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diff --git a/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/1-The_Nucleus.ipynb b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/1-The_Nucleus.ipynb new file mode 100644 index 0000000..1c8c00b --- /dev/null +++ b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/1-The_Nucleus.ipynb @@ -0,0 +1,684 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1: The Nucleus" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.11_1: Ion_accelerated_in_a_mass_spectrograph.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.11.1 : To find the speed, mass and mass number of the ion which is accelerated in a mass spectrograph : Page 40 (2011)\n", +"V = 1000; // Potential difference, volts\n", +"R = 0.122; // Radius of the circular path, m\n", +"B = 1500e-04; // Magnetic field, tesla\n", +"e = 1.602e-019; // Charge of the electron, C\n", +"amu = 1.673e-027; // Atomic mass unit, kg\n", +"v = (2*V)/(R*B); // Speed of the ion, m/s\n", +"M = 2*e*V/v^2; // Mass of the ion, kg\n", +"A = M/amu; // Mass number\n", +"printf('\n Speed > %5.3e m/s \n Mass > %5.3e kg \n Mass number > %5.2f ',v, M, A);\n", +"\n", +"// Result\n", +"// \n", +"// Speed > 1.093e+005 m/s \n", +"// Mass > 2.682e-026 kg \n", +"// Mass number > 16.03" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.11_2: Distance_between_isotopic_Ar_ions.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 1.11.2 : To determine distances between the isotopic Ar ions in Bainbridge mass spectrograph : Page 41 (2011)\n", +"amu = 1.673e-027; // Atomic mass unit, kg\n", +"E = 5e+04; // Electric field, V/m\n", +"B1 = 0.4; // Magnetic field, tesla\n", +"v = E/B1; // Velocity of ions, m/s\n", +"B = 0.8; // Magnetic field, tesla\n", +"e = 1.602e-019; //charge of electron,C\n", +"m_Ar = zeros(1,3); // Array of masses of three Ar ions, amu\n", +"m_Ar(1,1) = 36,m_Ar(1,2) = 38,m_Ar(1,3) = 40; // Masses of three isoptopes of Ar, amu\n", +"r_Ar = zeros(1,3); // Array of radii of three Ar ions, mm\n", +"for i = 1:1:3\n", +" r_Ar(1,i) = (m_Ar(1,i)*amu*v)/(B*e)*1e+03; // Radius of Ar ion orbit, mm\n", +" disp(r_Ar(1,i)); \n", +"end\n", +"d1 = 2*(r_Ar(1,2)-r_Ar(1,1)); // Distance b/w first and second line, mm\n", +"d2 = 2*(r_Ar(1,3)-r_Ar(1,2)); // Distance b/w second and third line, mm\n", +"printf('\nThe distance between successive lines due to three different isotopes : %3.1f mm and %3.1f mm', d1,d2);\n", +"\n", +"// Result\n", +"// The distance between successive lines due to three different isotopes : 6.5 mm and 6.5 mm" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.3_1: de_Broglie_relatio.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.3.1 Momentum determination for a neutron using de-Broglie relation : Page 31 (2011)\n", +"h = 6.626e-034; // Planck's constant, Js\n", +"e = 1.602e-019; // Charge on an electron, C\n", +"red_h = h/(2*%pi*e*1e+06); // Reduced Planck's constant, MeV\n", +"lambda = 5.0e-015; // de_Broglie wavelength of neutron, m\n", +"p = red_h/lambda; // Momentum of the neutron, MeV-s/m\n", +"printf('\nThe momentum of the neutron from de-Broglie relation : %5.3e MeV-s/m', p);\n", +"\n", +"// Result\n", +"// The momentum of the neutron from de-Broglie relation : 1.317e-007 MeV-s/m" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.3_2: Isotopes_Isotones_and_Isobars.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.3.2 : Grouping the nuclides as isotopes, isotones and isobars : Page 32 (2011)\n", +"E = cell(3,3); // Declare a cell array of empty matrices for nuclides information\n", +"E(1,1).entries = 'C'; // Assign element 'C' to (1,1) cell\n", +"E(2,1).entries = 'N'; // Assign element 'N' to (2,1) cell\n", +"E(3,1).entries = 'O'; // Assign element 'o' to (3,1) cell\n", +"E(1,2).entries = 6; // Assign atomic No. 6 to (1,2) cell\n", +"E(2,2).entries = 7; // Assign atomic No. 7 to (2,2) cell\n", +"E(3,2).entries = 8; // Assign atomic No. 8 to (3,2) cell\n", +"E(1,3).entries = [12,13,14,16]; // Assign mass numbers for 'C' to (1,3) cell\n", +"E(2,3).entries = [14,15,16,17]; // Assign mass numbers for 'N' to (2,3) cell\n", +"E(3,3).entries = [14,15,16,17]; // Assign mass numbers for 'O' to (3,3) cell\n", +"// Isotopes\n", +"printf('\nIsotopes:');\n", +"printf('\n=========');\n", +"for i = 1:1:3 // Search for the three elements one-by-one\n", +" printf('\n(Z = %d)\n',E(i,2).entries);\n", +" for j= 1:1:4\n", +" printf('\t%s(%d)',E(i,1).entries,E(i,3).entries(j));\n", +" end\n", +"end\n", +"// Isotones\n", +"printf('\n\nIsotones:');\n", +"printf('\n========');\n", +"for N = 6:1:9 // Search for the neutron numbers from 6 to 9\n", +" printf('\n(N = %d)\n',N);\n", +" for i = 1:1:3\n", +" for j= 1:1:4\n", +" if E(i,3).entries(j)- E(i,2).entries == N then // N = A-Z\n", +" printf('\t%s(%d)',E(i,1).entries,E(i,3).entries(j));\n", +" end\n", +" end\n", +" end \n", +"end\n", +"// Isobars\n", +"printf('\n\nIsobars:');\n", +"printf('\n=======');\n", +"for A = 14:1:17 // Search for the mass numbers from 14 to 17\n", +" printf('\n(A = %d)\n',A);\n", +" for i = 1:1:3\n", +" for j= 1:1:4\n", +" if E(i,3).entries(j) == A then\n", +" printf('\t%s(%d)',E(i,1).entries,E(i,3).entries(j));\n", +" end\n", +" end\n", +" end \n", +"end\n", +"//\n", +"// Result\n", +"//\n", +"// Isotopes:\n", +"// =========\n", +"// (Z = 6)\n", +"// C(12) C(13) C(14) C(16)\n", +"// (Z = 7)\n", +"// N(14) N(15) N(16) N(17)\n", +"// (Z = 8)\n", +"// O(14) O(15) O(16) O(17)\n", +"//\n", +"// Isotones:\n", +"// ========\n", +"// (N = 6)\n", +"// C(12) O(14)\n", +"// (N = 7)\n", +"// C(13) N(14) O(15)\n", +"// (N = 8)\n", +"// C(14) N(15) O(16)\n", +"// (N = 9)\n", +"// N(16) O(17)\n", +"//\n", +"// Isobars:\n", +"// =======\n", +"// (A = 14)\n", +"// C(14) N(14) O(14)\n", +"// (A = 15)\n", +"// N(15) O(15)\n", +"// (A = 16)\n", +"// C(16) N(16) O(16)\n", +"// (A = 17)\n", +"// N(17) O(17)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4_1: Rest_mass_energy_of_electron.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.4.1: To calculate the energy of electron at rest : Page 33 (2011)\n", +"m = 9.1e-031; // Mass of the electron, Kg\n", +"C = 3e+08; // Velocity of the light,m/s\n", +"E = m*C^2/1.6e-013; // Energy of the electron at rest, MeV\n", +"printf('\nEnergy of the electron at rest : %5.3f MeV', E)\n", +"\n", +"// Result\n", +"// Energy of the electron at rest : 0.512 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4_2: Nuclear_radius.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.4.2 : Estimation of the Nucleus type from its radius : Page 33 (2011)\n", +"r = 3.46e-015; // Radius of the nucleus, m\n", +"r0 = 1.2e-015; // Distance of closest approach of the nucleus, m\n", +"A = round((r/r0)^3); // Mass number of the nucleus\n", +"if A == 23 then\n", +" element = 'Na';\n", +"elseif A == 24 then\n", +" element = 'Mg';\n", +"elseif A == 27 then\n", +" element = 'Al';\n", +"elseif A == 28 then\n", +" element = 'Si';\n", +"end\n", +"printf('The mass number of the nucleus is %d and the nucleus is of %s', A, element);\n", +"\n", +"// Result\n", +"// The mass number of the nucleus is 24 and the nucleus is of Mg" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4_3: Nuclear_density.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.4.3 : Estimate the density of nuclear matter : Page 34 (2011)\n", +"m = 40*(1.66e-027); // Mass of the nucleus, kg \n", +"r0 = 1.2e-015; // Distance of the closest approach, m\n", +"A = 40; // Atomic mass of the nucleus\n", +"r = r0*A^(1/3); //Radius of the nucleus, m\n", +"V = 4/3*(%pi*r^3); // Volume of the nucleus, m^3\n", +"density = m/V; // Density of the nucleus, kg/m^3\n", +"printf('\nRadius of the nucleus: %3.1e m\nVolume of the nucleus: %5.3e m^3\nDensity of the nucleus: %3.1e kg/m^3',r,V,density);\n", +"\n", +"// Result\n", +"// Radius of the nucleus: 4.1e-015 m\n", +"// Volume of the nucleus: 2.895e-043 m^3\n", +"// Density of the nucleus: 2.3e+017 kg/m^3" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4_4: Density_of_uranium_235.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.4.4 : To determine the density of U-235 nucleus : Page 34 (2011)\n", +"m = 1.66e-027; // Mass of a nucleon, kg\n", +"A = 235; // Atomic mass of U-235 nucleus\n", +"M = A*m; //Mass of the U-235 nucleus, kg\n", +"r0 = 1.2e-015; // Distance of closest approach, m\n", +"r = r0*(A)^(1/3); // Radius of the U-235 nucleus\n", +"V = 4/3*(%pi*r^3); // Volume of the U-235 nucleus,m^3\n", +"d = M/V; // Density of the U-235 nucleus,kg/m^3\n", +"printf('\nThe density of U-235 nucleus : %4.2e kg per metre cube',d)\n", +"\n", +"// Result\n", +"// The density of U-235 nucleus : 2.29e+017 kg per metre cube" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4_5: Variation_of_nuclear_density_with_radius.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.4.5 : To calculate densities of O and Pb whose radii are given: Page 35 (2011)\n", +"m_O = 2.7e-026; // Mass of O nucleus, kg\n", +"r_O = 3e-015; // Radius of O nucleus, m\n", +"V_O = 4/3*(%pi*(r0)^3); // Volume of O nucleus, metre cube\n", +"d_O = m_O/V_O; // Density of O nucleus, kg/metre cube\n", +"m_Pb = 3.4e-025; // Mass of Pb nucleus, kg\n", +"r_Pb = 7.0e-015; // Radius of Pb nucleus, m\n", +"V_Pb = 4/3*(%pi*(r_Pb)^3); // Volume of Pb nucleus, metre cube\n", +"d_Pb = m_Pb/V_Pb; //Density of Pb nucleus,kg/metre cube\n", +"printf('\nThe density of oxygen nucleus : %4.2e in kg/metre cube',d_O);\n", +"printf('\nThe density of Pb nucleus : %4.2e in kg/metre cube',d_Pb);\n", +"\n", +"// Result\n", +"// The density of oxygen nucleus : 3.73e+018 in kg/metre cube\n", +"// The density of Pb nucleus : 2.37e+017 in kg/metre cube" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4_6: Distance_of_closest_approach.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.4.6 : Determination of distance of closest approach for alpha-particle : Page 35 (2011)\n", +"E = 5.48*1.6e-013; // Energy of alpha particle, J\n", +"e = 1.6e-019; // Charge of an electron, C\n", +"Z = 79; // Mas number of Au nucleus, \n", +"epsilon_0 = 8.85e-012; // Permittivity of free space, \n", +"D = (2*Z*e^2)/(4*%pi*epsilon_0*E); // Distance of closest approach, m\n", +"printf('\nThe distance of closest appproach of alpha particle : %4.2e m', D)\n", +"\n", +"// Result\n", +"// The distance of closest appproach of alpha particle : 4.15e-014 m" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.4_7: Radius_of_Pb_208.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.4.7 : Determination of radius of Pb-208 : Page 36 (2011)\n", +"A = 208; // Mass number of Pb-208\n", +"r0 = 1.2e-015; // Distance of closest approach, m\n", +"r = r0*((A)^(1/3)); // Radius of Pb-208, m\n", +"printf('\nThe radius of Pb-208 : %4.2e m', r)\n", +"\n", +"// Result\n", +"// The radius of Pb-208 : 7.11e-015 m" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5_1: Binding_energy_of_alpha_particle.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.5.1 : Calculation of binding energy of alpha particle and express in MeV and joule : Page 36 (2011)\n", +"amu = 931.49; // Atomic mass unit, MeV\n", +"M_p = 1.00758; // Mass of proton, amu\n", +"M_n = 1.00897; // Mass of neutron, amu\n", +"M_He = 4.0028; // Mass of He nucleus, amu\n", +"Z = 2; // Atomic number\n", +"N = 2; // Number of neutron\n", +"M_defect = Z*M_p+N*M_n-M_He; // Mass defect, amu\n", +"BE_MeV = M_defect*amu; // Binding energy, MeV\n", +"BE_J = M_defect*1.49239e-010; // Binding energy, J\n", +"printf('\nThe binding energy (in MeV): %5.2f', BE_MeV)\n", +"printf('\nThe binding energy (in J): %4.2e', BE_J)\n", +"\n", +"// Result\n", +"// The binding energy (in MeV): 28.22\n", +"// The binding energy (in J): 4.52e-012 " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5_2: Dissociation_energy_of_C12.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.5.2 : Calculation of energy required to break C-12 into 3-alpha particle : Page 37 (2011)\n", +"amu = 1.49239e-010; // Atomic mass unit, J\n", +"M_C = 12; // Mass of C-12, amu\n", +"M_a = 4.0026; // Mass of alpha particle, amu\n", +"M_3a = 3*M_a; // Mass of 3 alpha particle, amu\n", +"D = M_C-M_3a; // Difference in two masses, amu\n", +"E = D*amu; // Required energy,J\n", +"printf('\nThe energy required to break 3 alpha particles : %4.2e J',E)\n", +"\n", +"// Result\n", +"// The energy required to break 3 alpha particles : -1.16e-012 J " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5_3: Dissociation_energy_of_helium_nucleus.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.5.3 : Calculation of energy required to knock out nucleon from He nucleus : Page 37 (2011)\n", +"M_p = 1.007895; // Mass of proton, amu\n", +"M_n = 1.008665; // Mass of neutron, amu\n", +"M_He = 4.0026; // Mass of He-nucleus, amu\n", +"Z = 2; // Number of proton\n", +"N = 2; // Number of neutron\n", +"D_m = [(Z*M_p)+(N*M_n)-M_He]; // Mass defect, amu\n", +"amu = 931.49; // Atomic mass unit, MeV\n", +"E = D_m*amu; // Required energy, MeV\n", +"printf('\nThe energy required to knock out nucleons from the He nucleus = %5.2f MeV', E);\n", +"\n", +"// Result\n", +"// The energy required to knock out nucleons from the He nucleus = 28.43 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5_4: Binding_energy_of_Fe_56.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.5.4 : To calculate binding energy of Fe-56 : Page 38 (2011)\n", +"M_Fe = 55.934939; // Mass of Fe-56, amu\n", +"M_p = 1.007825; // Mass of proton, amu\n", +"M_n = 1.008665; // Mass of neutron, amu\n", +"Z = 26; // Atomic number of Fe-56\n", +"N = 30; // Number of neutron in Fe-56\n", +"amu = 931.49; // Atomic mass unit, MeV\n", +"BE = [(Z*M_p)+(N*M_n)-M_Fe]*amu; // Binding energy of Fe-56, MeV\n", +"printf('\nThe binding energy of Fe-56 : %6.4f MeV',BE)\n", +"\n", +"// Result\n", +"// The binding energy of Fe-56 : 492.2561 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5_5: Mass_defect_and_packing_fraction.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.5.5 : Calculation of mass defect and packing fraction from given data Page : 38 (2011)\n", +"amu = 931.49; // Atomic mass unit, MeV\n", +"M_p = 1.007825; // Mass of proton, amu\n", +"M_n = 1.008663; // Mass of neutron, amu\n", +"A = 2; // Mass number of deutron, amu\n", +"M_D = 2.014103; // Mass of deuteron nucleus, amu\n", +"M_Defect = (M_p+M_n-M_D)*amu; // Mass defect of the nucleus, MeV\n", +"P_fraction = (M_D - A)/A; // Packing fraction of nucleus\n", +"printf('\n Mass defect %4.2f MeV\n Packing fraction %7.5f',M_Defect,P_fraction);\n", +"\n", +"// Result\n", +"// Mass defect 2.22 MeV\n", +"// Packing fraction 0.00705" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.5_6: Average_binding_energy.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.5.6 : To calculate binding energy per nucleon of He-4 nucleus : Page 38 (2011)\n", +"m_p = 1.007825; // Mass of proton, amu\n", +"m_n = 1.008665; // Mass of neutron, amu\n", +"m_He = 4.002634; // Mass of He-4 nucleus, amu\n", +"amu = 931.47; // Atomic mass unit, MeV\n", +"A = 4, // Mass number of He-4 nucleus\n", +"BE = [2*m_p+2*m_n-m_He]*amu; // Binding energy of He-4 nucleus, MeV\n", +"Av_BE = BE/A; // Average binding energy or binding energy per nucleon, MeV\n", +"printf('\nThe binding energy per nucleon : %4.2f MeV', Av_BE);\n", +"\n", +"// Result\n", +"// The binding energy per nucleon of He-4 is \n", +"// The binding energy per nucleon : 7.07 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_1: Orbital_angular_momentum_of_coupled_nucleons.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.6.1 : Orbital angular momentum of coupled nucleons : Page 39 (2011)\n", +"l1 = 1; // Orbital qunatum number for p-state nucleon\n", +"l2 = 2; // Orbital qunatum number for d-state nucleon\n", +"// Display the value of L within the for loop\n", +"disp('The possible L values will be');\n", +"for i = abs(l1-l2):1:abs(l1+l2) // Coupling of l-orbitals \n", +" printf('\t %1d',i);\n", +"end\n", +"\n", +"// Result\n", +"// The possible L values will be \n", +"// 1 2 3 " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1.6_2: Total_angular_momentum_of_proton.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa1.6.2 : Total angular momentum of proton : Page 40 (2011)\n", +"// Get the l value from the user\n", +"l = 3; // Orbital qunatum number for f-state proton\n", +"s = 1/2; // Magnitude of spin quantum number\n", +"// Display the value of j within the for loop\n", +"disp('The j values will be between');\n", +"for i = abs(l-s):1:abs(l+s) // l-s Coupling \n", +" printf('\t %3.1f',i);\n", +"end\n", +"\n", +"// Result\n", +"// The j values will be between \n", +"// 2.5 3.5 " + ] + } +], +"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/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/2-Nuclear_Models.ipynb b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/2-Nuclear_Models.ipynb new file mode 100644 index 0000000..5dcce74 --- /dev/null +++ b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/2-Nuclear_Models.ipynb @@ -0,0 +1,375 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2: Nuclear Models" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2_1: Binding_energy_and_percentage_discrepancy.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa2.2.1 To calculate the binding energy of Ca(20,40) and %_age discrepancy : Page 66 (2011)\n", +"// For Ca(20,40), actual binding energy is ...... \n", +"m_p = 1.007825; // Mass of proton, amu\n", +"m_n = 1.008665; // Mass of neutron, amu\n", +"Z = 20; // Number of protons\n", +"N = 20; // Number of neutrons\n", +"M_n = 39.962591; // Mass of the nucleus, amu\n", +"B_actual = (M_n-Z*m_p-N*m_n)*931.49; // Actual binding energy, MeV\n", +"// For Ca(20,40), Binding energ as per semiemperical mas formula......\n", +"Z = 20; // Number of protons\n", +"a_v = 15.5; // Volume constant, MeV\n", +"a_s = 16.8; // Surface constant, MeV\n", +"a_a = 23.0; // Asymmetric constant, MeV\n", +"a_c = 0.7; // Coulomb constant, MeV\n", +"a_p = 34.0; // Paring constant, MeV\n", +"A = 40; // Mass number\n", +"B_semi = [a_v*A-(a_s*A^(2/3))-(a_c*Z*(Z-1)/A^(1/3))-(a_a*(A-2*Z)^2/A)-(a_p*A^(-3/4))]; // Binding energy as per semiemperical mass formula\n", +"// Percentage discrepancy between actual and semiemperical mass formula values are.......\n", +"Per_des = -(B_semi+B_actual)/B_actual*100; // Percentage discrepancy \n", +"printf('\nActual binding energy = %6.2f MeV\nBinding energy as per semiemperical mass formula = %6.2f MeV\nPercentage discrepancy = %3.1f percent', B_actual, B_semi, Per_des);\n", +"\n", +"// Result\n", +"// Actual binding energy = -342.05 MeV\n", +"// Binding energy as per semiemperical mass formula = 343.59 MeV\n", +"// Percentage discrepancy = 0.4 percent" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2_2: Coulomb_energies_and_nucelon_masses_of_mirror_nuclei.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa2.2.2 To calculate the difference in coulomb energy and nucleons' mass difference for mirror nuclei and show in agreement with actual mass difference Page 67 (2011)\n", +" // Calculation of coulomb energy for mirror nuclei : N-7 and O-8\n", +" // For N-7 nucleus\n", +"a_c = 0.7; // Coulomb energy constant, MeV\n", +"Z_N = 7; // Atpmic no. \n", +"A = 15; // Atomic mass\n", +"E_C_N = a_c*Z_N*(Z_N-1)/(A^(1/3)); // Coulomb energy for N-7, MeV\n", +"// For O-8 nucleus\n", +"a_c = 0.7; // Coulomb energy constant, MeV\n", +"Z_O = 8; // Atpmic no. \n", +"A = 15; // Atomic mass\n", +"E_C_O = a_c*Z_O*(Z_O-1)/(A^(1/3)); // Coulomb energy for O-8, MeV\n", +"C_E_d = E_C_O-E_C_N; // Coulomb energy difference, MeV\n", +"m_p = 1.007276*931.49; // Mass of proton, MeV\n", +"m_n = 1.008665*931.49; // Mass of neutron, MeV\n", +"M_d = m_n-m_p; // Mass difference of nucleons, MeV \n", +"D_C_M = round(C_E_d-M_d); // Difference in coulomb energy and nucleon mass difference, MeV\n", +"M_O = 15.003070*931.49; // Mass of O-8, MeV\n", +"M_N = 15.000108*931.49; // Mass of N-7, MeV\n", +"D_A = round(M_O-M_N); // Actual mass difference, MeV\n", +"printf('\nDifference in Coulomb energy = %5.3f MeV\nNucleon mass difference = %6.4f MeV\nDifference in Coulomb energy and nucleon mass difference = %5.3f MeV\nActual mass difference = %5.3f MeV',C_E_d, M_d ,D_C_M, D_A);\n", +"if D_A == D_C_M then printf('\nResult is verified')\n", +"end\n", +"// Result\n", +"// Difference in Coulomb energy = 3.974 MeV\n", +"// Nucleon mass difference = 1.2938 MeV\n", +"// Difference in Coulomb energy and nucleon mass difference = 3.000 MeV\n", +"// Actual mass difference = 3.000 MeV\n", +"// Result is verified" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2_3: Neutron_binding_energy_for_isotopes_of_krypton.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa2.2.3 To calculate the energy required to remove a neutron from Kr-81, Kr-82, Kr-83 : Page 68 (2011)\n", +"// For Kr-80, \n", +"m_p = 1.007825; // Mass of proton, amu\n", +"m_n = 1.008665; // Mass of neutron, amu\n", +"Z = 36; // Number of protons\n", +"N_80 = 44; // Number of neutrons\n", +"M_n_80 = 79.91628; // Mass of Kr nucleus\n", +"BE_Kr_80 = (Z*m_p+N_80*m_n-M_n_80)*931.49; // Binding energy for Kr-80, MeV\n", +"// For Kr-81,\n", +"N_81 = 45; // Number of neutrons\n", +"M_n_81 = 80.91661; // Mass of Kr-81 nucleus\n", +"BE_Kr_81 = (Z*m_p+N_81*m_n-M_n_81)*931.49; // Binding energy for Kr-81 nucleus\n", +"// For Kr-82\n", +"N_82 = 46; // Number of neutrons\n", +"M_n_82 = 81.913482; // Mass of Kr nucleus\n", +"BE_Kr_82 = (Z*m_p+N_82*m_n-M_n_82)*931.49; // Binding energy for Kr-82,MeV\n", +"// For Kr-83 \n", +"N_83 = 47; // Number of protons\n", +"M_n_83 = 82.914134; // Mass of Kr-83 nucleus\n", +"BE_Kr_83 = (Z*m_p+N_83*m_n-M_n_83)*931.49; // Binding energy for Kr-83, MeV\n", +"E_sep_81 = BE_Kr_81-BE_Kr_80; // Energy seperation of neutron for Kr-81, MeV\n", +"E_sep_82 = BE_Kr_82-BE_Kr_81; // Energy seperation of neutron for Kr-82, MeV\n", +"E_sep_83 = BE_Kr_83-BE_Kr_82; // Energy seperation of neutron for Kr-83, MeV\n", +", \n", +"printf('\nEnergy seperation of neutron for Kr-81 = %4.2f MeV\nEnergy seperation of neutron for Kr-82 = %4.2f MeV\nEnergy seperation of neutron for Kr-83 = %5.2f MeV', E_sep_81, E_sep_82, E_sep_83);\n", +"\n", +"// Result\n", +"// Energy seperation of neutron for Kr-81 = 7.76 MeV\n", +"// Energy seperation of neutron for Kr-82 = 10.99 MeV\n", +"// Energy seperation of neutron for Kr-83 = 7.46 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2_4: Isotopic_stability.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa2.2.4 To determine the most stable isotope of A = 75 : Page 68 (2011)\n", +"a_v = 15.5; // Volume energy coefficient, MeV\n", +"a_s = 16.8; // Surface energy coefficient MeV\n", +"a_c = 0.7; // Coulomb energy coefficient, MeV\n", +"a_a = 23.0; // Asymmetric energy coefficient, MeV\n", +"a_p = 34.0; // Pairing energy coefficient, MeV\n", +"A = 75; // Given atomic mass \n", +"z = poly(0, 'z'); // z declares a polynomial\n", +"B = -a_c*z*(z-1)/A^(1/3)-a_a*(A-2*z)^2/A ; // Binding energy as per liquid drop model\n", +"dB = derivat(B); // Differentiate B w.r.t. z\n", +"z = roots(dB); // Isotope of A = 75\n", +"z_i = round(z); // Most stable isotope of A = 75\n", +"printf('\nMost stable isotope of A = 75 corresponds to Z = %d ', z_i)\n", +"\n", +"// Result\n", +"// Most stable isotope of A = 75 corresponds to Z = 33" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2_5: Stable_isotopes_for_different_mass_numbers.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa2.2.5 To determine the most stable isotopes for A = 27, A = 118, A = 238 : Page 69 (2011)\n", +"a_v = 15.5; // Volume energy, MeV \n", +"a_s = 16.8; // Surface energy, MeV \n", +"a_c = 0.7; // Coulomb energy, MeV\n", +"a_a = 23.0; // Asymmetric energy, MeV\n", +"a_p = 34.0; // Pairing energy, MeV\n", +"z = poly(0, 'z')\n", +"// For A = 27; \n", +"B_27 = -a_c*z*(z-1)/27^(1/3)-a_a*(27-2*z)^2/27 ; // Binding energy as per liquid drop model\n", +"dB_27 = derivat(B_27) // Differentiate B w.r.t. z\n", +"z_27 = roots(dB_27) // Isotope of A = 27\n", +"z_i_27 = round(z_27) // Most stable isotope of A = 27\n", +"// For A = 118 \n", +"B_118 = -a_c*z*(z-1)/118^(1/3)-a_a*(118-2*z)^2/118 ; // Binding energy as per liquid drop model\n", +"dB_118 = derivat(B_118) // Differentiate B w.r.t. z\n", +"z_118 = roots(dB_118) // Isotope of A = 118\n", +"z_i_118 = round(z_118) // Most stable isotope of A = 118\n", +"// For A = 238\n", +"B_238 = -a_c*z*(z-1)/238^(1/3)-a_a*(238-2*z)^2/238 ; // Binding energy as per liquid drop model\n", +"dB_238 = derivat(B_238); // Differentiate B w.r.t. z\n", +"z_238 = roots(dB_238); // Isotope of A = 238\n", +"z_i_238 = round(z_238); // Most stable isotope of A = 238\n", +"printf('\nMost stable isotopes for A = 27, A = 118, A = 238 corresponds to z = %d, %d and %d respectively', z_i_27, z_i_118, z_i_238);\n", +"\n", +"// Result\n", +"// Most stable isotopes for A = 27, A = 118, A = 238 corresponds to z = 13, 50 and 92 respectively\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2_6: Coulomb_energy_coefficient_of_mirror_nuclei.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa2.2.6 : To calculate the coulomb coefficient and estimate nuclear radius for mirror nuclei: Page no. 69 (2011)\n", +"// Mirror nuclei : Na-11 and Mg-12 \n", +"m_p = 1.007276; // Mass of proton, amu\n", +"m_n = 1.008665; // Mass of neutron, amu\n", +"M_Mg = 22.994124; // Atomic mass of Mg-12, amu\n", +"M_Na = 22.989768; // Atomic mass of Na-11, amu\n", +"A = 23; // Mass number\n", +"Z_Mg = 12; // Atomic number of Mg-12\n", +"e = 1.6e-019; // Charge of the electron, C\n", +"K = 8.98e+09; // Coulomb force constant\n", +"a_c = A^(1/3)/(2*Z_Mg-1)*[(M_Mg-M_Na)+(m_n-m_p)]*931.47; // Coulomb coefficient, MeV \n", +"r_0 = 3/5*K*e^2/(a_c*1.6e-013); // Nuclear radius, m\n", +"printf('\nCoulomb coefficient = %4.2f MeV\nNuclear radius = %3.1e m', a_c, r_0)\n", +"// Result \n", +"// Coulomb coefficient = 0.66 MeV\n", +"// Nuclear radius = 1.3e-015 m " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2_7: Coulomb_and_surface_energies_of_uranium.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa2.2.7 To calculate coulomb energy and surface energy for U(92,236) : Page 71 (2011)\n", +"Z = 92; // Atomic number of U-236\n", +"e = 1.6e-019; // Charge of an electron, C\n", +"A = 236; // Mass number of U-236\n", +"K = 8.98e+09; // Coulomb constant,\n", +"r_o = 1.2e-015; // Distance of closest approach, m\n", +"a_s = -16.8; // Surface constant\n", +"E_c = -(3*K*Z*(Z-1)*e^2)/(5*r_o*A^(1/3)*1.6e-013); // Coulomb energy, MeV\n", +"E_s = a_s*A^(2/3); // Surface energy, MeV \n", +"printf('\nCoulomb energy for U(92,236) = %5.1f MeV \nSurface energy for U(92,236) = %5.1f MeV ', E_c, E_s)\n", +"// Result\n", +"// Coulomb energy for U(92,236) = -973.3 MeV \n", +"// Surface energy for U(92,236) = -641.6 MeVS" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3_1: Mass_of_decayed_radioactive_material.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa2.3.1 To calculate the mass of decayed radioactive material: Page 126 (2011)\n", +"t_prime = 1600; // Half life of radioactive material, years\n", +"t = 2000; // Total time, years\n", +"lambda = 0.6931/t_prime; // Decay constant, years^(-1)\n", +"m0 = 1; // The mass of radioactive substance at t0, mg\n", +"m = m0* %e^(-(lambda*t)); // Ratio of total number of atoms and number of atoms disintegrat, mg\n", +"a = 1-m; // The amount of radioactive substance decayed, mg \n", +"printf('\nThe amount of radioactive substance decayed : %6.4f mg', a)\n", +"\n", +"// Result \n", +"// The amount of radioactive substance decayed : 0.5795 mg " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.3_4: Magnetic_moment_of_nuclei.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa2.3.4 : To calculate the magnetic moment of given nuclei : Page no. 74 : (2011)\n", +"// For Ne(10.19) nucleus \n", +"j_Ne_9 = 5/2; // Total angular momentum for Ne-19 nucleus\n", +"u_Ne_9 = j_Ne_9+2.29; // Magnetic moment of Ne-19 nucleus , nuclear magneton\n", +"// For Ne(10,20) nucleus\n", +"j_Ne_10 = 0;// Total angular momentum for Ne-20 nucleus\n", +"u_Ne_10 = j_Ne_10+2.29; // Magnetic moment of Ne-20 nucleus, nuclear magneton\n", +"// For Ne(10,21) nucleus\n", +"j_Ne_11 = 5/2;// Total angular momentum for Ne-21 nucleus\n", +"u_Ne_11 = j_Ne_11+2.29; // Magnetic moment of Ne-21 nucleus, nuclear magneton\n", +"printf('\nMagnetic moment of Ne-19 nucleus = %4.2f nuclear magneton\nMagnetic moment of Ne-20 nucleus = %4.2f nuclear magneton\nMagnetic moment of Ne-21 nucleus = %4.2f nuclear magneton', u_Ne_9, u_Ne_10, u_Ne_11);\n", +"// Result\n", +"// Magnetic moment of Ne-19 nucleus = 4.79 nuclear magneton\n", +"// Magnetic moment of Ne-20 nucleus = 2.29 nuclear magneton\n", +"// Magnetic moment of Ne-21 nucleus = 4.79 nuclear magneton" + ] + } +], +"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/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/3-Radioactivity.ipynb b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/3-Radioactivity.ipynb new file mode 100644 index 0000000..5d0e7bc --- /dev/null +++ b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/3-Radioactivity.ipynb @@ -0,0 +1,884 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3: Radioactivity" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2_10: Power_in_radioactive_decay.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa3.2.10 : To calculate the power produced by 10 mg of Po-210 : Page no. 130 (2011)\n", +"N = 2.87e+019; // Number of atoms in 10e-10kg of Po-210\n", +"t_h = 138*24*3600; // Half life of Po-210, s\n", +"D = 0.693/t_h; // Decay constant, s^-1\n", +"A = N*D; // Activity of K-40, dps\n", +"E = 5.3*1.6e-013; // Power produce by one dps, MeV\n", +"P = A*E; // Power produced by 1.667e+012 dps, W\n", +"printf('\nThe Power produced by 1.667e+012 dps : %3.1f W', P)\n", +"// Result\n", +"// The Power produced by 1.667e+012 dps : 1.4 W " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2_1: Curie_becquerel_relatio.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa3.2.1: To determine how many curie in 10^10 Bq : Page 124 (2011)\n", +"Bq = 1/3.7e+010; // Number of curie in one Bq, Ci\n", +"N = 10^10*Bq; // The number of curie in 10^10 Bq, Ci\n", +"printf('\nThe number of curie in 10^10 Bq : %4.2f Ci', N)\n", +"// Result\n", +"// The number of curie in 10^10 Bq : 0.27 Ci\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2_2: Activity_of_thorium.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa3.2.2: To calculate the activity of 10g of Th-232 : Page 125 (2011)\n", +"lambda_232 = 1.58e-018; // Decay constant, s^-1\n", +"N = 2.596e+022; // Number of atoms in 10g Th-232\n", +"A = N*lambda_232; // The activity of 10g of Th-232, dps\n", +"printf('\nThe activty of 10g of Th-232 : %5.3e dps', A)\n", +"// Result\n", +"// The activty of 10g of Th-232 : 4.102e+004 dps" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2_3: Mass_of_radiactive_sample.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa3.2.3: Calculation of mass of 1 Ci sample of radioactive sample : Page 125 (2011)\n", +"A = 3.7e+010; // Activity of 1Ci sample, dps\n", +"t = 1608; // Half life of radioactive substance, s\n", +"N = 6.023e+023/214; // Number of atoms in 1g of substance having atomic mass 214\n", +"lambda = 0.6931/t; // Decay constant, s^-1\n", +"m = A/(lambda*N); // The mass of radoiactive substance, g\n", +"printf('\nThe mass of radioactive substance : %4.2e g', m)\n", +"// Result\n", +"// The mass of radioactive substance : 3.05e-008 g " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2_4: Activity_of_1_kg_of_uranium.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa3.2.4: To calculate the activity of 1kg of U-238: Page 125 (2011)\n", +"t = 1.419e+017; // Half life of U-238, s\n", +"N = 6.023e+023/238; // Number of atoms in 1g of U-238\n", +"lambda = 0.6931/t; // Decay constant, s^-1\n", +"A = (lambda*N)*1000/(3.7e+010); // The activity of 1kg of U-238, Ci\n", +"printf('\nThe activity of 1kg of U-238 : %4.2e Ci', A)\n", +"// Result\n", +"// The activity of 1kg of U-238 : 3.34e-004 Ci" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2_6: Half_life_of_radioactive_material.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa3.2.6 Determination of half life of radioactive material Page 127 (2011)\n", +"t = 10; // Total period of radioactive material, days\n", +"lambda = log(6.6667)/10; //Decay constant, day^-1\n", +"t_h = 0.6931/(lambda); // Half life of radioactive substance, days\n", +"printf('\nThe half life of radioactive substance : %4.2f days', t_h)\n", +"// Result\n", +"// The half life of radioactive substance : 3.65 days\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2_7: Mass_of_Ra_226.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa3.2.7 : To calculate the mass of Ra-226 :Page no. 127 (2011)\n", +"t_h = 1620*31536000; // Half life of Ra-226, S\n", +"D = 0.6931/t_h; // Decay constant, S^-1\n", +"A_Ci = 3.7e+010; // Activity, Ci\n", +"N_Ci = A_Ci/D; // Number of atoms decayed\n", +"m = 0.226; // Mass of 6.023e+023 atoms, kg\n", +"M_Ci = m*N_Ci/6.023e+023; // Mass of 1-Ci sample of Ra-226, kg\n", +"A_rf = 10^6; // Activity, Rf\n", +"N_rf = A_rf/D; // Number of atoms decayed\n", +"M_rf = m*N_rf/6.023e+023; // Mass of 1-Rf sample of Ra-226, kg\n", +"printf('\n Mass of 1-Ci sample of Ra-226 = %5.3e kg and \n Mass of 1-Rf sample of Ra-226 = %4.2e kg ',M_Ci, M_rf )\n", +"// Result\n", +"// Mass of 1-Ci sample of Ra-226 = 1.023e-003 kg and \n", +"// Mass of 1-Rf sample of Ra-226 = 2.77e-008 kg\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2_8: Activity_and_weight_of_radiactive_material.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa3.2.8 To calculate the activity and weight of radioactive material : Page 128 (2011)\n", +"N_o = 7.721e+018; // Number of atoms in 3 mg of U-234\n", +"t_h = 2.5e+05; // Half life of U-234, years\n", +"T = 150000; // Total time, years\n", +"lambda = 0.6931/t_h; // Decay constant, year^-1\n", +"N = N_o*(%e^-(lambda*T)); // Number of atoms left after T years\n", +"m = 234000; // Mass of 6.023e+023 atoms of U-234, mg\n", +"M = m*N/(6.023e+023); // Weight of sample left after t years, \n", +"L = 8.8e-014; // Given decay constant, S^-1\n", +"A = N*L*10^6/(3.7e+010); // Activity, micro Ci\n", +"printf('\nThe weight of sample = %5.3f mg \n Activity = %5.2f micro Ci ', M, A)\n", +"// Result\n", +"// The weight of sample = 1.979 mg \n", +"// Activity = 12.12 micro Ci \n", +" \n", +"\n", +" " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2_9: Activity_of_K_40.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa3.2.9 : To calculate the activity of K-40 : Page no. 129 (2011)\n", +"N = 6.324e+020; // Number of atoms in 4.2e-05 kg of K-40\n", +"t_h = 1.31e+09*31536000; // Half life of K-40, s\n", +"D = 0.693/t_h; // Decay constant, s^-1\n", +"A = N*D/(3.7e+010)*10^6; // Activity of K-40, microCi\n", +"printf('\nThe activity of K-40 : %5.3f micro Ci', A )\n", +"// Result\n", +"// The activity of K-40 : 0.287 micro Ci \n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3_1: Emitted_particles_during_nuclear_disintegration.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 3.3.1 : Finding particles in the given reactions : page no. 131 (2011)\n", +"// Declare three cells (for three reactions)\n", +"R1 = cell(4,3);\n", +"R2 = cell(4,3);\n", +"R3 = cell(3,3);\n", +"// Enter data for first cell (Reaction)\n", +"R1(1,1).entries = 'Pb';\n", +"R1(1,2).entries = 82;\n", +"R1(1,3).entries = 211;\n", +"R1(2,1).entries = 'Bi';\n", +"R1(2,2).entries = 83;\n", +"R1(2,3).entries = 211;\n", +"R1(3,1).entries = 'Tl';\n", +"R1(3,2).entries = 81;\n", +"R1(3,3).entries = 207;\n", +"R1(4,1).entries = 'Pb';\n", +"R1(4,2).entries = 82;\n", +"R1(4,3).entries = 207;\n", +"// Enter data for second cell (Reaction)\n", +"R2(1,1).entries = 'U';\n", +"R2(1,2).entries = 92;\n", +"R2(1,3).entries = 238;\n", +"R2(2,1).entries = 'Th';\n", +"R2(2,2).entries = 90;\n", +"R2(2,3).entries = 234;\n", +"R2(3,1).entries = 'Pa';\n", +"R2(3,2).entries = 91;\n", +"R2(3,3).entries = 234;\n", +"R2(4,1).entries = 'U';\n", +"R2(4,2).entries = 92;\n", +"R2(4,3).entries = 234;\n", +"// Enter data for third cell (Reaction)\n", +"R3(1,1).entries = 'Bi';\n", +"R3(1,2).entries = 83;\n", +"R3(1,3).entries = 211;\n", +"R3(2,1).entries = 'Pa';\n", +"R3(2,2).entries = 84;\n", +"R3(2,3).entries = 211;\n", +"R3(3,1).entries = 'Pb';\n", +"R3(3,2).entries = 82;\n", +"R3(3,3).entries = 207;\n", +"// Declare a function returning the type of particle emitted\n", +"function particle = identify_particle(d_Z, d_A)\n", +" if d_Z == 2 & d_A == 4 then \n", +" particle = 'Alpha';\n", +" elseif d_Z == -1 & d_A == 0 then\n", +" particle = 'Beta minus';\n", +" elseif d_Z == 1 & d_A == 0 then\n", +" particle = 'Beta plus';\n", +" end\n", +"endfunction\n", +"// Display emitted particles for first reaction\n", +"printf('\n\n\nReaction-I:');\n", +"for i = 1:1:3\n", +" dZ = R1(i,2).entries-R1(i+1,2).entries;\n", +" dA = R1(i,3).entries-R1(i+1,3).entries;\n", +" p = identify_particle(dZ,dA);\n", +" printf('\n%s(%d) - (%s) --> %s(%d)', R1(i,1).entries, R1(i,2).entries, p, R1(i+1,1).entries, R1(i+1,2).entries); \n", +"end\n", +"// Display emitted particles for second reaction\n", +"printf('\n\n\nReaction-II:');\n", +"for i = 1:1:3\n", +" dZ = R2(i,2).entries-R2(i+1,2).entries;\n", +" dA = R2(i,3).entries-R2(i+1,3).entries;\n", +" p = identify_particle(dZ,dA);\n", +" printf('\n%s(%d) - (%s) --> %s(%d)', R2(i,1).entries, R2(i,2).entries, p, R2(i+1,1).entries, R2(i+1,2).entries); \n", +"end\n", +"// Display emitted particles for third reaction\n", +"printf('\n\n\nReaction-III:');\n", +"for i = 1:1:2\n", +" dZ = R3(i,2).entries-R3(i+1,2).entries;\n", +" dA = R3(i,3).entries-R3(i+1,3).entries;\n", +" p = identify_particle(dZ,dA);\n", +" printf('\n%s(%d) - (%s) --> %s(%d)', R3(i,1).entries, R3(i,2).entries, p, R3(i+1,1).entries, R3(i+1,2).entries); \n", +"end\n", +"// Result\n", +"//\n", +"// Reaction-I:\n", +"// Pb(82) - (Beta minus) --> Bi(83)\n", +"// Bi(83) - (Alpha) --> Tl(81)\n", +"// Tl(81) - (Beta minus) --> Pb(82)\n", +"// Reaction-II:\n", +"// U(92) - (Alpha) --> Th(90)\n", +"// Th(90) - (Beta minus) --> Pa(91)\n", +"// Pa(91) - (Beta minus) --> U(92)\n", +"// Reaction-III:\n", +"// Bi(83) - (Beta minus) --> Pa(84)\n", +"// Pa(84) - (Alpha) --> Pb(82)" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3_2: Energy_of_Pb_decay.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 3.3.2 To calculate mass number of Pb isotope and energy emitted : Page no : 132 (2011)\n", +"M_U = 238.050786; // Atomic mass of U-238, amu\n", +"M_Pb = 205.9744550; // Atomic mass of Pb-205, amu\n", +"M_He = 4.002603; // Atomic mass of He-4, amu\n", +"M_e = 5.486e-04; // Atomic mass of electron, amu\n", +"M = M_Pb+(8*M_He)+(6*M_e); // Total mass of products, amu\n", +"D = M_U-M; // Decrease in mass, amu\n", +"E = D*931.47; // Energy evolved, MeV\n", +"printf('\nTotal mass of products = %1.7f amu \n Decrease in mass = %9.7f amu and \n Energy evolved = %4.1f MeV', M, D, E)\n", +"// Result\n", +"// Total mass of products = 237.9985706 amu \n", +"// Decrease in mass = 0.0522154 amu and \n", +"// Energy evolved = 48.6 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4_1: Atomic_and_mass_numbers_of_daughter_nuclei.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Finding atomic No. and mass No. of daughter nuclei in the given reactions : Page No. 133(2011)\n", +"// Declare cell (for given reaction)\n", +"R1 = cell(5,4);\n", +"// Enter data for cell (Reaction-I)\n", +"R1(1,1).entries = 'A';\n", +"R1(1,2).entries = 90;\n", +"R1(1,3).entries = 238;\n", +"R1(1,4).entries = 'Alpha';\n", +"R1(2,1).entries = 'B';\n", +"R1(2,4).entries = 'Beta minus';\n", +"R1(3,1).entries = 'C';\n", +"R1(3,4).entries = 'Alpha';\n", +"R1(4,1).entries = 'D';\n", +"R1(4,4).entries = 'Beta minus';\n", +"R1(5,1).entries = 'E'; \n", +"// Declare a function returning the type of particle emitted\n", +"function [Z, A] = daughter_nucleus(particle_emitted)\n", +" if particle_emitted == 'Alpha' then \n", +" Z = 2, A = 4;\n", +" elseif particle_emitted == 'Beta minus' then \n", +" Z = -1, A = 0; \n", +" elseif particle_emitted == 'Beta plus' then \n", +" Z = 1, A = 0;\n", +" end\n", +"endfunction\n", +"// Display emitted particles for first reaction\n", +"printf('\n\n\nReaction-I:');\n", +"for i = 1:1:4\n", +" [Z, A] = daughter_nucleus(R1(i,4).entries);\n", +" R1(i+1,2).entries = R1(i,2).entries-Z;\n", +" R1(i+1,3).entries = R1(i,3).entries-A; \n", +" printf('\n%s(%d,%d) - (%s) --> %s(%d,%d)', R1(i,1).entries, R1(i,2).entries, R1(i,3).entries, R1(i,4).entries, R1(i+1,1).entries, R1(i+1,2).entries, R1(i+1,3).entries)\n", +" ; \n", +"end\n", +"// Result \n", +"// \n", +"// Reaction-I:\n", +"// A(90,238) - (Alpha) --> B(88,234)\n", +"// B(88,234) - (Beta minus) --> C(89,234)\n", +"// (89,234) - (Alpha) --> D(87,230)\n", +"// D(87,230) - (Beta minus) --> E(88,230) " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4_2: Number_of_half_lives_of_Rn_222.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 3.4.2 : To determine the number of Rn-222 half lives elapsed when it reaches 99% of its equilibrium concentration : Page no. 133 : (2011)\n", +" D = log(2); // Decay constant, s^-1\n", +" t = log(100); // Half life, s\n", +" n = t/D; // Number of half-lives \n", +"printf('\n Number of half-lives : %4.2f ', n)\n", +"// Result\n", +"// Number of half-lives : 6.64" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4_3: Decay_constant_for_alpha_and_beta_decays.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 3.4.3 : To calculate the decay constant for alpha and beta decays : Page no. 133 : (2011)\n", +" H_t = 60.5*60; // Total half life period, s\n", +" T_d = 0.693/H_t; // Total decay constant, s^-1\n", +" A_d = 34/100*T_d; // Decay constant for alpha decays, s^-1\n", +" B_d = 66/100*T_d; // Decay constant for beta decay, s^-1\n", +"printf('\n Alpha decay = %4.2e s^-1 \n Beta decay = %4.2e s^-1', A_d, B_d)\n", +"// Result\n", +"// Alpha decay = 6.49e-005 s^-1 \n", +"// Beta decay = 1.26e-004 s^-1 " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4_4: Half_life_of_uranium_234.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 3.4.4 : To calculate the half life of U(92,234): Page no. 134 : (2011)\n", +"A_r = 1.8e+04; // Atomic ratio of U(92,238) and U(92,234)\n", +"T_238 = 2.5e+05; // Half life of U(92,238), years\n", +"T_234 = A_r*T_238; // Half life of U(92,234), years\n", +"printf('\n Half life of U(92,234): %3.1e years', T_234)\n", +"// Result\n", +"// Half life of U(92,234): 4.5e+009 years" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4_5: Decayed_amount_of_radioactive_matter.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa3.2.5 To calculate the mass of decayed radioactive material: Page 126 (2011)\n", +"t_h = 1600; // Half life of radioactive material, years\n", +"t = 2000; // Totaltime, years\n", +"lambda = 0.6931/t_h; // Decay constant, years^-1\n", +"m0 = 1; // The mass of radioactive substance at t0, mg\n", +"m = m0* %e^(-(lambda*t)); // Ratio of total number of atoms and number of atoms disintegrat, mg\n", +"A = 1-m; // The amount of radioactive substance decayed, mg\n", +"printf('\nThe amount of radioactive substance decayed : %6.4f mg',A)\n", +"// Result\n", +"// The amount of radioactive substance decayed : 0.5795 mg" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5_2: Kinetic_energy_of_alpha_particle.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 3.5.2 : To calculate the K.E. of alpha particle in following decay Pu-239 to U-235+He-4\n", +"M_239 = 239.052158; // Atomic mass of Pu-239, amu\n", +"M_235 = 235.043925; // Atomic mass of U-235, amu\n", +"M_4 = 4.002603; // Atomic mass of He-4, amu\n", +"Q = (M_239-M_235-M_4)*931.47; // Difference in masses, MeV\n", +"A = 241; // Mass number \n", +"K_alpha = Q*(A-4)/A; // Kinetic energy of alpha particle, MeV\n", +"printf('\nKinetic energy of alpha particle %5.2f MeV', K_alpha)\n", +"// Result\n", +"// Kinetic energy of alpha particle 5.16 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5_3: Height_of_barrier_faced_by_alpha_particle.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 3.5.3 : To calculate the height of barrier faced by alpha particle of Ra-226 : Page no. : 136 (2011)\n", +"Z = 88; // Atomic number of Ra-226 nucleus, \n", +"A = 226; // Atomic mass of Ra-226 nucleus\n", +"R_0 = 1.3e-015; // Distance of closest approach, m\n", +"E_0 = 8.854e-012; // Permittivity of free space, C^2/Nm^2\n", +"e = 1.6e-019; // Charge of an electron, C\n", +"B = 2/(1.6e-013)*(Z-2)*e^2/(4*%pi*E_0*R_0*A^(1/3)); // The barrier height faced by alpha particle, MeV\n", +"printf('\nThe barrier height faced by alpha particle : %4.1f MeV', B)\n", +"// Result\n", +"// The barrier height faced by alpha particle : 31.2 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5_4: Height_of_coulomb_barrier.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 3.5.4 : To calculate the height of coulomb barrier faced by alpha particle : Page no. : 136 (2011)\n", +"Z_1 = 2; //Atomic number of He-4, \n", +"Z_2 = 7; // Atomic number of N-14,\n", +"A_1 = 4; // Atomis mass of He-4 nucleus \n", +"A_2 = 14; // Atomic mass of N-14 nucleus\n", +"R_0 = 1.5e-015; // Distance of closest approach, m\n", +"E_0 = 8.854e-012; // Permittivity of free space, C^2/Nm^2\n", +"e = 1.6e-019; // Charge of an electron, C\n", +"B = Z_1/(1.6e-013)*Z_2*e^2/(4*%pi*E_0*R_0*(A_1^(1/3)+A_2^(1/3))); // The coulomb barrier faced by alpha particle, MeV\n", +"printf('\nThe coulomb barrier faced by alpha particle : %4.2f MeV', B)\n", +"// Result\n", +"// The coulomb barrier faced by alpha particle : 3.36 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5_5: KE_of_a_proton_to_penetrate_the_barrier.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 3.5.5 : To calculate the K.E. of a proton to penetrate the barrier of H nucleus : Page no. : 137 (2011)\n", +"R_0 = 1.2; // Distance of closest approach, m\n", +"E_b = 197/(R_0*137); // The K.E. of proton to penetrate the berrier of H nucleus, Mev\n", +"printf('\nThe K.E. of proton to penetrate the berrier of H nucleus : %3.1f MeV', E_b)\n", +"// Result\n", +"// The K.E. of proton to penetrate the berrier of H nucleus : 1.2 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6_1: Mass_of_daughter_nucleus.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 3.6.1 : To determine the mass of daughter nucleus for given reaction : Page no. 138 : (2011)\n", +" M_C = 14.007685; // Mass of C-14 nucleus, amu\n", +" E_e = 0.156/931.47; // Kinetic energy of emitted electron, amu\n", +" M_N = M_C-E_e; // Mass of N-14 nucleus, amu\n", +"printf('\n Mass of N-14 nucleus : %9.6f amu', M_N)\n", +"// Result\n", +"// Mass of N-14 nucleus : 14.007518 amu" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6_3: Number_of_proton_decayed_per_year_from_water.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa. 3.6.3 : To determine the number of proton decayed per year from H2O in a reservior : Page no. 139 : (2011)\n", +"N_p = 6.70e+033;// Number of protons \n", +"T_p = 10^32; // Mean life of proton, years\n", +"D_p = N_p/T_p*0.5; // Number of proton decays per year, decays/year \n", +"printf('\n Number of proton decays per year,: %4.1f decays/year', D_p)\n", +"// Result\n", +"// Number of proton decayed per year: 33.5 decays/year" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.7_1: Energy_of_gamma_photons_from_excited_Ni_60.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa. 3.7.1 : To determine the energies of two gamma rays emitted during de-excitation of Ni-60: Page no. 141 : (2011)\n", +"E_2 = 2505; // Second excited state of Ni-60, KeV\n", +"E_1 = 1332; // First excited state of Ni-60, KeV\n", +"E_0 = 0; // Ground state of Ni-60 , KeV\n", +"E_G_2 = E_2-E_1; // Energy of gamma rays emitted when transition from 2 to 1, KeV\n", +"E_G_1 = E_1-E_0; // Energy of gamma rays emitted when transition from 1 to 0, KeV\n", +"printf('\n Energies of two gamma rays emitted : %d KeV and %d KeV', E_G_2, E_G_1)\n", +"// Result\n", +"// Energy of two gamma rays emitted : 1173 KeV and 1332 KeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.7_2: Conversion_energies_for_K_and_L_shell_electrons.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa. 3.7.2 : To determine the energies conversion for K and L-shell electrons for reaction Cs(55,137) = Ba(56,137)+e(-1,0): Page no. 141 : (2011)\n", +"E = 662; // Energy available with the nucleus, KeV\n", +"I_b_K = 37.4; // Binding energy for K-shell, KeV\n", +"I_b_L = 6.0; // Binding energy for L-shell, KeV\n", +"E_c_K = E-I_b_K; // Energy conversion for K-shell, KeV\n", +"E_c_L = E-I_b_L; // Energy conversion for L-shell, KeV\n", +"printf('\n Energies conversion for K and L-shell electrons : %5.1f KeV and %d KeV', E_c_K, E_c_L)\n", +"// Result\n", +"// Energies conversion for K and L-shell electrons : 624.6 KeV and 656 KeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.9_1: Age_of_uranium_mineral.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa. 3.9.1 : To calculate the age of uranium mineral: Page no. 143 : (2011)\n", +"t_h = 4.5e+09; // Half life of mineral, years\n", +"D_c = 0.6931/t_h; // Decay constant of minerals, years^-1\n", +"N_1 = 6.023e+023/238; // Number of nuclei in 1g of Uranium\n", +"N = 6.023e+023*0.093/206; // Number of nuclei in 0.093g of lead\n", +"t = log(1+N/N_1)/D_c; // Age of the mineral, years\n", +"printf('\n Age of the mineral : %6.4e years ', t)\n", +"// Result\n", +"// Age of the mineral : 6.6261e+008 years" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.9_2: Age_of_boat_from_its_half_life.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa. 3.9.2 : To determine the age of boat whose half life is given : Page no. 145 : (2011)\n", +"t_h = 5760; // Half life of boat, years\n", +"D_c = 0.6931/t_h; // Decay constant of boat, years^-1\n", +"N_1 = 16; // Number of atoms decay per min. per gram initially \n", +"N = 5; // Number of atoms decay per min per gram presently\n", +"t = log(N_1/N)*1/D_c; // Age of the boat, years\n", +"printf('\n Age of the boat : %d years ', t)\n", +"// Result\n", +"// Age of the boat : 9666 years " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.9_4: radioactive_disintegration_of_Pu_239.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa. 3.9.4 : To calculate the number of nuclei at t = 0, initial activity and age of Pu-239 which emit alpha particle : Page no. 145 : (2011)\n", +"t_h = 24000*365*24*3600; // Half life of Pu-239, s\n", +"D_c = 0.6931/t_h; // Decay constant of Pu-239, s^-1\n", +"N = 6.023e+023*10/239; // Number of nuclei at t = 0, nuclei \n", +"A_0 = D_c*N; // Initial activity, disintegrations/sec\n", +"A = 0.1; // Activity after time t, disintegrations/sec\n", +"t = log(A_0/A)*1/D_c; // Age of the Pu-239, years\n", +"printf('\nThe number of nuclei at t = 0, = %4.2e nuclei \nInitial activity = %4.2e disintegrations/s and \nAge of Pu-239 = %4.2e years ', N, A_0, t)\n", +"// Result\n", +"// The number of nuclei at t = 0, = 2.52e+022 nuclei \n", +"// Initial activity = 2.31e+010 disintegrations/s and \n", +"// Age of Pu-239 = 2.86e+013 years" + ] + } +], +"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/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/4-Nuclear_Reactions.ipynb b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/4-Nuclear_Reactions.ipynb new file mode 100644 index 0000000..a73f227 --- /dev/null +++ b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/4-Nuclear_Reactions.ipynb @@ -0,0 +1,932 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: Nuclear Reactions" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3_1: Cross_section_of_lithium.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.3.1: To calculate the cross section of Li(3,7) : Page 179(2011)\n", +"t = 10^-5; // Thickness of Li(3,7), m\n", +"d = 500; // Density, Kg/m^3\n", +"N = 6.023e+026; // Number of nuclei in 7-Kg of Li-7\n", +"M = 7 ; // Molar mass of Li\n", +"n = d*N*t/M; // Number of Li(3,7) nuclei/area\n", +"N_p = 10^8; // Number of neutron produced/s\n", +"N_0 = 10^13; // Number of incident particle striking/unit area of target\n", +"C_s = N_p/(N_0*n*10^(-028)); // Cross section, b\n", +"printf('\n Cross section : %5.3f b', C_s)\n", +" // Result \n", +"// Cross section : 0.232 b " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.3_2: Neutron_absorption_ratio.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.3.2: To calculate the fraction of neutron absorbed by Cd sheet of given thickness : Page 180 (2011)\n", +"t = 0.2e-03; // Thickness of Cd sheet, m\n", +"d = 8.64e+03; // Density, Kg/m^3\n", +"N = 6.023e+026; // Number of nuclei in 7-Kg of Li-7\n", +"M = 112 ; // Atomic mass of Cd-113, amu\n", +"C_s = 20000e-028; // Cross section of neutron for Cd-113, m^2\n", +"n = 0.12*d*N/M; // Number of Cd atoms/volume, atoms/m^3\n", +"F_inc_absorb = [1-%e^(-n*C_s*t)]*100; // Fraction of neutron absorbed \n", +"printf('\n Fraction of neutron absorbed by Cd sheet : %4.2f percent',F_inc_absorb )\n", +"// Result\n", +"// Fraction of neutron absorbed by Cd sheet : 89.25 percent " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.4_1: Nuclear_reactions.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa test : Checking the possibility of occurence of reactions : page no. 181 (2011)\n", +"// Declare three cells (for three reactions)\n", +"R1 = cell(4,4);\n", +"R2 = cell(5,4);\n", +"R3 = cell(4,4);\n", +"// Enter data for first cell (Reaction)\n", +"R1(1,1).entries = 'Al'; // Element\n", +"R1(1,2).entries = 13; // Atomic number\n", +"R1(1,3).entries = 27; // Mass number\n", +"R1(1,4).entries = 0; // Lepton number \n", +"R1(2,1).entries = 'He';\n", +"R1(2,2).entries = 2;\n", +"R1(2,3).entries = 4;\n", +"R1(2,4).entries = 0;\n", +"R1(3,1).entries = 'Si';\n", +"R1(3,2).entries = 14;\n", +"R1(3,3).entries = 30;\n", +"R1(2,4).entries = 0;\n", +"R1(4,1).entries = 'n';\n", +"R1(4,2).entries = 0;\n", +"R1(4,3).entries = 1;\n", +"R1(2,4).entries = 0;\n", +"// Enter data for second cell (Reaction)\n", +"R2(1,1).entries = 'U';\n", +"R2(1,2).entries = 92;\n", +"R2(1,3).entries = 235;\n", +"R2(1,4).entries = 0;\n", +"R2(2,1).entries = 'n';\n", +"R2(2,2).entries = 0;\n", +"R2(2,3).entries = 1;\n", +"R2(2,4).entries = 0;\n", +"R2(3,1).entries = 'Ba';\n", +"R2(3,2).entries = 56;\n", +"R2(3,3).entries = 143;\n", +"R2(3,4).entries = 0;\n", +"R2(4,1).entries = 'Kr';\n", +"R2(4,2).entries = 36;\n", +"R2(4,3).entries = 90;\n", +"R2(4,4).entries = 0;\n", +"R2(5,1).entries = '2n';\n", +"R2(5,2).entries = 0;\n", +"R2(5,3).entries = 1;\n", +"R1(5,4).entries = 0;\n", +"// Enter data for third cell (Reaction)\n", +"R3(1,1).entries = 'P';\n", +"R3(1,2).entries = 15;\n", +"R3(1,3).entries = 32;\n", +"R3(1,4).entries = 0;\n", +"R3(2,1).entries = 'S';\n", +"R3(2,2).entries = 16;\n", +"R3(2,3).entries = 32;\n", +"R3(2,4).entries = 0;\n", +"R3(3,1).entries = 'e';\n", +"R3(3,2).entries = -1;\n", +"R3(3,3).entries = 0;\n", +"R3(3,4).entries = 0;\n", +"R3(4,1).entries = 'v_e';\n", +"R3(4,2).entries = 0;\n", +"R3(4,3).entries = 0;\n", +"R3(4,4).entries = 0;\n", +"// Declare a function returning equality status of nucleon number\n", +"function f = check_nucleon(nr_sum,np_sum)\n", +" if nr_sum == np_sum then\n", +" f = 1;\n", +" else \n", +" f = 0;\n", +" end\n", +"endfunction\n", +"// Declare a function returning equality status of proton number\n", +"function f = check_proton(pr_sum,pp_sum)\n", +" if pr_sum == pp_sum then\n", +" f = 1;\n", +" else \n", +" f = 0;\n", +" end \n", +"endfunction\n", +"// Declare a function returning equality status of lepton number\n", +"function f = check_lepton(lr_sum,lp_sum)\n", +" if lr_sum == lp_sum then\n", +" f = 1;\n", +" else \n", +" f = 0;\n", +" end \n", +"endfunction\n", +" \n", +"// Reaction-I\n", +"printf('\n\n\nReaction-I:\n\n');\n", +" pr_sum = R1(1,2).entries+R1(2,2).entries;\n", +" pp_sum = R1(3,2).entries+R1(4,2).entries;\n", +" nr_sum = R1(1,3).entries+R1(2,3).entries;\n", +" np_sum = R1(3,3).entries+R1(4,3).entries;\n", +" lr_sum = R1(1,4).entries+R1(2,4).entries;\n", +" lp_sum = R1(3,4).entries+R1(4,4).entries; \n", +" if (check_nucleon(nr_sum,np_sum)&check_proton(pr_sum,pp_sum)&check_lepton(lr_sum,lp_sum) == 1) then\n", +" printf('The Reaction\n')\n", +" printf('\t%s(%d) + %s(%d) --> %s(%d)+%s(%d)\nis possible', R1(1,1).entries, R1(1,3).entries, R1(2,1).entries, R1(2,3).entries, R1(3,1).entries, R1(3,3).entries, R1(4,1).entries, R1(4,3).entries);\n", +" elseif (check_proton(pr_sum,pp_sum) == 0) then\n", +" printf('The Reaction\n')\n", +" printf('\t%s(%d) + %s(%d) --> %s(%d)+%s(%d)\nis impossible', R1(1,1).entries, R1(1,3).entries, R1(2,1).entries, R1(2,3).entries, R1(3,1).entries, R1(3,3).entries, R1(4,1).entries, R1(4,3).entries);\n", +" R1(4,1).entries = 'H'; R1(4,3).entries = 1;\n", +" printf('\nThe correct reaction is:\n')\n", +" printf('\t%s(%d) + %s(%d) --> %s(%d)+%s(%d)\n', R1(1,1).entries, R1(1,3).entries, R1(2,1).entries, R1(2,3).entries, R1(3,1).entries, R1(3,3).entries, R1(4,1).entries, R1(4,3).entries); \n", +" end\n", +"// Display for reaction-II\n", +" printf('\n\n\nReaction-II:\n\n');\n", +" pr_sum = R2(1,2).entries+R2(2,2).entries;\n", +" pp_sum = R2(3,2).entries+R2(4,2).entries+R2(5,2).entries;\n", +" nr_sum = R2(1,3).entries+R2(2,3).entries;\n", +" np_sum = R2(3,3).entries+R2(4,3).entries+R2(5,3).entries;\n", +" lr_sum = R2(1,4).entries+R2(2,4).entries;\n", +" lp_sum = R2(3,4).entries+R2(4,4).entries+R2(5,4).entries; \n", +" if (check_nucleon(nr_sum,np_sum)&check_proton(pr_sum,pp_sum)&check_lepton(lr_sum,lp_sum) == 1) then\n", +" printf('The Reaction\n')\n", +" printf('\t%s(%d) + %s(%d) --> %s(%d)+%s(%d)+%s(%d)\nis possible', R2(1,1).entries, R2(1,3).entries, R2(2,1).entries, R2(2,3).entries, R2(3,1).entries, R2(3,3).entries, R2(4,1).entries, R2(4,3).entries, R2(5,1).entries, R2(5,3).entries);\n", +" elseif (check_nucleon(nr_sum,np_sum) == 0) then\n", +" printf('The Reaction\n')\n", +" printf('\t%s(%d) + %s(%d) --> %s(%d)+%s(%d)+%s(%d)\nis impossible', R2(1,1).entries, R2(1,3).entries, R2(2,1).entries, R2(2,3).entries, R2(3,1).entries, R2(3,3).entries, R2(4,1).entries, R2(4,3).entries, R2(5,1).entries, R2(5,3).entries);\n", +" R2(5,1).entries = '3n';\n", +" printf('\nThe correct reaction is:\n')\n", +" printf('\t%s(%d) + %s(%d) --> %s(%d)+%s(%d)+%s(%d)\n', R2(1,1).entries, R2(1,3).entries, R2(2,1).entries, R2(2,3).entries, R2(3,1).entries, R2(3,3).entries, R2(4,1).entries, R2(4,3).entries, R2(5,1).entries, R2(5,3).entries); \n", +" end \n", +"// Reaction-III\n", +" printf('\n\n\nReaction-III:\n\n');\n", +" pr_sum = R3(1,2).entries+R3(2,2).entries;\n", +" pp_sum = R3(3,2).entries+R3(4,2).entries;\n", +" nr_sum = R3(1,3).entries+R3(2,3).entries;\n", +" np_sum = R3(3,3).entries+R3(4,3).entries;\n", +" lr_sum = R3(1,4).entries+R3(2,4).entries;\n", +" lp_sum = R3(3,4).entries+R3(4,4).entries; \n", +" if (check_nucleon(nr_sum,np_sum)&check_proton(pr_sum,pp_sum)&check_lepton(lr_sum,lp_sum) == 1) then\n", +" printf('The Reaction\n')\n", +" printf('\t%s(%d) + %s(%d) --> %s(%d)+%s(%d)\nis possible', R3(1,1).entries, R3(1,3).entries, R3(2,1).entries, R3(2,3).entries, R3(3,1).entries, R3(3,3).entries, R3(4,1).entries, R2(4,3).entries);\n", +" elseif (check_lepton(nr_sum,np_sum) == 0) then\n", +" printf('The Reaction\n')\n", +" printf('\t%s(%d) + %s(%d) --> %s(%d)+%s(%d)\nis impossible', R3(1,1).entries, R3(1,3).entries, R3(2,1).entries, R3(2,3).entries, R3(3,1).entries, R3(3,3).entries, R3(4,1).entries, R3(4,3).entries);\n", +" R3(4,1).entries = 'v_e_a'\n", +" printf('\nThe correct reaction is:\n')\n", +" printf('\t%s(%d) + %s(%d) --> %s(%d)+%s(%d)\n', R3(1,1).entries, R3(1,3).entries, R3(2,1).entries, R3(2,3).entries, R3(3,1).entries, R3(3,3).entries, R3(4,1).entries, R3(4,3).entries); \n", +" end \n", +" \n", +"// Reaction-I:\n", +"// The Reaction\n", +"// Al(27) + He(4) --> Si(30)+n(1)\n", +"// is impossible\n", +"// The correct reaction is:\n", +"// Al(27) + He(4) --> Si(30)+H(1)\n", +"// Reaction-II:\n", +"// The Reaction\n", +"// U(235) + n(1) --> Ba(143)+Kr(90)+2n(1)\n", +"// is impossible\n", +"// The correct reaction is:\n", +"// U(235) + n(1) --> Ba(143)+Kr(90)+3n(1)\n", +"// Reaction-III:\n", +"// The Reaction\n", +"// P(32) + S(32) --> e(0)+v_e(0)\n", +"// is impossible\n", +"// The correct reaction is:\n", +"// P(32) + S(32) --> e(0)+v_e_a(0)\n", +" " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5_10: Energy_of_gamma_rays.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.5.10: To determine the energy of gamma ray for reaction :: P.no. 186 (2011)\n", +"// H(1,2)+G = H(1,1)+ n(0,1) is the given reaction\n", +"M_H_2 = 2.014735; // Mass of H-2, amu\n", +"M_H_1 = 1.008142 ; // Mass of H-1, amu\n", +"M_n_1 = 1.008987; // Mass of M_n_1, amu\n", +"Q = -5.4; // Q-value, MeV\n", +"E_g = (M_H_1*931.47+M_n_1*931.47)-(M_H_2*931.47); //Energy of the gama rays, MeV\n", +"printf('\nThe energy of the gama rays : %6.4f MeV ', E_g)\n", +"// Result \n", +"// The energy of the gama rays : 2.2299 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5_1: Q_value_for_reactio.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.5.1: To calculate Q-value for given reaction : Page 182 (2011)\n", +"M_n = 1.00866501; // Mass of neutron, amu\n", +"M_Hp = 2.014102; // Mass of proton, amu\n", +"M_Hd = 3.016049; // Mass of deutron, amu\n", +"M_He = 4.002603; // Mass of alpha particle, amu\n", +"Q = [M_Hp+M_Hd-M_He-M_n]*931.49; // Q-value, MeV\n", +"printf('\nThe Q-value for the reaction : %4.1f MeV', Q)\n", +"// Result\n", +"// The Q-value for the reaction : 17.6 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5_2: Energy_emitted_in_nuclear_reaction.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.5.2: To calculate Q-value for the reaction : Page 183 (2011)\n", +"M_Cf = 252.081621; // Mass of califronium, amu\n", +"M_Cm = 248.072343; // Mass of curium, amu\n", +"M_He = 4.002603; // Mass of alpha particle, amu\n", +"Q = [M_Cf-M_Cm-M_He]*931.49; // Q-value, MeV\n", +"printf('\nThe Q-value for the reaction : %4.2f MeV', Q)\n", +"// Result \n", +"// The Q-value for the reaction : 6.22 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5_3: Threshold_energy_and_Q_value_for_nuclear_reaction.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.5.3: To calculate Q-value and threshold energy for the given reaction : Page 183 (2011)\n", +"// Pb_208(Fe_56, Fe_54)Pb_210\n", +"M_Pb_208 = 207.976641; // Mass of Pb-208, amu\n", +"M_Fe_56 = 55.934939; // Mass of Fe-56, amu\n", +"M_Pb_210 = 209.984178; // Mass of Pb-210, amu\n", +"M_Fe_54 = 53.939612; // Mass of Fe-54, amu\n", +"Q = [M_Pb_208+M_Fe_56-M_Pb_210-M_Fe_54]*931.49; // Q-value, MeV\n", +"E_th = -Q*(M_Fe_56+M_Pb_208)/M_Pb_208; // Threshold energy, MeV \n", +"printf('\nThe Q-value for the reaction = %5.2f MeV \n Threshold energy = %5.2f MeV ', Q,E_th)\n", +"// Result \n", +"// The Q-value for the reaction = -11.37 MeV \n", +"// Threshold energy = 14.43 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5_4: Mass_of_neutron_from_nuclear_reaction.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.5.4: To calculate the mass of neutron for given reaction : P.No. 184 (2011)\n", +"// H(1,1)+n(0,1) = H(1,2)+G is the reaction\n", +"M_H_2 = 2.014735; // Mass of H-2, amu\n", +"M_H_1 = 1.008142 ; // Mass of H-1, amu\n", +"E_g = 2.230; // Energy of gamma rays, MeV\n", +"M_n_1 = [(M_H_2*931.47+E_g)-(M_H_1*931.47)]/931.47; //Mass of neutron, amu\n", +"printf('\nThe mass of the neutron : %8.6f MeV ', M_n_1)\n", +"// Result\n", +"// The mass of the neutron : 1.008987 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5_5: Q_value_sign_for_nuclear_reaction.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 4.5.5 : Checking given reaction condition : page no. 184 (2011)\n", +"// Li-6 + n-1 > He-4 + H-3 is the given reaction \n", +"M_Li = 6.0151234; // Atomic mass of Li, amu\n", +"M_n = 1.0086654; // Atomic mass of neutron, amu\n", +"M_He = 4.0026034; // Atomic mass of He, amu\n", +"M_H = 3.0160294; // Atomic mass of H, amu\n", +"r_sum = M_Li+M_n; // Sum of reactant, amu\n", +"p_sum = M_He+M_H; // Sum of product, amu\n", +"// Declare a function returning equality status of nucleon number\n", +"function Q = check_Qvalue(r_sum,p_sum)\n", +" if r_sum >= p_sum then\n", +" Q = 1;\n", +" else \n", +" Q = 0;\n", +" end\n", +"endfunction\n", +"// Reaction\n", +" if (check_Qvalue(r_sum,p_sum) == 1) then\n", +" printf('\n Reaction : \n\n\t Li(6)+n(1) ----> He(4)+H(3)')\n", +" printf('\n\n\t\tThis reaction is exoergic')\n", +" elseif (check_Qvalue(r_sum,p_sum) == 0) then\n", +" printf('\n Reaction : \n\n\t Li(6)+n(1) ----> He(4)+H(3)')\n", +" printf('\n\n\t\tThis reaction is endoergic')\n", +" end \n", +"// Reaction : \n", +"// Li(6)+n(1) ----> He(4)+H(3)\n", +" // This reaction is exoergic" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5_6: Spontaneity_of_Q_value_for_nulclear_reaciton.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 4.5.5 : Checking whether the reaction is spontaneous or exoergic : page no. 185 (2011)\n", +"// Cf-252 > Zr-98 +Ce-145 + 9*n-1 is the given reaction\n", +"M_Cf = 252.081621; // Atomic mass of Cf, amu\n", +"M_Zr = 97.912735; // Atomic mass of Zr, amu\n", +"M_Ce = 144.917230; // Atomic mass of Ce, amu\n", +"M_n = 3.0160294; // Atomic mass of neutron, amu\n", +"r_sum = M_Cf+M_Zr; // Sum of reactant, amu\n", +"p_sum = M_Ce+M_n; // Sum of product, amu\n", +"// Declare the function which check the Q-value \n", +"function Q = check_Qvalue(r_sum,p_sum)\n", +" if r_sum >= p_sum then\n", +" Q = 1;\n", +" else \n", +" Q = 0;\n", +" end\n", +"endfunction\n", +"// Reaction\n", +" if (check_Qvalue(r_sum,p_sum) == 1) then\n", +" printf('\n Reaction : \n\n\t Cf(256) ----> Zr(98)+Ce(145)+9*n(1)')\n", +" printf('\n\n\t\tThis reaction is spontaneous')\n", +" elseif (check_Qvalue(r_sum,p_sum) == 0) then\n", +" printf('\n Reaction : \n\n\t Cf(256) ----> Zr(98)+Ce(145)+9*n(1)')\n", +" printf('\n\n\t\tThis reaction is not spontaneous')\n", +" end \n", +" // Reaction : \n", +" // Cf(256) ----> Zr(98)+Ce(145)+9*n(1)\n", +" // This reaction is spontaneous " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5_7: Nuclear_reaction_Q_value.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.5.7: To calculate Q-value for given reaction : Page 185 (2011)\n", +"// O(8,16) > N(7,15)+ H(1,1) is the given reaction\n", +"M_N_15 = 15.000108; // Mass of N-15, amu\n", +"M_O_16 = 16; // Mass of O-16, amu\n", +"M_H_1 = 1.007825; // Mass of H-1, amu\n", +"Q = [M_O_16-M_N_15-M_H_1]*931.49; // Q-value, MeV\n", +"printf('\nThe Q-value for the reaction : %3.1f MeV ', Q)\n", +"// Result \n", +"//The Q-value for the reaction : -7.4 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5_8: Threshold_energy_for_given_reaction.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.5.8: To determine the threshold energy for given reaction : P.no. 185 (2011)\n", +"// Na(11,23)+ n > F(9,20)+ He(2,4) is the reaction\n", +"M_Na_23 = 22.99097; // Mass of Na-23, amu\n", +"M_n_1 =1.00866 ; // Mass of n-1, amu\n", +"Q = -5.4; // Q-value, MeV\n", +"E_th = -Q*(M_Na_23+M_n_1)/M_Na_23; // Threshold energy, MeV\n", +"printf('\nThe threshold energy for the reaction : %4.2f MeV ', E_th)\n", +"// Result \n", +"// The threshold energy for the reaction : 5.64 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.5_9: Q_value_of_nuclear_reaction.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.5.9: To calculate Q-value for the reaction : Page 187 (2011)\n", +"// He(2,4)+ N(7,14) = O(8,17)+ H(1,1)is the given reaction\n", +"M_N_14 = 14.00755; // Mass of N-14, amu\n", +"M_He_4 = 4.00388; // Mass of He-4, amu\n", +"M_O_17 = 17.00452; // Mass of O-17, amu\n", +"M_H_1 = 1.00815; // Mass of H-1, amu\n", +"Q = [M_N_14+M_He_4-M_O_17-M_H_1]*931.49; // Q-value, MeV\n", +"printf('\nThe Q-value for the reaction : %4.2f MeV ', Q)\n", +"// Result \n", +"//The Q-value for the reaction : -1.16 MeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7_1: Energy_and_power_released_during_fission_of_U_235.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.7.1: To calculate the energy and power released during fission of U-235 : Page 189 (2011)\n", +"m = 0.001; // Mass of U-235 lost during fission, Kg\n", +"c = 3e+08; // Velocity of light, m/s\n", +"E = m*c^2; // Energy released during fission, J\n", +"E_t = E/(4e+09*1000); // Energy requires TNT, Kt \n", +"printf('\n Energy released during fission = %1.0e J \n Destructive power of bomb = %4.1f Kt of TNT', E, E_t)\n", +"// Result \n", +" // Energy released during fission = 9e+013 J \n", +"// Destructive power of bomb = 22.5 Kt of TNT " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7_2: Fission_rate_induced_in_the_uranium_foil_by_neutron.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.7.2: To determine the fission rate induced in the foil by neutron : Page 190 (2011)\n", +"t = 0.15; // Thickness of the foil, Kg\n", +"N = 6.023e+026; // Number of nuclei in 1Kg of U-235, nuclei\n", +"N_1 = N/235*t; // Number of nuclei in 0.15Kg of U-235, nuclei\n", +"A = 2e-026; // Area present in each nucleus, m^2\n", +"I = 10^6; // Intensity ,s^-1 \n", +"F_r = N_1*A; // Rate of fissions induced in the foil by the neutrons, s^-1\n", +"printf('\n Rate of fissions induced in the foil by the neutrons: %5.3e per sec', F_r)\n", +"// Result \n", +"// Rate of fissions induced in the foil by the neutrons: 7.689e-003 per sec" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7_3: Power_in_fission_process.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.7.3: To determine the fission power produced by one microgram of Fm-256 : Page 190 (2011)\n", +"N = 6.023e+023/256*10^-6; // Number of nuclei in 1ug of Fm-256\n", +"t_h = 158*60; // Half life of Fm-256, s\n", +"D_c = log(2)/t_h; // Decay constant, s^-1\n", +"F_r = N*D_c; // Fission rate, fissions/s\n", +"E = 220*1.6e-013; // Energy released during fission of one nucleus, J\n", +"P = E*F_r; // Power released in fission of 1 microgram of Fm-256, W\n", +"printf('\n Power released in fission of 1 microgram of Fm-256 = %d W', P)\n", +"// Result \n", +"// Power released in fission of 1 microgram of Fm-256 = 6 W " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7_4: Power_released_in_fission.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.7.4: To determine the power produced by 100 milligram of Cf-252 : Page 191 (2011)\n", +"N = 6.023e+023/252*0.1; // Number of nuclei in 100mg of Cf-252\n", +"t_h = 2.62*365*24*3600; // Half life of Cf-252, s\n", +"D_c = log(2)/t_h; // Decay constant, s^-1\n", +"F_r = N*D_c; // Fission rate, fissions/s\n", +"E = 210*1.6e-013; // Energy released during fission of one nucleus, J\n", +"P = E*F_r; // Power released in fission of 100 milligram of Cf-252, W\n", +"printf('\n Power released in fission of 100 milligram of Cf-252: %4.1f W', P)\n", +"// Result \n", +"// Power released in fission of 100 milligram of Cf-252: 67.4 W " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.7_5: Fission_counts_and_mass_reduction_of_fissile_material.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.7.5: To determine the number of nuclear fission and decrease in mass during explosion at hiroshima : Page 191 (2011)\n", +"E = 200*1.6e-013; // Energy released during fission of one nucleus, J\n", +"E_t = 20000*4.18e+09; // Energy released in detonation of 20000 tons of TNT, J\n", +"N_f = E_t/E; // Number of fission occured during eplosion, fissions\n", +"c = 3e+08; // Velocity of light, m/s\n", +"m = E_t/(c)^2*10^6; // Decrease in mass during explosion, mg\n", +"m_r = round(m)\n", +"printf('\n Number of fissions occured during explosion = %4.2e fissions \n Decrease in mass during explosion = %d mg ', N_f, m_r)\n", +"// Result \n", +"// Number of fissions occured during explosion = 2.61e+024 fissions \n", +"// Decrease in mass during explosion = 929 mg" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8_1: Energy_liberated_in_fusion_reaction.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.8.1: To calculate the energy liberated during fusion reaction: Page 194 (2011)\n", +"// 5*H(1,2)= He(2,3)+He(2,4)+H(1,2)+2*n(0,1)+25MeV is the given reaction\n", +"N = 6.023e+026/2*10; // Number of atoms in 10Kg of H-2, atoms\n", +"E = 25/5*1.6e-013; // Energy liberate during fusion of 1 atom of H-2, J\n", +"E_l = E*N; // Energy liberate during fusion of 10 Kg of H-2, J\n", +"printf('\n Energy liberated during fusion of 10 Kg of H-2 = %4.2e J', E_l)\n", +"// Result \n", +"// Energy liberated during fusion of 10 Kg of H-2 = 2.41e+015 J" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8_2: Energy_produced_by_helium_carbon_fusion.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.8.2: To calculate the energy produced by the fusion reaction He(2,4)+C(6,12)= O(8,16) : Page 194 (2011)\n", +"M_r = 16.002603; // Mass of the reactant, amu\n", +"M_p = 15.994915; // Mass of reactant, amu\n", +"M_d = 7.688e-03; // Difference in masses, amu\n", +"E_p = M_d*931.49; // Energy produced, MeV\n", +"printf('\n Energy produced by the fusion reaction :%4.2f MeV', E_p)\n", +"// Result \n", +"// Energy produced by the fusion reaction :7.16 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8_3: Energy_released_and_temperature_required_for_fusion_of_gases.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.8.3: To calculate the energy released and temperature required for fusion of given gases : Page 194 (2011)\n", +"// Firstly calculate for B-10\n", +"Z_B = 5; // Atomic number of B-10\n", +"r_B = 5.17; // Seperation of two nuclei, fm\n", +"K = 1.38e-023; // Boltzmann's constant\n", +"F = 1/137; // Fine structure constant\n", +"E = 197.5*1.6e-013; // Energy, J\n", +"V_c_B = F*Z_B^2*E/r_B; // Coulomb barrier for B-10, J\n", +"T_B = 2/3*V_c_B/K; // Temperature required to overcome the barrier for B-10, K\n", +"// Now calculate for Mg-24\n", +"Z_Mg = 12; // Atomic number of Mg-24\n", +"r_Mg = 6.92; // Seperation of two nuclei, fm\n", +"K = 1.38e-023; // Boltzmann's constant\n", +"F = 1/137; // Fine structure constant\n", +"E = 197.5*1.6e-013; // Energy, J\n", +"V_c_Mg = F*Z_Mg^2*E/r_Mg; // Coulomb barrier for Mg-24, J\n", +"T_Mg = 2/3*V_c_Mg/K; // Temperature required to overcome the barrier for Mg-24, K\n", +"printf('\nFor B-10 \n Energy released = %4.2e J \n Temperature required = %4.1e K \nFor Mg-24 \n Energy released = %4.2e J \n Temperature required = %4.2e K', V_c_B,T_B, V_c_Mg, T_Mg)\n", +"// Result \n", +" // For B-10 \n", +" // Energy released = 1.12e-012 J \n", +" // Temperature required = 5.4e+010 K \n", +"// For Mg-24 \n", +"// Energy released = 4.80e-012 J \n", +"// Temperature required = 2.32e+011 K " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8_4: Life_time_of_su.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.8.4: To calculate the life time of sun for given reaction : Page 196 (2011)\n", +"// 4*H(1,1)= He(2,4)+2*e(1,0)+2*v+G is the reaction\n", +"E_r = 3.9e+026; // Energy releasd in 1s, J\n", +"N = 1.2e+057; // Number of hydrogen atoms in the sun, atoms\n", +"M_d = 0.027599;// Mass difference, amu\n", +"E = M_d*931.47; // In terms of energy, MeV\n", +"E_t = N/4*E*1.6e-013; // Total energy available in the sun, J\n", +"t = E_t/(E_r*365*24*3600*10^9); // Life time of the sun, billion years\n", +"printf('\n Life time of the sun : %5.1f billion years', t)\n", +"// Result \n", +"// Life time of the sun : 100.3 billion years" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8_5: Particle_identification_in_the_nuclear_reaction.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 4.8.5 : Identifying the nucleus and energy released in the given reaction : page no. 197 (2011)\n", +"// Declare three cells (for three reactions)\n", +"R = cell(4,3);\n", +"// Enter data for first cell (Reaction)\n", +"R(1,1).entries = 'H'; // Element\n", +"R(1,2).entries = 1; // Atomic number\n", +"R(1,3).entries = 2; // Mass number\n", +"R(2,1).entries = 'H';\n", +"R(2,2).entries = 1;\n", +"R(2,3).entries = 3;\n", +"R(3,1).entries = 'n'\n", +"R(3,2).entries = 0;\n", +"R(3,3).entries = 1;\n", +"R(4,1).entries = 'He'\n", +"R(4,2).entries = 2;\n", +"R(4,3).entries = 3;\n", +"// Declare a function returning equality status of nucleon number\n", +" \n", +" p_sum = R(1,2).entries+R(2,2).entries;\n", +" if (p_sum == 2) then\n", +" \n", +" printf('\n The particle is : %s(%d,%d) ',R(4,1).entries,R(4,2).entries,R(4,3).entries ) \n", +" end \n", +"// Calculate the energy released\n", +"m_n = 1.008665; // Mass of neutron, amu\n", +"m_d = 2.014102; // Mass of deutron, amu\n", +"m_He = 3.0160293; // Mass of He-3, amu\n", +"E = [2*m_d-(m_n+m_He)]*931.47; // Energy released in this reaction, MeV\n", +"printf('\n The energy released in this reaction : %4.2f MeV', E )\n", +"// Result \n", +"// The particle is : He(2,3) \n", +" // The energy released in this reaction : 3.27 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4.8_6: Mass_defect_and_q_value_for_fusion_reaction.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa4.8.6: To calculate the mass defect and Q-value for the fusion reactions : Page 197 (2011)\n", +"// Reaction-1 = H(1,2)+H(1,2)= He(2,3)+n(0,1)\n", +"m_p = 1.007825; // Mass of proton, amu\n", +"m_n = 1.008665; // Mass of neutron, amu\n", +"m_H = 2.014102; // Mass of H(1,2), amu\n", +"m_He = 3.016029; // Mass of He(2,3), amu\n", +"m_d_1 = 2*m_H-m_He-m_n; // Mass defect for reaction first, amu\n", +"Q_1 = m_d_1*931.47; // Q-value for reaction first, MeV\n", +"// Reaction-2 = H(1,2)+H(1,2)= H(1,3)+p(1,1)\n", +"m_p = 1.007825; // Mass of proton, amu\n", +"m_n = 1.008665; // Mass of neutron, amu\n", +"m_H = 2.014102; // Mass of H(1,2), amu\n", +"m_H_3 = 3.016049; // Mass of H(1,3), amu\n", +"m_d_2 = 2*m_H-m_H_3-m_p; // Mass defect for reaction second, amu\n", +"Q_2 = m_d_2*931.47; // Q-value for reaction second, MeV\n", +"printf('\nFor first reaction \n Mass defect = %7.5f amu \n Q-value = %7.5f amu \nFor second reaction \n Mass defect = %7.5f MeV \n Q-value = %4.2f MeV ', m_d_1,Q_1, m_d_2, Q_2)\n", +"// Result \n", +"// For first reaction \n", +"// Mass defect = 0.00351 amu \n", +"// Q-value = 3.26946 amu \n", +"// For second reaction \n", +"// Mass defect = 0.00433 MeV \n", +"// Q-value = 4.03 MeV " + ] + } +], +"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/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/5-Interaction_of_Radiations_with_Matter.ipynb b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/5-Interaction_of_Radiations_with_Matter.ipynb new file mode 100644 index 0000000..4f3409a --- /dev/null +++ b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/5-Interaction_of_Radiations_with_Matter.ipynb @@ -0,0 +1,729 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: Interaction of Radiations with Matter" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.2_1: Energy_lost_during_collision.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.2.1: To calculate the energy and no. of collision required to stop collision : P.no. 223 (2011)\n", +"m = 511; // Mass of electron, KeV\n", +"M = 938*10^3; // Mass of incident charged particle, KeV\n", +"E = 10*10^3; // Energy of proton, KeV\n", +"E_l = 4*m*E/M; // Energy lost during collison, KeV\n", +"n = E/E_l; // Number of collisions, \n", +"N = round(n)\n", +"printf('\n The energy lost during collision = %5.2f KeV \n Number of collision required = %d collisions',E_l, N )\n", +"// Result\n", +"// The energy lost during collision = 21.79 KeV \n", +"// Number of collision required = 459 collisions " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.5_1: Half_value_thickness_of_aluminium.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.5.1: To calculate the half value thickness of Al for given radiation : P.no. 225 (2011)\n", +"x = 0.2; // Thickness of Al material , m\n", +"I_r = 3/100; // Intensity ratios, \n", +"x_h = log(2)*x/log(1/I_r); // Half value thickness of Al, m\n", +"printf('\n Half value thickness of Al : %6.4f m',x_h )\n", +"// Result\n", +"// Half value thickness of Al : 0.0395 m " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.5_2: Thickness_of_lead.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.5.2: To calculate the thickness of Pb: P.no. 226 (2011)\n", +"u = 0.75; // Absorption coefficient , cm^-1\n", +"I_r = 1/100; // Intensity ratios, \n", +"x = log(1/I_r)*u; // Thckness of Pb, cm\n", +"printf('\n Thickness of Pb : %5.3f cm',x )\n", +"// Result\n", +"// Thickness of Pb : 6.140 m " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.5_3: Percentage_loss_of_intensity_of_gamma_rays.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.5.3: To calculate the percentage loss of intensity of gamma rays : P.no. 226 (2011)\n", +"x_h = 5; // Half thickness of an absorber, mm\n", +"u = log(2)/x_h; // Absorption coefficient, mm^-1\n", +"x = 20; // Thickness of an absorber, mm\n", +"I_r = %e^(-u*x); // Intensity ratios, \n", +"P_loss = I_r*100; // Percentage loss in intensity, percent\n", +"printf('\n Percentage loss in intensity : %4.2f percent',P_loss )\n", +"// Result\n", +"// Percentage loss in intensity : 6.25 percent " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.6_1: Velocity_of_ejected_photoelectro.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.6.1: To calculate the velocity of ejected photoelectron : P.no. 230 (2011)\n", +" C = 3e+08; // Speed of light, m/s\n", +" h = 6.626e-034; // Planck's constant, Js\n", +" lambda = 2500e-010; // wavelength of light, m\n", +"e = 1.602e-019; // Charge of electron, C\n", +" w = 1.9; // Work function, J\n", +"m = 9.1e-031; // Mass of the electron, kg \n", +"E_c = h*C/(lambda*e); // Calculated energy, J\n", +"E_e = E_c-w; // Energy of photoelectron, J\n", +"v = sqrt((2*E_e*e)/m); // Velocity of photoelectron, m/s\n", +"printf('\nThe velocity of photoelectron : %4.2e m/s ', v )\n", +"// Result\n", +"// The velocity of photoelectron : 1.04e+006 m/s \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.6_2: Rate_of_photoelectron_emission.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.6.2: To calculate the kinetic energy of photoelectron and rate at which photoelectron emitted : P.no. 231 (2011)\n", +" C = 3e+08; // Speed of light, m/s\n", +" h = 6.626e-034; // Planck's constant, Js\n", +" lambda = 250e-09; // Wavelength of light, m\n", +" w = 2.30; // Work function, eV\n", +"A = 2e-04; // Area of the surface, m^2 \n", +"I = 2; // Intensity of light, W/m^2\n", +"e = 1.6e-019; // Charge of the electron, C\n", +"E_p = h*C/(lambda*e); // Energy of photoelectron, eV\n", +"E_max = E_p-w; // Maximum kinetic energy of photoelectron, eV\n", +"n_p = I*A/(E_p*e); // Number of photons reaching the surface per second, photons/s\n", +"R_p = 0.2/100*n_p; // Rate at which photoelectrons are emitted, photoelectrons/s\n", +"printf('\n The maximum kinetic energy = %4.2f eV \n The rate at which photoelectrons are emitted = %4.2e photoelectrons/s ', E_max, R_p)\n", +"// Result\n", +"// The maximum kinetic energy = 2.67 eV \n", +" // The rate at which photoelectrons are emitted = 1.01e+012 photoelectrons/s\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.6_3: Kinetic_energy_of_photoelectron.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.6.3: To calculate the wavelength of light whose kinetic energy is given : P. No. 232 (2011)\n", +" C = 3e+08; // Speed of light, m/s\n", +" h = 6.626e-034; // Planck's constant, Js\n", +" T_lambda = 190e-09; // Threhold wavelength of light, m\n", +"e = 1.6e-019; // Charge of the electron, C\n", +"E_max = 1.1; // Maximum kinetic energy of photoelectron, eV\n", +"w = h*C/(T_lambda*e); // Work function, eV \n", +"E_t = E_max+w; // threshold energy, eV\n", +"lambda = h*C/(E_t*e); // Wavelength of light used, m\n", +"printf('\nThe wavelength of light used : %5.3e m', lambda)\n", +"// Result\n", +"// The wavelength of light used : 1.626e-007 m\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7_1: Compton_shift.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.7.1: To calculate the Compton shift : P.no. 233 (2011)\n", +"h = 6.62e-034; // Value of Planck's constant, J\n", +"m_e = 9.11e-031; // Mass of the electron,Kg\n", +"c = 3e+08; // Velocity of light, m/s\n", +"A = 65; // Angle between scattered radiation and incident radiation, degree\n", +"C_s = h/(m_e*c)*(1-cosd(A)); // Compton shift, m\n", +"printf('\nCompton shift : %4.2e m',C_s )\n", +"// Result\n", +"// Compton shift : 1.40e-012 m" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7_2: Wavelength_of_the_scattered_gamma_rays.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.7.2: To calculate the wavelength of the scattered gamma rays: P.no. 233 (2011)\n", +"h = 6.626e-034; // Value of Planck's constant, J\n", +"m_e = 9.11e-031; // Mass of the electron,Kg\n", +"c = 3e-04; // Velocity of light, m/s\n", +"A = 135; // Angle between scattered radiation and incident radiation, degree\n", +"W_i = 1.87; // Wavelength of incident radiation, pm\n", +"W_s = W_i + [h*(1-cosd(A))]/(m_e*c); // Wavelength of scattered radiation, pm\n", +"printf('\nWavelength of scattered radiation : %4.2f pm',W_s )\n", +"// Result\n", +"// Wavelength of scattered radiation : 6.01 pm " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7_3: Wavelength_of_the_incident_beam_of_X_rays.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.7.3: To calculate the wavelength of the incident beam of X-rays : P.no. 234 (2011)\n", +"h = 6.626e-034; // Value of Planck's constant, J\n", +"m_e = 9.11e-031; // Mass of the electron,Kg\n", +"c = 3e-04; // Velocity of light, pm/s\n", +"A = 90; // Angle between scattered radiation and incident radiation, degree\n", +"W_s = 3.8; // Wavelength of scattered radiation, pm\n", +"W_i = [W_s - h/(m_e*c)*(1-cosd(A))]; // Wavelength of incident beam of Xrays, pm\n", +"printf('\nWavelength of incident beam of X-rays : %4.2f pm', W_i )\n", +"// Result\n", +"// Wavelength of incident beam of X-rays : 1.38 pm " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7_4: Frequency_of_the_scattered_photon.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 5.7.4 : To calculate the frequency of the scattered photon Page.no. 234 (2011)\n", +"h = 6.626e-034; // Value of Planck's constant, J\n", +"m_e = 9.11e-031; // Mass of the electron,Kg\n", +"c = 3e+08; // Velocity of light, pm/s\n", +"A = 60; // Angle between scattered radiation and incident radiation, degree\n", +"v_0 = 3.2e+019; // Frequency of the incident photon, Hz\n", +"V = 1/v_0 + h/(m_e*c^2)*(1-cosd(A)); \n", +"v =(1/V); // Frequency of the scattered photon, Hz\n", +"printf('\n Frequency of the scattered photon: %4.2e Hz', v )\n", +"// Result\n", +"// Frequency of the scattered photon: 2.83e+019 Hz" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7_5: Energy_of_scattered_photon_and_recoil_electron.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 5.7.5 : To calculate the energy of the scattered photon and the energy of recoil electron : P.no. 235 (2011)\n", +"h = 6.626e-034; // Value of Planck's constant, J\n", +"m_e = 9.11e-031; // Mass of the electron,Kg\n", +"c = 3e+08; // Velocity of light, pm/s\n", +"A = 180; // Angle between scattered radiation and incident radiation, degree\n", +"E_i = 1836; // Energy of the incident electron, KeV\n", +"E = 1/E_i + 1/511*(1-cosd(A)); \n", +"E_s = round(1/E); // Energy of the sscattered photon, KeV\n", +"E_r = E_i-E_s; // Energy of the recoil electron, KeV\n", +"printf('\n Energy of the scattered photon = %d KeV \n Energy of the recoil electron = %d KeV ', E_s, E_r )\n", +"// Result\n", +"// Energy of the scattered photon = 224 KeV \n", +" // Energy of the recoil electron = 1612 KeV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.7_6: Scattering_angle_of_X_rays.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa 5.7.6 : To calculate the scattering angle of X-rays Page.no. 235 (2011)\n", +"E_s = 180; // Energy of the scattered X-rays, KeV\n", +"E_i = 200; // Energy of the incident X-rays, KeV\n", +"a = acosd(1-[{1/E_s-1/E_i}*511]); // \n", +"A = round(a); // Scattering angle of X-rays, degree\n", +"printf('\n Scattering angle of X-rays: %d degree', A )\n", +"// Result\n", +"// Scattering angle of X-rays: 44 degree " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5.8_1: Kinetic_energy_of_electron_and_positron.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa5.8.1: To calculate the kinetic energy of electron and positron :P.no. 236 (2011)\n", +" M_e = 0.511; // Rest mass of electron, MeV\n", +" M_p = 0.511; // Rest mass of positron, MeV\n", +" E_c = M_e+M_p; // Energy consumed, Mev\n", +" E_g = 5.0; // Given energy, MeV\n", +" E_l = E_g-E_c; // Energy left, Mev\n", +" E_k = E_l/2; // Kinetic energy of electron and positron, MeV\n", +" printf('\n The kinetic energy of electron and positron : %5.3f Mev', E_k)\n", +"// Result\n", +"// The kinetic energy of electron and positron : 1.989 Mev\n", +" \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +], +"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/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/6-Particle_Accelerators.ipynb b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/6-Particle_Accelerators.ipynb new file mode 100644 index 0000000..98f59d0 --- /dev/null +++ b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/6-Particle_Accelerators.ipynb @@ -0,0 +1,509 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: Particle Accelerators" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.10_1: Magnetic_field_of_the_electron.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.10.1 : To calculate the value of magnetic field of the electron whose energy is given Page 274(2011)\n", +"q = 1.602e-019; // Charge of an electron, C\n", +"r = 0.28; // Radius of stable orbit,m\n", +"E = 70*1.6e-013; // Energy of the electron, j\n", +"c = 3e+08; // Velocity of light, m/s\n", +"B = E/(e*r*c); // Magnetic field, T \n", +"printf('\nThe magnetic field of the electron : %4.2f T', B)\n", +"// Result\n", +"// The magnetic field of the electron : 0.83 T " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.10_2: Radius_of_proton_orbit_in_synchrotron.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.10.2 : To calculate the radius of proton orbit in synchrotron of given energy Page 275(2011)\n", +"c= 3e+08; // Speed of light in vacuum, m/s\n", +"q = 1.602e-019; // Charge on proton, coulomb\n", +"amu = 931; // Energy equivalent of 1 amu, MeV\n", +"m = 938; // Rest mass of a proton, MeV\n", +"KE = 12e+03; // Kinetic energy of proton, MeV\n", +"B = 1.9; // Magnetic field, T\n", +"E = m + KE; // Total energy of proton, MeV\n", +"// As E = m*amu, solving for m, the mass of proton\n", +"m = E/amu*1.672e-027; // Proton mass in motion, kg \n", +"v = 0.9973*c; // Velocity of the proton, m/s\n", +"r = m*v/(B*q); // Radius of the proton, m \n", +"printf('\nRadius of the proton orbit : %4.2f m', r)\n", +"// Result\n", +"// Radius of the proton orbit: 22.84 m" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.2_1: Kinetic_energy_of_protons.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.2.1 : To calculate the kinetic energy of protons : Page 264 (2011)\n", +"q = 1; // Number of proton, \n", +"V = 800; // Voltage applied to the dome, kV\n", +"E = q*V; // The kinetic energy of proton,keV\n", +"printf('\nThe kinetic energy of proton : %d keV', E);\n", +"// Result\n", +"// The kinetic energy of proton : 800 keV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3_1: Protons_in_Van_de_Graff_accelerator.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.3.1 : To calculate the kinetic energy of protons in Van de Graff accelerator: Page 265 (2011)\n", +"q = 1; // Number of proton, \n", +"V = 7; // Voltage applied to the dome, MV\n", +"E = q*V; // The kinetic energy of proton,MeV\n", +"printf('\nThe kinetic energy of proton : %d MeV', E);\n", +"// Result\n", +"// The kinetic energy of proton : 7 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.3_2: Reactions_at_different_particle_energies.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.3.2 : To calculate the kinetic energy of protons and no. of possibile reactions: Page 265 (2011)\n", +"V = 5; // Voltage of accelerator, MV\n", +"// Declare three cells (for three reactions): Page no. : 133(2011)\n", +" R1 = cell(3,2)\n", +" R2 = cell(10,2)\n", +"// Enter data for first cell (Reaction)\n", +"R1(1,1).entries = 'p';\n", +"R1(1,2).entries = 1;\n", +"R1(2,1).entries = 'd';\n", +"R1(2,2).entries = 1;\n", +"R1(3,1).entries = 'He';\n", +"R1(3,2).entries = 2;\n", +"E_p = (R1(1,2).entries)*V\n", +"E_d = (R1(2,2).entries)*V\n", +"E_He = (R1(3,2).entries)*V\n", +" // Enter data for second cell (Reaction)\n", +" R2(1,1).entries = 'p'\n", +" R2(1,2).entries = 1\n", +" R2(2,1).entries = 'N'\n", +" R2(2,2).entries = 14\n", +" R2(3,1).entries = 'O'\n", +" R2(3,2).entries = 15\n", +" R2(4,1).entries = 'y'\n", +" R2(4,2).entries = 0\n", +" R2(5,1).entries = 'd'\n", +" R2(5,2).entries = 1\n", +" R2(6,1).entries = 'n'\n", +" R2(6,2).entries = 0\n", +" R2(7,1).entries = 'He'\n", +" R2(7,2).entries = 3\n", +" R2(8,1).entries = 'C'\n", +" R2(8,2).entries = 13\n", +" R2(9,1).entries = 'He'\n", +" R2(9,2).entries = 4\n", +" R2(10,1).entries = 'C'\n", +" R2(10,2).entries = 12\n", +" printf('\nProtons energy = -%d MeV \n Deuterons energy = -%d MeV \n Double charged He-3 = -%d MeV', E_p, E_d, E_He)\n", +" printf('\n Possible reaction at these energies are')\n", +" printf('\n %s + %s(%d) ---> %s(%d)+ %s', R2(1,1).entries,R2(2,1).entries,R2(2,2).entries,R2(3,1).entries,R2(3,2).entries,R2(4,1).entries)\n", +"printf('\n %s + %s(%d) ---> %s(%d) + %s ', R2(5,1).entries,R2(2,1).entries,R2(2,2).entries,R2(3,1).entries,R2(3,2).entries,R2(6,1).entries)\n", +"printf('\n %s(%d) +%s(%d) ---> %s(%d)+ %s', R2(7,1).entries,R2(7,2).entries,R2(8,1).entries,R2(8,2).entries,R2(3,1).entries,R2(3,2).entries,R2(6,1).entries)\n", +" printf('\n %s(%d) + %s(%d) ---> %s(%d) +%s', R2(9,1).entries,R2(9,2).entries,R2(10,1).entries,R2(10,2).entries,R2(3,1).entries,R2(3,2).entries,R2(6,1).entries)\n", +"// Result\n", +"// Protons energy = -5 MeV \n", +"// Deuterons energy = -5 MeV \n", +"// Double charged He-3 = -10 MeV\n", +"// Possible reaction at these energies are\n", +"// p + N(14) ---> O(15)+ y\n", +"// d + N(14) ---> O(15) + n \n", +"// He(3) +C(13) ---> O(15)+ n\n", +"// He(4) + C(12) ---> O(15) +n" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.4_1: Protons_passing_through_the_carbon_stripper_foil.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.4.1 : To calculate the kinetic energy of protons passing through the carbon stripper foil : Page 266 (2011)\n", +"q = 2; // Number of proton, \n", +"V = 15; // Voltage applied to the dome, MV\n", +"E = q*V; // The kinetic energy of proton,MeV\n", +"printf('\nThe kinetic energy of proton : %d MeV', E);\n", +"// Result\n", +"// The kinetic energy of proton : 30 MeV" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5_1: Electron_at_relativistic_energy.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.5.1 : To calculate the difference between the electron's speed and speed of light. Page 265 (2011)\n", +"v = 2.999999997e+08; // Velocity of the electron, m/s\n", +"c = 3e+08; // Velocity of light,m/s\n", +"D = c-v; // difference between electron's speed and speed of light,m/s\n", +"printf('\nThe difference between electron speed and speed of light : %3.1f m/s', D);\n", +"// Result\n", +"// The difference between electron speed and speed of light : 0.3 m/s " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5_2: Protons_accelerating_through_drift_tubes.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.5.2 : To calculate the length of the first and last drift tubes which accelerate the protons whose frequency and energies are given. Page 268 (2011)\n", +"f = 200e+06; // Frequency of applied the voltage, Hz\n", +"V_0 = 750e+03; // Applied potential difference, V\n", +"q = 1.6e-019; // Charge of proton, C\n", +"m = 1.67e-027; // Mass of proton, Kg\n", +"n_1 = 1; // For first tube\n", +"L_1 = sqrt(2*n_1*q*V_0/m)/(2*f); // Length of the first tube, m\n", +"n_n = 128; // For last tube\n", +"L_n = 1/(2*f)*sqrt(2*n_n*q*V_0/m); // Length of the last tube,m\n", +"printf('\n Length of the first tube = %4.2f m \n Length of the last tube = %4.2f m ', L_1,L_n);\n", +"// Result\n", +"// Length of the first tube = 0.03 m \n", +"// Length of the last tube = 0.34 m " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.5_3: Electron_speed_at_relativistic_energies.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.5.3 : To calculate the velocity of the electrons using relativistic considerations . Page 269 (2011)\n", +"K_E = 1.17; // Kinetic energy of the electron, MeV\n", +"E_r = 0.511; // Rest mass energy of the electron, MeV\n", +"v = [1-1/(K_E/E_r+1)^2]; // Velocity of the electron, m/s\n", +"printf('\nVelocity of the electron : %4.2fc', v)\n", +"// Result\n", +"// Velocity of the electron : 0.91c " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7_1: Proton_accelerating_in_a_cyclotron.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.7.1 : To calculate the maximum energy, oscillator frequency and number of revolutions of proton accelerated in a cyclotron. Page 270(2011)\n", +"V = 20e+03; // Potential difference across the dees, V\n", +"r = 0.28; // Radius of the dees, m \n", +"B = 1.1; // Magnetic field, tesla\n", +"q = 1.6e-019; // Charge of the proton, C\n", +"m = 1.67e-027; // Mass of the proton, Kg\n", +"E_max = B^2*q^2*r^2/(2*m*1.6e-013); // Maximnum energy acquired by protons,MeV\n", +"f = B*q/(2*%pi*m*10^06); // Frequecy of the oscillator,MHz\n", +"N = E_max*1.6e-013/(q*V); // Number of revolutions, \n", +"disp(N)\n", +"printf('\n Maximum energy acquired by proton = %4.2f MeV \n Frequency of the oscillator = %4.2f MHz \n Number of revolutions = %d revolutions ', E_max,f,N)\n", +"// Result\n", +"// Maximum energy acquired by proton = 4.54 MeV \n", +"// Frequency of the oscillator = 16.77 MHz \n", +"// Number of revolutions = 227 revolutions" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7_2: Frequency_of_deutron_accelerated_in_a_cyclotron.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.7.2 : To calculate the frequency of deutron accelerated in a cyclotron. Page 271(2011)\n", +"B= 2.475; // Magnetic field, tesla\n", +"q = 1.6e-019; // Charge of the deutron, C\n", +"m = 2*1.67e-027; // Mass of the deutron, Kg\n", +"f = B*q/(2*%pi*m*10^06); // Frequency of the deutron,MHz\n", +"printf('\nFrequency of the deutron: %4.2f MHz ', f)\n", +"// Result\n", +"// Frequency of the deutron: 18.87 MHz" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7_3: Relation_between_magnetic_field_and_cyclotron_frequency.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.7.3 : To calculate the magnetic field applied to cyclotron whose frequency is given. Page 271(2011)\n", +"q = 1.6e-019; // Charge of the proton, C\n", +"r = 0.60; // radius of the dees, m\n", +"m = 1.67e-027; // Mass of the proton, Kg\n", +"f = 10^6; // Frequecy of the proton,Hz\n", +"B = 2*%pi*m*f/q; // Magnetic field applied to cyclotron, tesla\n", +"printf('\nMagnetic field applied to cyclotron : %6.4f tesla ', B)\n", +"// Result\n", +"// Magnetic field applied to cyclotron : 0.0656 tesla" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.7_4: Frequency_of_alternating_field.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.7.4 : To calculate the frequency of alternating field applied to dees. Page 272(2011)\n", +"q = 1.6e-019; // Charge of the proton, C\n", +"m = 1.67e-027; // Mass of the proton, Kg\n", +"B = 1.4; // Magnetic field , tesla\n", +"f = B*q/(2*%pi*m*10^06); // Frequency of the applied field, tesla\n", +"printf('\n Frequency of the applied field : %4.2f MHz', f)\n", +"// Result\n", +"// Frequency of the applied field : 21.35 MHz " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.8_1: Energy_gained_by_an_electron_in_the_magnetic_field.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.8.1. : To calculate the energy gained per turn of an electron present in given magnetic field. Page 273(2011)\n", +"e = 1.6e-019 ; // Charge of an electron, C\n", +"f = 60; // Frequency of variation magnetic field, Hz\n", +"B_0 = 1; // Magnetic field , tesla\n", +"r_0 = 1; // Radius of doughnut, m\n", +"E = 4*e*2*%pi*f*r_0^2/(1.6e-019); // Energy gained by electron per turn, eV\n", +"E_g = round(E)\n", +"printf('\n Energy gained by electron per turn: %d eV', E_g)\n", +"// Result\n", +"// Energy gained by electron per turn: 1508 eV " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.9_1: Ratio_of_highest_to_the_lowest_frequency_of_accelerating_proton.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.9.1 : To determine the ratio of highest to the lowest frequency of cyclotron accelerating protons whose energy is given. Page 273(2011)\n", +"K = 500; // Kinetic energy of the proton, MeV\n", +"E_r = 938; // Rest mass energy of the proton, MeV\n", +"R_f = E_r/(K+E_r); // The ratio of highest to the lowest frequency, \n", +"printf('\nThe ratio of highest to the lowest frequency : %4.2f ', R_f)\n", +"// Result\n", +"// The ratio of highest to the lowest frequency : 0.65 " + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6.9_2: W_B_ration_of_completely_stripped_nitrogen.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa6.9.2 : To calculate the w/B ratio for a completely stripped nitrogen to move in a stable orbit : Page 274(2011)\n", +"E_k = 1200; // Kinetic energy of the proton, MeV\n", +"q = 7; // Number of proton in nitrogen\n", +"E_r = 13040 // Rest mass energy of the electron, MeV\n", +"E = (E_k+E_r)*1.6e-013; // Total energy,j\n", +"c = 3e+08; // Velocity of light, m/s\n", +"R_w_B = q*1.6e-019*c^2/E; // Ratio of w/B, m^2/W \n", +"printf('\nThe ratio of w/B : %4.2e m^2/W ', R_w_B)\n", +"// Result\n", +"// The ratio of w/B : 4.42e+007 m^2/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/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/7-Radiation_Detectors.ipynb b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/7-Radiation_Detectors.ipynb new file mode 100644 index 0000000..f981eb2 --- /dev/null +++ b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/7-Radiation_Detectors.ipynb @@ -0,0 +1,1392 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: Radiation Detectors" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.2_1: Energy_of_alpha_particle.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.2.1: To calculate the energy of alpha particle :P.no. 308 (2011)\n", +" E_p = 30; // Energy required for one pair, eV\n", +" n = 150000; // Number of pairs \n", +" E_a = n*E_p/10^6; // Energy of alpha particle, Mev\n", +" printf('\n The energy of alpha particle : %3.1f Mev', E_a)\n", +"// Result\n", +"// The energy of alpha particle : 4.5 Mev\n", +" \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3_1: Pulse_height_of_ionising_particle.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.3.1: To calculate the pulse height of ionising particle :P.no. 308 (2011)\n", +" E = 5.48e+06; // Energy of alpha particle, eV\n", +" C = 50e-012; // Capacitance of the chamber, F\n", +" R = 10^6; // Resistance, ohm\n", +" E_p = 35; // Energy required to produced an ion pair, eV\n", +" n = E/E_p; // Number of ion pair produced\n", +" e = 1.6e-019; // Charge of an electron, C\n", +" V =( n*e)/C; // Pulse height, V\n", +" I = V/R; // current produced, A\n", +" printf('\n The pulse height = %4.3e V \n Current produced = %5.3e A', V,I) \n", +"// Result\n", +"// The pulse height = 5.010e-004 V \n", +" //Current produced = 5.010e-010 A" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.3_2: Charge_deposited_on_detector_plate.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.3.2: To calculate the kinetic energy and amount of charge collected on plate :P.no. 309 (2011)\n", +" E_p = 35; // Energy required to produced an ion pair, eV\n", +" n = 10^5; // Number of ion pair produced\n", +" e = 1.6e-019; // Charge of an electron, C\n", +" E_k = E_p*n/10^6; // Kinetic energy of the proton, MeV\n", +" A = n*e; // The amount of charge collected on each plate, C \n", +" printf('\n The kinetic energy of the proton = %3.1f MeV \n The amount of charge collected on each plate = %3.1e C ', E_k, A)\n", +"// Result\n", +" // The kinetic energy of the proton = 3.5 MeV \n", +"// The amount of charge collected on each plate = 1.6e-014 C \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.4_1: Height_of_voltage_pulses.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.4.1: To calculate the charge flow in a counter and height of voltage pulses :P.no. 310 (2011)\n", +" E_p = 30; // Energy required to produced an ion pair, eV\n", +" M = 1000; // Multiplication factor \n", +" e = 1.6e-019; // Charge of an electron, C\n", +" t = 10^-3; // Time, s\n", +" R = 10^5; // Resistance, ohm\n", +" E_k = 20*10^6; // Kinetic energy of the proton, eV\n", +" n = E_k/E_p; // Number of ion pairs produced\n", +" n_a = n*M; // Number of ion-pair after multiplication\n", +" Q = n_a*e; // Charge carried by these ion, C \n", +" I = Q/t; // The current through 100-ohm resistance, A\n", +" A = I*R; // ,The amplitude of voltage pulse, V \n", +" printf('\n The current through 100-ohm resistance = %6.4e A \n The amplitude of voltage pulse = %6.4e V ', I, A)\n", +"// Result\n", +"// The current through 100-ohm resistance = 1.0667e-007 A \n", +"// The amplitude of voltage pulse = 1.0667e-002 V \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.4_2: Electric_field_at_the_surface_of_wire.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.4.2: To calculate the electric field at the surface of wire :P.no. 310 (2011)\n", +" V = 1500; // Potential difference, V\n", +" a = 0.0001; // Radius of the wire, m\n", +" b = 0.02; // Radius of the cylinderical tube, m\n", +" r = 0.0001; // Distance of electric field from the surface, m\n", +" E_r = V/(r*log(b/a)); // the electric field at the surface, V/m \n", +" printf('\n The electric field at the surface : %4.2e V/m', E_r)\n", +"// Result\n", +"// The electric field at the surface : 2.83e+006 V/m\n", +" \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5_1: Electric_filed_in_G_M_counter.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.5.1: To calculate the electric field at the surface of wire of G.M. counter :P.no. 311 (2011)\n", +" V = 2000; // Potential difference, V\n", +" a = 0.01; // Radius of the wire, cm\n", +" b = 2; // Radius of the cylinderical tube, cm\n", +" r = 0.01; // Radius of the wire, m\n", +" E_r = V/(r*log(b/a)); // the electric field at the surface, V/m \n", +" printf('\n The electric field at the surface : %d V/cm', E_r)\n", +"// Result\n", +"// The electric field at the surface : 37747 V/cm \n", +" \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5_2: Life_of_G_M_counter.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.5.2: To calculate the life of G.M. counter :P.no. 312 (2011)\n", +" n_t = 10^9; // Total number of counts \n", +" n_d = 2000*3*60; // Count recorded per day\n", +" n_y = n_d*365; // Counts recorded in 365-days\n", +" t = n_t/n_y; // The life of G.M. counter, year\n", +"printf('\nThe life of G.M. counter : %4.2f year', t)\n", +"// Result\n", +"// The life of G.M. counter : 7.61 year \n", +" // \n", +" \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5_3: Amplitude_of_voltage_pulses_in_G_M_counter.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.5.3: To calculate the voltage pulse of G.M. counter :P.no. 312 (2011)\n", +" E_p = 30; // Energy required for one electron pair, eV\n", +" E = 10e+06 ; // Energy lost by alpha particle, eV\n", +" n = E/E_p; // Number of ion-pairs produced\n", +" M = 5000; // Multiplication factor\n", +" C = 50e-012; // Capacitance, F\n", +" n_M = n*M; // Number of ion-pairs after multiplication\n", +" e = 1.6e-019; // Charge of an electron, C\n", +" Q = n_M*e; // Charge present in each ion\n", +" A = Q/C; // Amplitude of voltage pulse, V\n", +" printf('\n Amplitude of voltage pulse : %3.1f V', A)\n", +"// Result\n", +" // Amplitude of voltage pulse : 5.3 V\n", +" \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.5_4: Estimating_true_count_rate_of_G_M_counter.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.5.4: To estimate the true count rate of G.M. counter :P.no. 312 (2011)\n", +" n = 30000; // Count per minute \n", +" n_o = n/60; // Observed count rate, count/s\n", +" t = 2e-04; // Dead time, s \n", +" n_t = round(n_o/(1-n_o*t)); // The true count rate, count/s\n", +" printf('\n The true count rate : %d counts/s', n_t)\n", +"// Result\n", +" // The true count rate : 556 counts/s \n", +" \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6_1: Energy_resolution_of_gamma_rays.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.6.1: To calculate the energy resolution of gamma rays emitted by Na-22 for channel first and second :P.no. 313 (2011)\n", +"// For 511 KeV gamma rays (for channel first)\n", +"F_W_H_M_1 = 97; // Frequency width at half maximum for channel first\n", +"P_pos_1 = 1202; // Peak position for channel first\n", +"Res_KeV_1 = F_W_H_M_1/P_pos_1*511; // Resolution in KeV for channel first\n", +"// For 1275 KeV gamma rays (for channel second) \n", +"F_W_H_M_2 = 82; // Frequency width at half maximum for channel second\n", +"P_pos_2 = 1202; // Peak position for channel second\n", +"Res_KeV_2 = round(F_W_H_M_2/P_pos_2*1275); // Resolution in KeV for channel second\n", +" printf('\n Resolution for channel first = %d KeV \n Resolution for channel second = %d KeV ',Res_KeV_1, Res_KeV_2)\n", +" // Result\n", +"// Resolution for channel first = 41 KeV \n", +"// Resolution for channel second = 87 KeV \n", +" \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6_2: Amplitude_of_output_voltage_pulse.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.6.2 : To calculate the amplitude of output voltage pulse for NaI(Tl) :P.no. 314 (2011)\n", +"e = 1.6e-019; // Charge of an electron, C\n", +"n = 4.2e+08; // Number of photoelectrons\n", +"C = 200e-012; // Capacitance, F\n", +"A = n*e/C; // Amplitude of output voltage pulse, V\n", +"printf('\n Amplitude of output voltage pulse : %4.2f V ',A)\n", +"// Result\n", +"// Amplitude of output voltage pulse : 0.34 V \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.6_3: Resolution_of_scintillation_detector.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.6.3 : To calculate the %-resolution and resolution in KeV for scintillation detector for Cs-137 :P.no. 315 (2011)\n", +"F_W_H_M = 0.72; // Full width at half maximum, V\n", +"P_p = 6.0; // Peak position, V\n", +"E = 662; // Energy of photopeak, KeV\n", +"%_resolution = F_W_H_M/P_p*100; // Percentage resolution in percent\n", +"Res_KeV = %_resolution/100*E; // Resolution in KeV for Cs-137\n", +"printf('\n The percentage resolution = %d percent \n Resolution in KeV = %4.1f KeV ', %_resolution, Res_KeV)\n", +"// Result\n", +"// The percentage resolution = 12 percent \n", +"// Resolution in KeV = 79.4 KeV \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.7_1: Silicon_pulse_detector.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.7.1 : To calculate the thickness of depletion layer of silicon detector and amplitude of voltage pulse :P.no. 316 (2011)\n", +"E_r = 12; // Relative permittivity \n", +"E_o = 8.85e-012; // Permittivity of free space\n", +"E = E_r*E_o; // Absolute dielectric constant\n", +"C = 100e-012; // Capacitance of the dielectric, F\n", +"A = 1.6e-04; // Area of the detector, m^2\n", +"e = 1.602e-019; // Charge of an electrin, C\n", +"E_p = 3.2; // Energy required to create an ion pair, eV\n", +"E_s = 12e+06; // Energy required to stopped ion pair, eV\n", +"n = E_s/E_p; // Number of ion-pair produced\n", +"Q = n*e; // Charge of these ion pair, C\n", +"d = A*E/(C*10^-6); // The thickness of the depletion layer, micron\n", +"A = Q/C*1000; // The amplitude of voltage pulse, mV\n", +"printf('\n The thickness of the depletion layer = %d micron \n The amplitude of voltage pulse: = %6.4f mV ', d, A)\n", +"// Result\n", +"// The thickness of the depletion layer = 169 micron \n", +"// The amplitude of voltage pulse: = 6.0075 mV \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.7_2: Detector_characteristics.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa7.7.2 : To calculate the capacitance and the amplitude of voltage pulse across the detector :Page 316 (2011)\n", +"E_r = 12; // Relative permittivity \n", +"E_o = 8.85e-012; // Permittivity of free space\n", +"E = E_r*E_o; // Absolute dielectric constant\n", +"A = 2e-04; // Area of the detector, m^2\n", +"e = 1.602e-019; // Charge of an electron, C\n", +"d = 100e-06; // The thickness of the depletion layer, m\n", +"C = E*A/d; // The capacitance of the dielectric, F\n", +"E_p = 3.0; // Energy required to create an ion pair, eV\n", +"E_s = 5.48e+06; // Energy required to stopped ion pair, eV\n", +"n = E_s/E_p; // Number of ion-pair produced\n", +"Q = n*e; // Charge of these ion pair, C\n", +"A = Q/C*1000; // The amplitude of voltage pulse, mV\n", +"printf('\n The capacitance of dielectric = %5.3e F \n The amplitude of voltage pulse = %5.3f mV ', C, A)\n", +"// Result\n", +"// The capacitance of dielectric = 2.124e-010 F \n", +"// The amplitude of voltage pulse = 1.378 mV \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +], +"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/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/8-Particle_Physics.ipynb b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/8-Particle_Physics.ipynb new file mode 100644 index 0000000..9e078bd --- /dev/null +++ b/Introduction_To_Nuclear_And_Particle_Physics_by_V_K_Mittal/8-Particle_Physics.ipynb @@ -0,0 +1,360 @@ +{ +"cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: Particle Physics" + ] + }, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.5_1: Average_kinetic_energy_of_pion.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa8.5.1: To calculate the average kinetic energy of each pion:P.No.360 (2011)\n", +"// Proton and antiproton annihilate to produced three pions\n", +"E_p = 938; // Energy of proton, MeV\n", +"E_pi = 139.5; // Energy of pions, MeV\n", +"E_pi_0 = 134.9; // Energy of pi_0_ion, MeV\n", +"E_KE = [2*E_p-(2*E_pi+E_pi_0)]/3; // The average kinetic energy of each pions, MeV\n", +"printf('\n The average kinetic energy of each pions : %5.1f MeV', E_KE)\n", +"// Result\n", +"// The average kinetic energy of each pions : 487.4 MeV\n", +" \n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.5_2: Inherent_uncertainity_in_mass_of_the_particle.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa8.5.2: To calculate the inherent uncertainity in mass of the given particle : P.no. 360 (2011)\n", +" // Here r_1 and r_2 are two decay rates are given\n", +" // Declare the cell\n", +" R1 = cell(1,2)\n", +" R1(1,1).entries = 'r_1'\n", +" R1(1,2).entries = 'r_2'\n", +" printf('\n The inherent uncertainity in mass of particle = h(%s + %s) ', R1(1,1).entries, R1(1,2).entries) \n", +"// Result\n", +"// The inherent uncertainity in mass of particle = h(r_1 + r_2)\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"\n", +"" + ] + } +, +{ + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.7_3: Sub_nuclear_reactions.sci" + ] + }, + { +"cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], +"source": [ +"// Scilab code Exa8.7.3: Determine the possibility of the given reaction:P.no. 362 (2011)\n", +"// Declare cell for the given reaction\n", +"R1 = cell(7,5)\n", +"// Enter data for the cell\n", +"R1(1,1).entries = 'p'\n", +"R1(1,2).entries = 1 \n", +"R1(1,3).entries = 1\n", +"R1(1,4).entries = 0\n", +"R1(1,5).entries = 1/2\n", +"R1(2,1).entries = 'K_+'\n", +"R1(2,2).entries = 1\n", +"R1(2,3).entries = 0\n", +"R1(2,4).entries = 1\n", +"R1(2,5).entries = 1/2\n", +"R1(3,1).entries = 'S_+'\n", +"R1(3,2).entries = 1\n", +"R1(3,3).entries = 1\n", +"R1(3,4).entries = -1\n", +"R1(3,5).entries = 1\n", +"R1(4,1).entries = 'pi_-'\n", +"R1(4,2).entries = -1\n", +"R1(4,3).entries = 0\n", +"R1(4,4).entries = 0\n", +"R1(4,5).entries = 1\n", +"R1(5,1).entries = 'S_0'\n", +"R1(5,2).entries = 0\n", +"R1(5,3).entries = 1\n", +"R1(5,4).entries = -1\n", +"R1(5,5).entries = 0\n", +"R1(6,1).entries = 'p_-'\n", +"R1(6,2).entries = -1\n", +"R1(6,3).entries = -1\n", +"R1(6,4).entries = 0\n", +"R1(6,5).entries = 1/2\n", +"R1(7,1).entries = 'n_0'\n", +"R1(7,2).entries = 0\n", +"R1(7,3).entries = 0\n", +"R1(7,4).entries = 0\n", +"R1(7,5).entries = 0\n", +"\n", +"function f = check_Isotopic_no(Ir_sum,Ip_sum)\n", +" if Ir_sum == Ip_sum then\n", +" f = 1;\n", +" else \n", +" f = 0;\n", +" end\n", +"endfunction\n", +"1\n", +"// Declare a function returning equality status of proton number\n", +"function f = check_strangeness(sr_sum,sp_sum)\n", +" if sr_sum == sp_sum then\n", +" f = 1;\n", +" else \n", +" f = 0;\n", +" end \n", +"endfunction\n", +"function f = check_charge(cr_sum,cp_sum)\n", +" if cr_sum == cp_sum then\n", +" f = 1;\n", +" else \n", +" f = 0;\n", +" end \n", +"endfunction\n", +"// Declare a function returning equality status of lepton number\n", +" \n", +"// Reaction-I\n", +"printf('\n\n\nReaction-I:\n\n');\n", +" Ir_sum = R1(1,5).entries+R1(1,5).entries;\n", +" Ip_sum = R1(2,5).entries+R1(3,5).entries;\n", +" if (check_Isotopic_no(Ir_sum,Ip_sum) == 0) then\n", +" printf('The Reaction\n')\n", +" printf('\t%s + %s --> %s + %s \nis not possible', R1(1,1).entries, R1(1,1).entries, R1(2,1).entries, R1(3,1).entries)\n", +"// Reaction-II\n", +" printf('\n\n\nReaction-II')\n", +" sr_sum = R1(1,4).entries+R1(4,4).entries;\n", +" sp_sum = R1(5,4).entries+R1(7,4).entries;\n", +" if (check_strangeness(sr_sum,sp_sum)== 0) then\n", +" printf('\n\nThe Reaction\n')\n", +" printf('\t%s + %s --> %s + %s \nis not possible', R1(1,1).entries, R1(4,1).entries, R1(5,1).entries, R1(7,1).entries)\n", +"// Reaction-III\n", +" printf('\n\n\nReaction-III:\n\n');\n", +" cr_sum = R1(1,2).entries+R1(1,2).entries;\n", +" cp_sum = R1(1,2).entries+R1(1,2).entries+R1(1,2).entries+R1(6,2).entries; \n", +" if (check_charge(cr_sum,cp_sum) == 1) then\n", +" printf('The Reaction\n')\n", +" printf('\t%s + %s --> %s + %s + %s \nis possible', R1(1,1).entries, R1(1,1).entries, R1(1,1).entries, R1(1,1).entries, R1(6,1).entries) \n", +" end\n", +" // Reaction-I:\n", +"\n", +"// The Reaction\n", +"// p + p --> K_+ + S_+ \n", +"// is not possible\n", +"\n", +"\n", +"// Reaction-II\n", +"\n", +"// The Reaction\n", +"// p + pi_- --> S_0 + n_0 \n", +"// is not possible\n", +"\n", +"\n", +"// Reaction-III:\n", +"\n", +"// The Reaction\n", +"// p + p --> p + p + p_- \n", +"// is possible \n", +" \n", +" \n", +" " + ] + } +], +"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 +} |