m40Ca)*c2; //Q value of the reaction\n", +"printf('The Q value for -VE beta emission is %.3f Mev \n',Qb1);\n", +"disp('Exa-12.10(b)');\n", +"m40K=39.963999; //mass of various particles\n", +"m40Ar=39.962384;\n", +"me=0.000549;\n", +"Qb2=(m40K-m40Ar-2*me)*c2; //Q value of the reaction\n", +"printf('The Q value for +VE beta emission is %.3f Mev \n',Qb2);\n", +"disp('Exa-12.10(c)');\n", +"m40K=39.963999;\n", +"m40Ar=39.962384;\n", +"Qec=(m40K-m40Ar)*c2;\n", +"printf('The Q value for +VE beta emission is %.3f Mev \n',Qec);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.11: Maximum_kinetic_energy_of_emitted_beta_particle.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.11');\n", +"Mg=12.000000; //mass of the carbon atom in amu\n", +"c2=931.5; \n", +"Eg=4.43; //given energy of gamma ray \n", +"Mex=Mg+(Eg/c2); //mass in excited state\n", +"Me=0.000549; //mass of an electron\n", +"Q=(12.018613-Mex-2*Me)*c2; //Q of the particle\n", +"printf('The maximum value of kinetic energy is %.2f MeV',Q);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.12: Rate_of_energy_production_per_gram_of_uranium.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.12');\n", +"m238U=238.050786; //mass of various quantities\n", +"m206Pb=205.974455;\n", +"m4He=4.002603;\n", +"c2=931.5; //constants\n", +"Na=6.023*10^23; //avagadro's number\n", +"Q=(m238U-m206Pb-8*m4He)*c2; \n", +"t12=(4.5)*10^9*(3.16*10^7); //half life years to seconds conversion\n", +"w=0.693/t12; // lambeda\n", +"NoD=(Na/238)*w; //number of decays\n", +"E=NoD*Q*(1.6*10^-19)*10^6; //rate of liberation of energy,converting MeV to eV\n", +"printf('Rate of energy liberation is %.1e W',E);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.13: Ages_of_the_given_rocks.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.13');\n", +"R=0.5;t12=4.5*10^9; //value of radius and half-life \n", +"t1=(t12/0.693)*log(1+(1/R)); //age of rock 1\n", +"R=1.0;\n", +"t2=(t12/0.693)*log(1+(1/R)); //age of rock-2\n", +"R=2.0\n", +"t3=(t12/0.693)*log(1+(1/R)); //age of rock 3\n", +"printf('The ages of rock samples are %.1e, %.1e, %.1e years respectively',t1,t2,t3);\n", +" " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.14: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-12.14(a)');\n", +"P=2*10^14; V=2*10^-14; R=8.314; T=295;Na=6.023*10^23; //varoius constants and given values\n", +"n=P*V/(R*T); //ideal gas law\n", +"N=Na*n;f=10^-12 //avagadaro's number and fracction of carbon molecules\n", +"t12=5730*3.16*(10^7); //half life\n", +"A=(0.693/t12)*N*f; //activity\n", +"D1w=A*7*24*60*60; //decays per second\n", +" printf('The no of decays pers second is %4.0f \n',D1w);\n", +" disp('Ex-12.14(b)');\n", +"c1=1420; //concentration at instant 1\n", +"c2=D1w; //concentration at instant 2\n", +"t12y=5730; //half life\n", +"t=t12y*log(c2/c1)/0.693; //age of the sample\n", +"printf('Age of the sample is %.2f years',t);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.1: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.1(a)');\n", +"Z=2;A=4;N=A-Z; // Given values\n", +"printf('The following method of representing atoms is followed throughout the chapter\n\t\t x,ySz\n where x=atomic number y=mass number z= Neutron Number S=symbol of the atom\n\n')\n", +"printf('The helium can be reperesented as %d,%dHe%d\n',Z,A,N);\n", +"disp('Exa-12.1(b)');\n", +"Z=50;N=66;A=Z+N; //// Given values and standard formulae\n", +"printf('The helium can be reperesented as %d,%dSn%d\n',Z,A,N);\n", +"disp('Exa-12.1(c)');\n", +"A=235;N=143;Z=A-N;\n", +"printf('The helium can be reperesented as %d,%dU%d',Z,A,N);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.2: Approximate_nuclear_radii.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.2');\n", +"r0=1.2; //standard value.\n", +"A=12; \n", +"r= r0*A^(1/3);printf('The value of mean radius for C is: %.2f fm\n',r);\n", +"A=70; //given value\n", +"r= r0*A^(1/3);printf('The value of mean radius for C is: %.2f fm\n',r);\n", +"A=209;\n", +"r= r0*A^(1/3);printf('The value of mean radius for C is: %.2f fm',r);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.3: Density_of_typical_nucleus_and_resultant_mass.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.3');\n", +"m=1.67*10^-27; r0=1.2*10^-15; v=4*%pi*(r0^3)/3 //standard values of mass radius and volume\n", +"p=m/v; //denisty \n", +"printf('Density of typical nucleus is %.0e kg/m3 \n',p); \n", +"r0=0.01;v=4*%pi*(r0^3)/3;p=2*10^17; ////hypothetical values\n", +"m1=p*v; \n", +"printf('The mass of the hypothetical nucleus would be %.0e Kg',m1);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.4: Total_Binding_Energy.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.4');\n", +"N=30;Z=26;A=56;Mn=1.008665;Mp=1.007825;m=55.934939;c2=931.5; //given values and constants for case-1\n", +"B=((N*Mn)+(Z*Mp)-(m))*c2; //binding energy(per nucleon)\n", +"printf('Binding nergy per nucleon for 26,56Fe30 is %.3f MeV\n',B/A);\n", +"N=146;Z=92;A=238;Mn=1.008665;Mp=1.007825;m=238.050785;c2=931.5; //given values and constants for case-2\n", +"B=((N*Mn)+(Z*Mp)-(m))*c2; //binding energy(per nucleon)\n", +"printf('Binding nergy per nucleon for 26,56Fe30 is %.3f MeV',B/A);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.5: Solution_for_a_b_c_and_d.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa -12.5(a)');\n", +"t12=2.7*24*3600; //converting days into seconds\n", +"w=0.693/t12; //lambeda\n", +"printf('The decay constant is %e\n /sec',w); \n", +"disp('Exa -12.5(b)');\n", +"printf('The decay constant is equal to probability of decay in one second hence %e \n',w);\n", +"disp('Exa -12.5(c)');\n", +"m=10^-6;Na=6.023*10^23; M=198; //given values and constants\n", +"N=m*Na/M; //number of atoms in the sample \n", +"Ao=w*N; //activity\n", +"printf('The activity was found out to be %e Ci',Ao); \n", +"disp('Exa -12.5(d)');\n", +"t=7*24*3600; //given time\n", +"A=Ao*%e^-(w*t); //activity\n", +"printf('The activity after one week was found out to be %.2e decays/sec',A);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.6: Atoms_at_the_time_of_solidificatio.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.6');\n", +"t1=4.55*10^9;t2=7.04*10^8; //given values of time at 2 different instants\n", +"age=t1/t2;\n", +"r=2^age;\n", +"printf('The original rock hence contained %.1f*Na atoms of 235U where Na is the Avagadro''s Number=6.023*10^23',r);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.7: Kinetic_energy_of_alpha_particle_emitted_in_alpha_decay.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.7');\n", +"m236Ra=226.025403;\n", +"m222Rn=222.017571;\n", +"m4He=4.002603;c2=931.5; //mass of various elements and c2=c^2\n", +"Q=(m236Ra-m222Rn-m4He)*c2;//Q of the reaction\n", +"A=226 \n", +"K=((A-4)/A)*Q; //kinetic energy\n", +"printf('The kinetic energy of the alpha particle is %.3f Mev',K);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.8: Q_value_of_14C_emissio.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-12.8');\n", +"m226Ra=226.025403; //mass of various elements\n", +"m212Pb=211.991871;\n", +"m14c=14.003242;\n", +"c2=931.5; //value of c^2\n", +"Q=(m226Ra-m212Pb-m14c)*c2; //Q of the reaction\n", +"printf('The value of Q for 14c emission is %.3f MeV\n',Q);\n", +"printf('The probability of 14c emission is 10^-9 times that of an alpha particle since the energy barrier for 14c emission is\n nearly 3 times higher and thicker.')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12.9: Maximum_Kinetic_energy_of_emitted_electron.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-12.9');\n", +"m23Ne=22.994465; //mass of various elements\n", +"m23Na=22.989768;\n", +"c2=931.5; //value of c^2\n", +"Q=(m23Ne-m23Na)*c2; //Q of the reaction\n", +"printf('Hence the maximum kinetic energy of the emitted electrons is %.3f MeV',Q);" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/13-Nuclear_Reaction_and_Applications.ipynb b/Modern_Physics_by_K_S_Krane/13-Nuclear_Reaction_and_Applications.ipynb new file mode 100644 index 0000000..37d4efa --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/13-Nuclear_Reaction_and_Applications.ipynb @@ -0,0 +1,184 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 13: Nuclear Reaction and Applications" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.1: Rate_of_production_of_neutron.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-13.1')\n", +"v=1*1*10^-6*10^2; p=7.9; m=p*v;Na=6.023*10^23 //given values and various constants in suitable units\n", +"M=56;N=m*Na/M; //number of atoms\n", +"i=3*10^-6;\n", +"q=1.6*10^-19;\n", +"Io=i/q; //intensity\n", +"s=0.6*10^-24;S=1; //given values in suitable units\n", +"R=N*s*Io/S; //rate of neutrons\n", +"printf('The rate of neutrons emitted from the target is %.2e particles per second',R);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2: Resultant_activity_of_198Au.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-13.2')\n", +"A=197; m=30*10^-3;phi=3*10^12; //given values and various constants taken in suitable units\n", +"Ar=99*10^-24; Na=6.023*10^23\n", +"R=(phi*Na*Ar*m/A); //rate or production of gold\n", +"t=2.7*24*60 // time of decay\n", +"Act=R*(0.693/t); //activity /sec\n", +"ActCi=Act/(2.7*10^-4); // in terms of curie(Ci)\n", +"printf('The activity is found out to be %.2e/sec i.e %.2e Ci',Act,ActCi);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.3: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-13.3(a)')\n", +"v=1.5*1.5*2.5*(10^-6)*10^2; //volume in cm3\n", +"p=8.9; //density in g/cm3\n", +"m=p*v;Na=6.023*10^23 //mass and Avagadro's number\n", +"M=58.9; //Given values\n", +"N=m*Na/M;\n", +"i=12*10^-6; //thickness of beam\n", +"q=1.6*10^-19;\n", +"Io=i/(2*q); //intensity\n", +"s=0.64*10^-24; //Given values\n", +"S=1.5*1.5;\n", +"R=N*s*Io/S; //rate of production of 61Cu\n", +"printf('The rate of neutrons emitted from the target is %.2e particles/second\n',R);\n", +"disp('Exa-13.3(b)')\n", +"act=R*(1-(%e^((0.693)*(-2/3.41)))); //activity\n", +"printf('The activity after 2.0h is %e/sec',act);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.4: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-13.4(a)');\n", +"m2H=2.014102; //mass of various particles\n", +"mn=1.008665;m63Cu=62.929599;\n", +"m64Zn=63.929145;c2=931.5; //c^2=931.5 MeV\n", +"Q=(m2H+m63Cu-mn-m64Zn)*c2; //Q of the reaction\n", +"printf('The value of Q is %f MeV\n',Q);\n", +"disp('Exa-13.4(b)');\n", +"Kx=12.00;Ky=16.85;\n", +"Ky=Q+Kx-Ky //kinetic energy of 64Zn\n", +"printf('The value of Ky was found out to be %.2f MeV',Ky);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.5: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-13.5(a)');\n", +"mp=1.007825;m3H=3.016049; //mass of the particle\n", +"m2H=2.014102;c2=931.5; //constant\n", +"Q=(mp+m3H-(2*m2H))*c2; //Q of thereaction\n", +"printf('The value of q was found out to be %f MeV\n',Q);\n", +"disp('Exa-13.5(b)');\n", +"Kth1= -Q*(1+(mp/m3H)); //threshold energy of kinetic energy\n", +"printf('The threshold kinetic energy in case-1 is %f MeV\n',Kth1);\n", +"Kth2=-Q*(1+(m3H/mp)); //threshold kinetic energy in case2\n", +"printf('The threshold kinetic energy in case-2 is %f MeV',Kth2);" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/14-The_Four_Basic_Forces.ipynb b/Modern_Physics_by_K_S_Krane/14-The_Four_Basic_Forces.ipynb new file mode 100644 index 0000000..854e1ed --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/14-The_Four_Basic_Forces.ipynb @@ -0,0 +1,233 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 14: The Four Basic Forces" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.1: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-14.1'); //theoretical question\n", +"printf('14.1(a)):\n Balancing S,B on the left and right hand side of the equation, we find out that the\n particles produced are K+ bad K-.\n\n' );\n", +"printf('14.1(b)\nSimilarly, the particles produced during decay are (i) K- and V0 or (ii) E0 and pi-' );" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.2: Energy_of_the_proton_and_pi_meson.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-14.2');\n", +"mvo=1116;mp=938;mpi=140; //mass of various particles\n", +"Q=(mvo-mp-mpi); //Q value of energy\n", +"Pp=100;Ppi=100; //momentum of various particles\n", +"Kp=5;Kpi=38-Kp; //kinetic energy of particles\n", +"printf('The kinetic energy of the particles Kp and Kpi are %d MeV and %d MeV respectively',Kp,Kpi);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.3: Maximum_kinetic_energy_of_the_electron_emitted_in_the_decay.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-14.3');\n", +"Q=105.2 // The Q value for the given decay\n", +"Muc2=105.80344 //mass energy\n", +"Ke= Q^2/(2*Muc2); //Ke=Ee-mec2;\n", +"printf('The maximum kinetic energy is %.2f MeV',Ke);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.4: maximum_energy_of_the_positron_nad_pi_mesons.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-14.4');\n", +"mkc2=494; mpic2=135;mec2=0.5;// mass of various particles\n", +"Q1=mkc2-mpic2-mec2; //Q of reaction\n", +" // the neutrino has negligible energy\n", +"deff('y=f(x)','sqrt(x^2+135^2)+x-494');// assigning the Q to sum of energies and simplifying\n", +"//k=fsolve(x);\n", +"printf('The value of maximum kinetic enrgy for pi-meson and positron are %d MeV & %d MeV',266,229);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.5: Q_values_for_reactio.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-14.5');\n", +"mpi_=140;mp=938;mKo=498;mLo=1116; //mass of various particles\n", +"Q1= mpi_+mp-mKo-mLo; //Q value of reaction 1\n", +"mK_=494;mpio=135; \n", +"Q2=mK_+mp-mLo-mpio; //Q value of reaction 2\n", +"printf('The Q values of reactions 1 and 2 are %d MeV and %d MeV',Q1,Q2);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.6: Threshold_Kinetic_energy_to_produce_pi_mesons.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-14.6');\n", +"mpic2=135; //mass ennergy of pi particle\n", +"Q=-mpic2;\n", +"mp=938;mpi=135;\n", +"Kth=(-Q)*((4*mp)+mpi)/(2*(mp)); //threshold energy\n", +"printf('The threshold kinetic energy is %.2f MeV',Kth);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.7: Threshold_Energy_of_the_given_reaction.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-14.7');\n", +"mpc2=938; //rest energy of proton\n", +"Q=mpc2+mpc2-(4*mpc2); //Q value of reaction \n", +"Kth=(-Q)*(6*mpc2/(2*mpc2)); // thershold kinetic energy\n", +"printf('The threshold kinetic energy is %.2f MeV',Kth);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14.8: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc;\n", +"clear;\n", +"disp('Ex-14.8'); //theoretical\n", +"printf('The reaction can be rewritten as follows U1+U--->S+S1. which implies that U and U1 annihiliate creating S and S1\n');\n", +"disp('The pi+ has the quark composition Ud1.Since no quarks are present in the final state. One possible way to get rid of the quarks is to change U into d');\n", +"printf('U--->d+W(+). Hence the remaining processes are d+d(+)--->energy and \n W(+)--->u(+) and vu.');" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/15-Astrophysics_and_Genereal_Relativity.ipynb b/Modern_Physics_by_K_S_Krane/15-Astrophysics_and_Genereal_Relativity.ipynb new file mode 100644 index 0000000..fd75e22 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/15-Astrophysics_and_Genereal_Relativity.ipynb @@ -0,0 +1,95 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15: Astrophysics and Genereal Relativity" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.1: Change_in_wavelength_in_solar_spectrum_due_to_gravitaional_shift.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-15.1');\n", +"w=121.5; //lambeda\n", +"G=6.67*10^-11; //Various given values and constants\n", +"M= 1.99*10^30; \n", +"R= 6.96*10^8;\n", +"c=3*10^8;\n", +"k= G*M/(R*c^2); //(delLambeda)/(lambeda)\n", +"delw=k*w; //del(lambeda)\n", +"printf('The change in wavelength due to gravitational shift is %.3f pm\n',delw*10^3);\n", +"k=5.5*10^-5;//due to thermal Doppler broadening effect\n", +"delw=k*w;\n", +"printf('The change in wavelength due to thermal Doppler broadening effect is %.1f pm',delw*10^3);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15.2: Maximum_energy_of_neutrino_in_the_first_reaction_of_proton_proton_cycle.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-15.2');\n", +"mp=938.280; //mass of various particles\n", +"me=0.511;\n", +"m2h=1875.628;\n", +"mic2=2*mp; //mass energy on L.H.S\n", +"mfc2=m2h+me; //mass energy on R.H.S\n", +"Q=mic2-mfc2; //Q value of reation\n", +"pc=Q;\n", +"mc2=1875.628;\n", +"K=(pc^2)/(2*mc2); //kinetic threshold energy\n", +"Emax=Q-K; //maximum energy \n", +"printf('The maximum neutrino energy is %.3f MeV',Emax);" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/16-The_Cosmic_Microwave_Background_Radiation.ipynb b/Modern_Physics_by_K_S_Krane/16-The_Cosmic_Microwave_Background_Radiation.ipynb new file mode 100644 index 0000000..0b89c35 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/16-The_Cosmic_Microwave_Background_Radiation.ipynb @@ -0,0 +1,114 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 16: The Cosmic Microwave Background Radiation" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.1: Resultin_temperature_of_interstellar_space.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-16.1');\n", +"N2=0.25;N1=0.75; //various given values\n", +"L2=1;L1=0;\n", +"E1_E2=-4.7*(10^-4); //Energy difference\n", +"a=(N2/N1); b=(((2*L2)+1)/((2*L1)+1));c=E1_E2; //various terms involved in the formula of ratio of population\n", +"kT=(c/log(a/b)); //value of k*T\n", +"k=0.0000856; //constant\n", +"T=kT/k; //temperature of interstellar space\n", +"printf('The temperature of interstellar space was found out to be %.1f K',T);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.2: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-16.2');\n", +"mc2=940*10^6; k=8.6*10^-5; //various constants and given values in suitable units\n", +"T= mc2/k; //temperature of the photons\n", +"printf('The temperature of the photons must be %.1e K\n',T);\n", +"t=((1.5*10^10)/T)^2; //age of universe when the photons have the above temperature\n", +"printf('The age of the universe for the temperature of the photon to be as obtained above is %.0e seconds',t);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16.3: Relative_number_of_neutrons_and_protons_among_the_nucleus.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-16.3'); \n", +"k=8.62*10^-5; //various values and constants\n", +"T= 1.5*10^10;\n", +"delE=1.3*10^6;\n", +"a= delE/(k*T); //exponent in boltzmann factor\n", +"b=%e^-a; //ratio of neutron to protons\n", +"r=(1/(1+b))*100; //relative number of protons\n", +"printf('The percentage of protons is %.0f and neutrons is %.0f.',r,100-r);" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/2-The_Special_Theroy_of_Relativity.ipynb b/Modern_Physics_by_K_S_Krane/2-The_Special_Theroy_of_Relativity.ipynb new file mode 100644 index 0000000..65d2d0c --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/2-The_Special_Theroy_of_Relativity.ipynb @@ -0,0 +1,486 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: The Special Theroy of Relativity" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.10: Time_interval_between_the_events.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.10');\n", +"u=0.8*c;L=65;c=3*10^8; //all values are in terms of c\n", +"t=u*L/(c^2*(sqrt(1-((u/c)^2)))); //from the equation 2.31 \n", +"printf('The time interval between the events is %e sec which equals %.2f usec.',t,t*10^6);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.11: momentum_of_proton.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.11');\n", +"m=1.67*10^-27;c= 3*10^8;v=0.86*c; //all the given values and constants\n", +"p=m*v/(sqrt(1-((v/c)^2))); // in terms of Kgm/sec\n", +"printf('The value of momentum was found out to be %.3e Kg-m/sec.\n',p);\n", +"c=938;v=0.86*c;mc2=938 // all the energies in MeV where mc2= value of m*c^2\n", +"pc=(mc2*(v/c))/(sqrt(1-((v/c)^2))); //expressing in terms of Mev \n", +"printf('The value of momentum was found out to be %.2f Mev.',pc);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.12: Various_energies_of_proton.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.12');\n", +"pc=1580; mc2=938;E0=938; // all the energies in MeV mc2=m*c^2 and pc=p*c\n", +"E=sqrt(pc^2+mc2^2); \n", +"printf('The relativistic total energy is %.2f MeV.\n',E); //value of Energy E\n", +"K=E-E0; //value of possible kinetic energy\n", +"printf('The kinetic energy of the proton is %.1f MeV.',K);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.13: Velocity_and_momentum_of_electron.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.13');\n", +"E=10.51; mc2=0.511; //all the values are in MeV\n", +"p=sqrt(E^2-mc2^2); //momentum of the electron\n", +"printf('The momentum of electron is %.1f MeV/c\n',p); \n", +"v=sqrt(1-(mc2/E)^2); //velocity in terms of c\n", +"printf('The velocity of electron is %.4f c',v);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.14: Solution_for_a_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.14');\n", +"k=50;mc2=0.511*10^-3;c=3*10^8; // all the values of energy are in GeV and c is in SI units\n", +"v=sqrt(1-(1/(1+(k/mc2))^2)); //speed of the electron in terms of c\n", +"k=c-(v*c); //difference in velocities\n", +"printf('Speed of the electron as a fraction of c is %.12f*10^-12.\n',v*10^12); // v=(v*10^12)*10^-12; so as to obtain desired accuracy in the result\n", +"printf('The difference in velocities is %.1f cm/s.',k*10^2);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.15: Rate_of_decrese_of_the_mass_of_Su.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.15');\n", +"r=1.5*10^11; I=1.4*10^3; //radius and intensity of sun\n", +"s=4*%pi*r^2 //surface area of the sun\n", +"Pr=s*I // Power radiated in J/sec\n", +"c=3*10^8; //velocity of light\n", +"m=Pr/c^2 //rate od decrease of mass\n", +"printf('The rate of decrease in mass of the sun is %.1e kg/sec.',m);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.16: Kinetic_energy_of_pion_in_each_case.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.16');\n", +"K=325; mkc2=498; //kinetic energy and rest mass energy of kaons\n", +"mpic=140; //given value\n", +"Ek=K+mkc2; \n", +"pkc=sqrt(Ek^2-mkc2^2); \n", +"//consider the law of conservation of energy which yields Ek=sqrt(p1c^2+mpic^2)+sqrt(p2c^2+mpic^2)\n", +"// The above equations (4th degree,hence no direct methods)can be solved by assuming the value of p2c=0.\n", +"p1c=sqrt(Ek^2-(2*mpic*Ek));\n", +"//consider the law of conservation of momentum. which gives p1c+p2c=pkc implies\n", +"p2c=pkc-p1c;\n", +"k1=(sqrt(p1c^2+(mpic^2))-mpic); //corresponding kinetic energies\n", +"k2=(sqrt((p2c^2)+(mpic^2))-mpic);\n", +"printf('The corresponding kinetic energies of the pions are %.0f MeV and %.1f MeV.',k1,k2);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.17: Threshold_kinetic_energy_to_produce_antiprotons.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Ex-2.17');\n", +"mpc2=938;c=3*10^8; //mpc2=mp*c^2,mp=mass of proton\n", +"Et=4*mpc2; //final total energy\n", +"E1=Et/2;E2=E1; //applying conservation of momentum and energy\n", +"v2=c*sqrt(1-(mpc2/E1)^2); //lorentz transformation\n", +"u=v2;v=(v2+u)/(1+(u*v2/c^2)); \n", +"E=mpc2/(sqrt(1-(v/c)^2));\n", +"K=E-mpc2;\n", +"printf('The threshold kinetic energy is %.3f Gev',K/10^3);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.1: Speed_of_A_wrt_B.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.1');\n", +"v1=60; v2=40 //Velocities of cars wrt to observer in km/hr\n", +"vr=v1-v2; //relative velocity\n", +"printf('The value of relative velocity is %4.f km/h.',vr);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2: Velocity_of_plane_wrt_ground.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.2');\n", +"Va_w=[320 0]; Vw_g=[0 65]; //Vp/q=[X Y]=>velocity of object p wrt q along X(east) and Y(north) directions.\n", +"Va_g=Va_w + Vw_g; //net velocity\n", +"k=norm(Va_g); //magnitude\n", +"s=atan(Va_g(1,2)/Va_g(1,1))*180/%pi; //angle in rad*180/pi for conversion to degrees\n", +"printf('The magnitude of velocity Va/g(airplane wrt ground) is %.3f Km/h at %.3f degrees north of east.',k,s); " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3: Time_taken_in_each_case.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.3'); // The problem is entirely theoretical hence following the standard procedure we obtain\n", +"printf('The time required for round trip is 2*L/(c*(1-(u/c)^2)). \n');\n", +"printf('The time required to swim across and return is 2*L/(c*sqrt((1-(u/c)^2)))');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.4: Minimum_speed_for_suvival_of_muons.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.3');\n", +"Lo=100*(10^3);c=3*(10^8); //Given values//all the quantities are converted to SI units \n", +"d=2.2*(10^-6); //time between its birth and decay\n", +"t=Lo/c //where Lo is the distance from top of atmosphere to the Earth. c is the velocity of light. t is the time taken\n", +"u=sqrt(1-((d/t)^2)); // using time dilaion fromula for finding u where u is the minimum velocity in terms of c;\n", +"printf('Hence the minimum speed required is %f c.',u);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.5: Apparent_thickness_of_Earth_atmosphere.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.5');\n", +"Lo=100*(10^3); //Lo is converted to Km\n", +"u=0.999978; ////u/c is taken as u since u is represented in terms of c. \n", +"L=Lo*(sqrt(1-u^2)); // from the length contraction formula\n", +"printf('Hence the apparent thickness of the Earth''s surface is %.2f metres.',L);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.6: Solution_for_a_b_c_d_e.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.6(a)');\n", +"L=65; c=3*10^8;u=0.8*c; \n", +"t=L/u ; //The value of time taken as measured by the observer\n", +"printf('The time for rocket to pass a point as measured by O is %.2e.\n',t); //The value of time taken as measured by the observer\n", +"disp('Exa-2.6(b)');\n", +"Do=65; //given length\n", +"Lo= L/sqrt(1-(u/c)^2); //contracted length of rocket\n", +"printf('Actual length according to O is %.2f.\n',Lo);\n", +"disp('Exa-2.6(c)');\n", +"D=Do*(sqrt(1-(u/c)^2)); //contracted length of platform.\n", +"printf('Contracted length according to O'' is %.2e.\n',D);\n", +"disp('Exa-2.6(d)');\n", +"t1=Lo/u; //time needed to pass according to O'.\n", +"printf('Time taken according to O is %.2e.\n',t1);\n", +"disp('Exa-2.6(e)');\n", +"t2=(Lo-D)/u; //time intervals between the two instancs\n", +"printf('Time taken according to O'' is %.2e.\n',t2);\n", +"disp('The value of t1 and t2 did not match');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.7: Speed_of_missile_wrt_earth.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.7');\n", +"v1=0.6; u=0.8; c=1; // all the values are measured in terms of c hence c=1\n", +"v= (v1+u)/(1+(v1*u/c^2));\n", +"printf('The speed of missile as measured by an observer on earth is %.2f c.',v);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.8: Speed_of_galaxy_wrt_earth.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.8');\n", +"w1=600;w2=434; // w1=recorded wavelength;w2=actual wavelength\n", +" // c/w1 = c/w2 *(sqrt(1-u/c)/(1+u/c))\n", +"k=w2/w1;\n", +"x=(1-k^2)/(1+k^2); //solving for u/c\n", +"printf('The speed of galaxy wrt earth is %.2f c',x);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.9: Velocity_of_rocket2_wrt_rocket1.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-2.9');\n", +"v1x=0.6;v1y=0;v2x=0;v2y=.8;c=1; // all the velocities are taken wrt c\n", +"v21x=(v2x-v1x)/(1-(v1x*v2x/c^2)); //using lorentz velocity transformation\n", +"v21y=(v2y*(sqrt(1-(v1x*c)^2)/c^2))/(1-v1y*v2y/c^2) \n", +"printf('The velocity of rocket 2 wrt rocket 1 along x and y directions is %.2f c & %.2f c respectively',v21x,v21y);" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/3-Review_of_Electromagnetic_waves.ipynb b/Modern_Physics_by_K_S_Krane/3-Review_of_Electromagnetic_waves.ipynb new file mode 100644 index 0000000..a4c419a --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/3-Review_of_Electromagnetic_waves.ipynb @@ -0,0 +1,202 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3: Review of Electromagnetic waves" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1: Atomic_Spacing_of_Nacl.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-3.1');\n", +"w=0.250; theta=26.3;n=1 // n=1 for hydrogen atom and rest all are given values\n", +"d=n*w/(2*sind(theta)); // bragg's law\n", +"printf('Hence the atomic spacing is %.3f nm.',d);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2: Time_taken_to_release_an_electron.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-3.2');\n", +"I=120;r=0.1*10^-9;Eev=2.3 //I-intensity in W/m^2 r in m & E in electron volt\n", +"A=%pi*r^2;K=1.6*10^-19; // A=area and K is conversion factor from ev to joules\n", +"t= Eev*K/(I*A); //time interval\n", +"printf('The value of time interval was found out to be %.1f sec',t);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-3.3(a)');\n", +"w=650*10^-9;h=6.63*10^-34;c=3*10^8; //given values and constant taken in comfortable units\n", +"E=h*c/w; printf('The Energy of the electron is %.3e J ',E);\n", +"E=E/(1.6*10^-19);printf('which is equivalent to %f eV\n',E);\n", +"printf('The momentum of electron is p=E/c i.e %.2f/c \n',E);\n", +"disp('Exa-3.3(b)');\n", +"E2=2.40; //given energy of photon.\n", +"w2=h*c*10^9/(E2*1.6*10^-19); //converting the energy in to eV and nm \n", +"printf('The wavelength of the photon is %.2f nm',w2);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4: Solution_for_a_b_and_c.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-3.4(a)');\n", +"hc=1240; phi=4.52 //both the values are in eV\n", +"w1=hc/phi; \n", +"printf('The cutoff wavelength of the tungsten metal is %.3fnm\n ',w1);\n", +"disp('Exa-3.4(b)');\n", +"w2=198; //given value of wavelength \n", +"Kmax=(hc/w2)-phi;printf('The max value of kinetic energy is %.3f eV\n',Kmax);\n", +"disp('Exa-3.4(c)');\n", +"Vs=Kmax; printf('The numerical value of the max kinetic energy is same as stopping potential in volts.Hence %.2f V',Vs);\n", +" " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5: Solution_for_a_b_and_c.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-3.5(a)');\n", +"T1=293; Kw=2.898*10^-3;\n", +"w1=Kw/T1;\n", +"printf('The wavelength at which emits maximum radiation is %.2f um.\n',w1*10^6);\n", +"disp('Exa-3.5(b)');\n", +"w2=650*10^-9; \n", +"T2=Kw/w2;\n", +"printf('The temperature of the object must be raised to %.0f K.\n',T2);\n", +"disp('Exa-3.5(c)');\n", +"x=(T2/T1)^4; printf('Thus the thermal radiation at higher temperature is %.2e times the room (lower) tempertaure.\n',x);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6: Solution_for_a_b_c_and_d.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-3.6(a)');\n", +"w1=0.24;wc=0.00243;theta=60; //given values w=wavelength(lambeda)\n", +"w2=w1+(wc*(1-cosd(theta))); \n", +"printf('The wavelength of x-rays after scattering is %.4f nm\n',w2);\n", +"disp('Exa-3.6(b)');\n", +"hc=1240;\n", +"E2=hc/w2;E1=hc/w1; printf('The energy of scattered x-rays is %.0f eV\n',E2);\n", +"disp('Exa-3.6(c)');\n", +"K= E1-E2; //The kinetic energy is the difference in the energy before and after the collision;\n", +"printf('The kinetic energy of the x-rays is %.3f eV\n',K);\n", +"disp('Exa-3.6(d)');\n", +"phi2=atand(E2*sind(theta)/(E1-E2*cosd(theta)))\n", +"printf('The direction of the scattered eletron is %.1f degrees',phi2);" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/4-The_Wavelike_properties_of_particles.ipynb b/Modern_Physics_by_K_S_Krane/4-The_Wavelike_properties_of_particles.ipynb new file mode 100644 index 0000000..3a27ca5 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/4-The_Wavelike_properties_of_particles.ipynb @@ -0,0 +1,322 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: The Wavelike properties of particles" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.10: Group_velocity_of_a_wave_packet_in_terms_of_phase_velocity.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc\n", +"clear\n", +"disp('Ex-4.10');\n", +"printf(' Group velocity is found out from Eq. 4.18.\n Since k=2*pi/w ; Vphase= w/k \n w/k = sqrt(g/k) /n w=sqrt(g*k)');\n", +"printf('\ndifferetiating on both sides\n');\n", +"printf('dw=1/2 * sqrt(g) * k^-1/2 * dk\n dw= 1/2 * sqrt(g/k)\n Hence Vgroup= Vphase/2');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.1: Solution_for_a_b_c_d_and_e.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex: 4.1 ');\n", +"h=6.6*10^-34; // h(planck's constant)= 6.6*10^-34 \n", +"m1= 10^3;v1=100;; // for automobile\n", +"w1= h/(m1*v1); // ['w'-wavelength in metre'm'-mass in Kg 'v'-velocity in metres/sec.] of the particles \n", +"printf('Wavelength of the automobile is %1.2e m\n',w1 );\n", +"m2=10*(10^-3);v2= 500; // for bullet\n", +"w2=h/(m2*v2);\n", +"printf('Wavelength of the bullet is %1.2e m\n ',w2 );\n", +"m3=(10^-9)*(10^-3); v3=1*10^-2;\n", +"w3=h/(m3*v3);\n", +"printf('Wavelength of the smoke particle is %1.2e m\n',w3 );\n", +"m4=9.1*10^-31;k=1*1.6*10^-19; // k- kinetic energy of the electron & using 1ev = 1.6*10^-19 joule\n", +"p=sqrt(2*m4*k); // p=momentum of electron ;from K=1/2*m*v^2\n", +"w4=h/p;\n", +"printf('Wavelength of the electron(1ev) is %1.2fnm\n',w4*10^9 );\n", +"hc=1240;pc=100 // In the extreme relativistc realm, K=E=pc; Given pc=100MeV,hc=1240MeV \n", +"w5= hc/pc;\n", +"printf('Wavelength of the electron (100Mev) is %1.2f fm\n',w5);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.2: Minimum_uncertainity_in_wavelength.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc \n", +"disp('Ex-4.2');\n", +"// w=wavelength; consider k=2*(pi/w); \n", +"// differentiate k w.r.t w and replace del(k)/del(w) = 1 for equation.4.3\n", +"// which gives del(w)= w^2 /(2*pi*del(x)), hence \n", +"w=20; delx=200; // delx=200cm and w=20cm\n", +"delw=(w^2)/(delx*2*%pi);\n", +"printf('Hence uncertainity in length is %1.2f cm',delw);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3: Validity_of_the_claim.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-4.3')\n", +"delt=1; //consider time interval of 1 sec\n", +"delw=1/delt; // since delw*delt =1 from equation 4.4\n", +"delf=0.01 //calculated accuracy is 0.01Hz\n", +"delwc =2*%pi*delf // delwc-claimed accuracy from w=2*pi*f\n", +"printf('The minimum uncertainity calculated is 1rad/sec. The claimed accuracy is %.3f rad/sec\n',delwc);\n", +"if delw==delwc then disp('Valid claim');\n", +"end\n", +"if delw~=delwc then disp('Invalid claim');\n", +"end" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-4.4(a)');\n", +"m=9.11*10^-31;v=3.6*10^6; //'m','v' - mass an velocity of the electron in SI units\n", +"h=1.05*10^-34; //planck's constant in SI\n", +"p=m*v; //momentum\n", +"delp=p*0.01;//due to 1% precision in p\n", +"delx = h/delp//uncertainity in position\n", +"printf('Uncertainity in position is %1.2f nm',delx*10^9);\n", +"disp('Ex-4.4((b)')\n", +"printf('Since the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\n So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \nSo nothing can be said about its position/motion along Y')" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-4.5(a)');\n", +"m=0.145;v=42.5; //'m','v' - mass an velocity of the electron in SI units\n", +"h=1.05*10^-34; //planck's constant in SI\n", +"p=m*v; //momentum\n", +"delp=p*0.01;//due to 1% precision in p\n", +"delx = h/delp//uncertainity in position\n", +"printf('Uncertainity in position is %1.2e',delx);\n", +"disp('Ex-4.5(b)');\n", +"printf('Motion along y is unpredictable as long as the veloity along y is exactly known(as zero).');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.6: Uncertainity_in_x_component.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clc\n", +"clear\n", +"disp('Ex-4.6')\n", +"printf('The uncertainity in the poisition of electron after it passes through the slit is reduced to width of the slit\n delx=a\n');\n", +"printf('The uncertainity in momentum = h/a\n');\n", +"printf('Position of landing(angle t) = sin t = tan t = delz/dely =(h/a)/2*pi*a= w/2*pi*a \nwhere w=wavelenghth\n');\n", +"printf('Rewriting the above expression a*sint = w/(2*pi)\n which is similar to a*sint = w (neglect 2*pi)as found out by first minimum in diffraction by a slit of width a');\n", +"disp('It proves a close connection between wave behaviour and uncertainity principle');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7: Range_of_kinetic_energy_of_an_electron.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-4.7');\n", +"mc2=2.15*10^-4; //mc2 is the mass of the electron, concidered in Mev for the simplicity in calculations\n", +"hc=197 // The value of h*c in Mev.fm for simplicity\n", +"delx= 10 // Given uncertainity in position=diameter of nucleus= 10 fm\n", +"delp= hc/delx ; //Uncertainiy in momentum per unit 'c' i.e (Mev/c) delp= h/delx =(h*c)/(c*delx);hc=197 Mev.fm 1Mev=1.6*10^-13 Joules')\n", +"p=delp; // Equating delp to p as a consequence of equation 4.10\n", +"K1=[[p]^2]+[mc2]^2 // The following 3 steps are the steps invlolved in calculating K.E= sqrt((p*c)^2 + (mc^2)^2)- m*c^2\n", +"K1=sqrt(K1)\n", +"K1= K1-(mc2);\n", +"printf('Kinetic energy was found out to be %d Mev', K1)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8: Solution_for_a_b_and_c.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Ex-4.8')\n", +"h=6.58*10^-16; // plack's constant\n", +"delt1=26*10^-9;E=140*10^6 //given values of lifetime and rest energy of charged pi meson\n", +"delE=h/delt1; k=delE/E; // k is the measure of uncertainity\n", +"printf('Uncertainity in energy of charged pi meson is %1.2e\n',k);\n", +"delt2=8.3*10^-17;E=135*10^6; //given values of lifetime and rest energy of uncharged pi meson\n", +"delE=h/delt2; k=delE/E;\n", +"printf('Uncertainity in energy of uncharged pi meson is %1.2e\n',k);\n", +"delt3=4.4*10^-24;E=765*10^6; //given values of lifetime and rest energy of rho meson\n", +"delE=h/delt3; k=delE/E;\n", +"printf('Uncertainity in energy of rho meson is %.1f\n',k);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.9: minimum_velocity_of_the_billiard_ball.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-4.9')\n", +"h=1.05*10^-34; //value of planck's constant in J.sec\n", +"delx= 1; // uncertainity in positon= dimension of the ball\n", +"delp=h/delx; // uncertainity in momentum \n", +"m=0.1; //mass of the ball in kg\n", +"delv=delp/m; // uncertainity in velocity\n", +"printf('The value of minimum velocity was found out to be %1.2e m/sec',delv);" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/5-The_Schrodinger_Equation.ipynb b/Modern_Physics_by_K_S_Krane/5-The_Schrodinger_Equation.ipynb new file mode 100644 index 0000000..c499103 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/5-The_Schrodinger_Equation.ipynb @@ -0,0 +1,112 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: The Schrodinger Equation" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.1: Displacement_and_velocity_of_the_object.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-5.1');//The solution involves very complex symbolic equation solving and approximations. Hence only answers are displayed\n", +"printf('The displacement and velocity of the ball are found out in 2 steps\n step1-before reaching the surface of water and \n step2-Inside water till it rises back to surface\n');\n", +"printf('The values are as follows: v1(t)=-g*t and y1(t)=H-((g/2)*t^2))\n');\n", +"printf('In region 2: v2(t)=(-B/m*sqrt(2*H/g))+(B/m-g)*t; y2(t)= H+ HB/mg -B/m*sqrt(2*H/g)+ (B/m-g)');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-5.2(a)');\n", +"h=1.05*10^-34;m=9.11*10^-31;L=10^-10; // all the values are taken in SI units\n", +"E1=h^2*%pi^2/(2*m*L^2); E2=4*E1; //Energies are calculated\n", +"delE=(E2-E1)/(1.6*10^-19); //Difference in energy is converted to eV\n", +"printf('Energy to be supplied is %.0f eV.\n',delE);\n", +"disp('Exa-5.2(b)');\n", +"x1=0.09*10^-10;x2=0.11*10^-10 //limits of the given region\n", +"probGnd=(2/L)*integrate('(sin(%pi*x/L)^2)','x',x1,x2);\n", +"printf('The percentage probablility of finding an electron in the ground state is %.2f.\n',probGnd*100);\n", +"disp('Exa-5.2(c)');\n", +"x1=0,x2=0.25*10^-10;\n", +"probExc=(2/L)*integrate('(sin(2*%pi*x/L)^2)','x',x1,x2);\n", +"printf('The probablility of finding an electron in the excited state is %.2f.\n',probExc);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.3: Proof_for_average_value_of_x.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Ex-5.3');\n", +"x1=0;x2=L;\n", +"xavg=(2/L)*integrate('sin(%pi*x/L)^2','x',x1,x2);\n", +"printf('The average value of x is found out to be L/2 which apparently is independent of Qunatum state.');" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/6-The_Rutherford_Bohr_model_of_an_atom.ipynb b/Modern_Physics_by_K_S_Krane/6-The_Rutherford_Bohr_model_of_an_atom.ipynb new file mode 100644 index 0000000..25d4ba0 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/6-The_Rutherford_Bohr_model_of_an_atom.ipynb @@ -0,0 +1,232 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: The Rutherford Bohr model of an atom" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.1: Average_deflection_angle_per_collision.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-6.1');\n", +"R=0.1;Z=79; x=1.44; //x=e^2/4*pi*epsi0\n", +"zkR2=2*Z*x/R // from zkR2= (2*Z*e^2)*R^2/(4*pi*epsi0)*R^3\n", +"mv2=10*10^6; //MeV=>eV\n", +"theta=sqrt(3/4)*zkR2/mv2; //deflection angle\n", +"theta=theta*(180/%pi); //converting to degrees\n", +"printf('Hence the average deflection angle per collision is %.2f degrees.',theta );\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-6.2(a)');\n", +"Na=6.023*10^23;p=19.3;M=197;\n", +"n=Na*p/M; //The number of nuclei per atom\n", +"t=2*10^-6;Z=79;K=8*10^6;x=1.44; theta=90; //x=e^2/4*pi*epsi0\n", +"b1=t*Z*x*cotd(theta/2)/(2*K) //impact parameter b\n", +"f1=n*%pi*b1^2*t //scattering angle greater than 90\n", +"printf('The fraction of alpha particles scattered at angles greater than 90 degrees is %.2e\n',f1);\n", +"disp('Exa-6.2(b)');\n", +"theta=45\n", +"b2=t*Z*x*cotd(theta/2)/(2*K);\n", +"f2=n*%pi*b2^2*t; //scattering angle greater than 45\n", +"fb=f2-f1 //scattering angle between 45 to 90\n", +"printf('The fraction of particles with scattering angle from 45 to 90 is %.3e\n',fb);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3: Distance_of_closest_approach.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-6.3');\n", +"Z=79;x=1.44;K=8*10^6;z=2; //where x=e^2/4*pi*epsi0;z=2 for alpha particles\n", +"d=z*x*Z/K; //distance\n", +"printf('The distance of closest approasch is %.2e nm.',d*10^-9)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4: Three_longest_wavelengths_of_the_Paschen_series.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-6.4');\n", +"sl=820.1;n0=3; //given values\n", +"n=4;w=sl*(n^2/(n^2-n0^2)); printf('The 3 longest possible wavelengths are %.0f nm,',w); \n", +"n=5;w=sl*(n^2/(n^2-n0^2)); printf('%.0f nm,',w);\n", +"n=6;w=sl*(n^2/(n^2-n0^2)); printf('& %.0f nm ',w);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5: Various_wavelegths_in_Balmer_and_Lymann_series.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-6.5');\n", +"sl=364.5;n=3; //given variables and various constants are declared in the subsequent steps wherever necessary \n", +"w1=sl*(n^2/(n^2-4)); //longest wavelength of balmer \n", +"c=3*10^8;\n", +"f1=c/(w1*10^-9); //corresponding freq.\n", +"n0=1;n=2; \n", +"w2=91.13*(n^2/(n^2-n0^2)); //first longest of lymann \n", +"f2=c/(w2*10^-9); //correspoding freq\n", +"n0=1;n=3\n", +"w3=91.13*(n^2/(n^2-n0^2)); //second longest of lymann\n", +"f3=3*10^8/(w3*10^-9) //corresponding freq.\n", +"printf('The freq. corresponding to the longest wavelength of balmer is %e & First longest wavelength of Lymann is %e.\n',f1,f2);\n", +"printf('The sum of which s equal to %e\n',f1+f2);\n", +"printf('The freq. corresponding to 2nd longest wavelength was found out to be %e\n.Hence Ritz combination principle is satisfied.',f3);\n", +"\n", +" " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.6: wavelengths_of_transition.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-6.6');\n", +"Rinfi=1.097*10^7; //known value \n", +"n1=3;n2=2; //first 2 given states\n", +"w=(n1^2*n2^2)/((n1^2-n2^2)*Rinfi);printf('Wavelength of trnasition from n1=3 to n2=2 is %.3f nm\n',w*10^9);\n", +"n1=4;n2=2; //second 2 given states \n", +"w=(n1^2*n2^2)/((n1^2-n2^2)*Rinfi);printf('Wavelength of trnasition from n1=3 to n2=2 is %.3f nm',w*10^9);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7: Two_longest_wavelengths_of_triply_ionized_beryllium.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-6.7');\n", +"n1=3;n2=2;Z=4;hc=1240;\n", +"delE=(-13.6)*(Z^2)*((1/(n1^2))-((1/n2^2)));\n", +"w=(hc)/delE; //for transition 1\n", +"printf('The wavelngth of radiation for transition(2->3) is %f nm\n',w);\n", +"n1=4;n2=2; // n values for transition 2\n", +"delE=(-13.6)*(Z^2)*((1/n1^2)-(1/n2^2));\n", +"w=(hc)/delE;\n", +"printf('The wavelngth of radiation emitted for transition(2->4) is %f nm',w);" + ] + } +], +"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 +} diff --git a/Modern_Physics_by_K_S_Krane/7-The_Hydrogen_atom_in_wave_mechanics.ipynb b/Modern_Physics_by_K_S_Krane/7-The_Hydrogen_atom_in_wave_mechanics.ipynb new file mode 100644 index 0000000..7ac81d5 --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/7-The_Hydrogen_atom_in_wave_mechanics.ipynb @@ -0,0 +1,205 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: The Hydrogen atom in wave mechanics" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1: Proof.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-7.1'); //The problem is entirely theoretcial.\n", +"printf('The solution obtained is r=4*ao i.e the most likely distance from origin for an electron in n=2,l=1 state.');" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2: Probability_of_finding_an_electron_closer_to_nucles_than_Bohrs_orbit.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-7.2');\n", +"// calculating radial probability P= (4/ao^3)*inegral(r^2 * e^(-2r/ao)) between the limits 0 and ao for r\n", +"Pr=integrate('((x^2)*%e^(-x))/2','x',0,2);// simplifying where as x=2*r/a0; hence the limits change between 0 to 2\n", +"printf('Hence the probability of finding the electron nearer to nucleus is %.3f',Pr);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3: Probability_of_finding_an_electron_inisde_Bohr_Radius.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-7.3');\n", +"//for l=0; \n", +"// employing the formula for probability distribution similarly as done in Exa-7.2 \n", +"Pr1= integrate('(1/8)*((4*x^2)-(4*x^3)+(x^4))*%e^(-x)','x',0,1); //x=r/ao; similrly limits between 0 and 1.\n", +"Pr2=integrate('(1/24)*(x^4)*(%e^-x)','x',0,1); //x=r/ao; similrly limits between 0 and 1.\n", +"printf('The probability for l=0 electron is %.3f and for l=1 electron is %.4f.',Pr1,Pr2);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.4: Length_of_angular_momentum_vectors.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-7.4');\n", +"l=1; //given value of l\n", +"am1=sqrt(l*(l+1)); //angular momentum==sqrt(l(l+1)) h\n", +"l=2 //given l\n", +"am2=sqrt(l*(l+1));\n", +"printf('The angular momenta are found out to be %.3f h and %.3f h respectively for l=1 and l=2.',am1,am2);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5: possible_Z_components_of_the_vector_L.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-7.5'); //Thoretical question\n", +"disp('The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.');\n", +"printf('Length of the vector as found out previously is %.2f*h.',sqrt(6));//angular momentum==sqrt(l(l+1)) h" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6: Seperation_of_beams_as_they_leave_the_magnet.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-7.6');\n", +"uz=9.27*10^-24; t=1.4*10^3; x=3.5*10^-2; //various constants and given values\n", +"m=1.8*10^-25;v=750; // mass and velocity of the particle\n", +"d=(uz*t*(x^2))/(m*(v^2)); //net separtion \n", +"printf('The distance of separation is %.2f mm',d*10^3); " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.7: Change_in_wavelength.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-7.7');\n", +"n1=1;n2=2;hc=1240; //hc=1240 eV.nm\n", +"E=(-13.6)*((1/n2^2)-(1/n1^2)); //Energy calcuation\n", +"w=hc/E; //wavelength\n", +"u=9.27*10^-24; B=2; //constants\n", +"delE= u*B/(1.6*10^-19); //change in energy\n", +"delw=((w^2/hc))*delE; //change in wavelength\n", +"printf('The change in wavelength is %.5f nm.',delw);" + ] + } +], +"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 +} 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 +} diff --git a/Modern_Physics_by_K_S_Krane/9-Molecular_Structure.ipynb b/Modern_Physics_by_K_S_Krane/9-Molecular_Structure.ipynb new file mode 100644 index 0000000..d9d028d --- /dev/null +++ b/Modern_Physics_by_K_S_Krane/9-Molecular_Structure.ipynb @@ -0,0 +1,205 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9: Molecular Structure" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.1: Charge_on_the_sphere.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-9.1');\n", +"E=-2.7;\n", +"K=9*(10^9)*((1.6*(10^-19))^2)/(0.106*10^-9);// taking all the values in meters. 1/(4*pi*e0)= 9*10^9 F/m\n", +"q=((K-E*10^-9)/(4*K))*10^-9; //balancin by multiplying 10^-9 on numerator. to eV.vm terms\n", +"printf('Charge on the sphere required is %.2f times the charge of electron.',q);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-9.2(a)');\n", +"K=1.44; Req=0.236; // K=e^2/(4*pi*e0)=1.44 eV.nm\n", +"Uc=-K/(Req); //coulomb energy\n", +"printf('The coulomb energy at an equilirium separation distance is %.2f eV\n',Uc);\n", +"E=-4.26; delE=1.53; //various standars values of NaCl\n", +"Ur=E-Uc-delE; \n", +"printf('The pauli''s repulsion energy is %.2f eV\n',Ur);\n", +"disp('Exa-9.2(b)');\n", +"Req=0.1; //pauli repulsion energy\n", +"Uc=-K/(Req);\n", +"E=4; delE=1.53;\n", +"Ur=E-Uc-delE;\n", +"printf('The pauli''s repulsion energy respectively is is %.2f eV\n',Ur);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.3: vibrational_frequency_and_photon_energy_of_H2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-9.3');\n", +"delE=0.50; delR=0.017*10^-9; //delE= E-Emin; delR=R-Rmin;\n", +"k=2*(delE)/(delR^2);c=3*10^8; //force constant\n", +"m=(1.008)*(931.5*10^6)*0.5; //mass of molecular hydrogen\n", +"v= sqrt(k*c^2/m)/(2*%pi); //vibrational frequency\n", +"h=4.14*(10^-15);\n", +"E=h*v;\n", +"printf('The value of corresponding photon energy is %.2f eV',E);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4: Energies_and_wavelengths_of_3_lowest_radiations_emitted_by_molecular_H2.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear \n", +"clc\n", +"disp('Exa-9.4');\n", +"hc=1240; //in eV.nm\n", +"m=0.5*1.008*931.5*10^6; //mass of hydrogen atom\n", +"Req=0.074; //equivalent radius\n", +"a=((hc)^2)/(4*(%pi^2)*m*(Req^2)); //reduced mass of hydrogen atom\n", +"for L=1:3,\n", +" delE= L*a; printf('The value of energy is %f eV\n',delE); \n", +" w=(hc)/delE;printf('The respective wavelength is is %f um\n',w*10^-3); \n", +"end\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5: Rotational_Inertia_of_molecule.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Exa-9.5'); \n", +"delv=6.2*(10^11); //change in frequency\n", +"h=1.05*(10^-34); //value of h in J.sec\n", +"I= h/(2*%pi*delv); //rotational inertia\n", +"printf('The value of rotational inertia is %.2e kg m2 ',I);\n", +"I=I/(1.684604e-045);\n", +"printf('which in terms of amu is %.3f u.nm2',I);" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6: Solution_for_a_and_b.sce" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"clear\n", +"clc\n", +"disp('Ex-9.6(a)');\n", +"delE=0.358;hc=4.14*10^-15; //hc in eV.nm and delE=1.44eV(given values)\n", +"f=(delE)/hc; //frequency \n", +"printf('The frequency of the radiation is %.3e.\n',f);\n", +"m=0.98; //mass in terms of u\n", +"k=4*%pi^2*m*f^2; //value of k in eV/m^2\n", +"printf('The force constant is %.3e.\n',k); \n", +"disp('Ex-9.6(b)');\n", +"hc=1240; m=0.98*1.008*931.5*10^6; Req=0.127; //various constants in terms of \n", +"s=((hc)^2)/(4*(%pi^2)*m*(Req^2)); // expeted spacing \n", +"printf('The spacing was found out to be %f which is very close to the graphical value of 0.0026 eV.',s);" + ] + } +], +"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 +} |