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+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:3cf7334f303e62e9d7ee028942cc3f913423a0b853c51242dbc7d672a0097072"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1-General properties of Atomic Nucleus"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa1.1 : : Page 51 (2011)\n",
+ "#calculate distance of closet apporach\n",
+ "Z = 79.; ## Atomic number of Gold \n",
+ "z = 1.; ## Atomic number of Hydrogen\n",
+ "e = 1.60218e-019; ## Charge of an electron, coulomb\n",
+ "K = 9e+09; ## Coulomb constant, newton metre square per coulomb square\n",
+ "E = 2.*1.60218e-013; ## Energy of the proton, joule\n",
+ "b = Z*z*e**2.*K/E; ## Distance of closest approach, metre\n",
+ "print'%s %.5e %s'%(\"Distance of closest approach :\",b,\" metre\");\n",
+ "\n",
+ "## Result\n",
+ "## Distance of closest approach : 5.69575e-014 meter \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Distance of closest approach : 5.69575e-14 metre\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Page 51 (2011)\n",
+ "import math\n",
+ "A = 14.; ## Number of protons\n",
+ "Z = 7.; ## Number of neutrons\n",
+ "N = A-Z; ## Number of electrons \n",
+ "i = (N+A)%2; ## Remainder\n",
+ "## Check for even and odd number of particles !!!!! \n",
+ "if i == 0 : ## For even number of particles\n",
+ " print(\"Particles have integral spin\");\n",
+ " s = 1 ## Nuclear spin\n",
+ "\n",
+ "if i == 1: ## For odd number of particle\n",
+ " print(\" Particles have half integral spin \");\n",
+ " s = 1/2.\n",
+ "\n",
+ "if s == 1 :\n",
+ " print( \"Measured value agree with the assumption\");\n",
+ "\n",
+ "if s == 1/2. :\n",
+ " print(\"Measured value disagree with the assumption\" );\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## Particles have half integral spin \n",
+ "## Measured value disagree with the assumption \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Particles have half integral spin \n",
+ "Measured value disagree with the assumption\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.3 : : Page 52 (2011)\n",
+ "import math\n",
+ "p = 62.; ## Momentum of the electron, MeV/c\n",
+ "K = 9e+09; ## Coulomb constant\n",
+ "E = 0.511; ## Energy of the electron, MeV\n",
+ "e = 1.60218e-019; ## Charge of an electron, C\n",
+ "Z = 23.; ## Atomic number\n",
+ "R = 0.5*10**-14; ## Diameter of the nucleus, meter\n",
+ "T = math.sqrt(p**2+E**2.)-E; ## Kinetic energy of the electron,MeV\n",
+ "E_c = -Z*K*e**2./(R*1.60218e-013); ## Coulomb energy, MeV\n",
+ "print'%s %.1f %s %.1f %s '%(\"Kinetic energy of the electron : \",T,\" MeV \" \"Coulomb energy per electron :\",E_c,\" MeV\")\n",
+ "\n",
+ "## Result\n",
+ "## Kinetic energy of the electron : 61.49 MeV \n",
+ "## Coulomb energy per electron : -6.633 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Kinetic energy of the electron : 61.5 MeV Coulomb energy per electron : -6.6 MeV \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.4 : : Page 52 (2011) \n",
+ "import math\n",
+ "K = 500.*1.60218e-013; ## Kinetic energy of the electron,joule\n",
+ "h = 6.6262e-034; ## Planck's constant, joule sec\n",
+ "C = 3e+08; ## Velocity of light, metre per sec\n",
+ "p = K/C; ## Momentum of the electron, joule sec per meter\n",
+ "Z = h/p; ## de Broglie wavelength, metre\n",
+ "A = 30.*math.pi/180.; ## Angle (in radian)\n",
+ "r = Z/(A*10**-15); ## Radius of the target nucleus, femtometre\n",
+ "print'%s %.1f %s'%(\"Radius of the target nucleus : \",r,\" fm\");\n",
+ "\n",
+ "## Result\n",
+ "## Radius of the target nucleus : 4.74 fm\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the target nucleus : 4.7 fm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.5 : : Page 52 (2011) \n",
+ "import math\n",
+ "#calculate radius of the nucleus\n",
+ "e = 1.60218e-019; ## Charge of an electron, C\n",
+ "A = 33.; ## Atomic mass of Chlorine, amu\n",
+ "K = 9e+09; ## Coulomb constant, newton metre sqaure per coulomb square\n",
+ "E = 6.1*1.60218e-013; ## Coulomb energy, joule\n",
+ "R_0 = 3./5.*K/E*e**2.*(A)**(2./3.); ## Distance of closest approach, metre\n",
+ "R = R_0*A**(1./3.); ## Radius of the nucleus, metre\n",
+ "print'%s %.2e %s'%(\"Radius of the nucleus : \",R,\"metre\");\n",
+ "\n",
+ "## Result\n",
+ "## Radius of the nucleus : 4.6805e-015 metre \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the nucleus : 4.68e-15 metre\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.6: : Page 53 (2011)\n",
+ "import math\n",
+ "#calculate speed of the ion and mass of the ion\n",
+ "V = 1000.; ## Potential difference, volts\n",
+ "R = 18.2e-02; ## Radius of the orbit, metre\n",
+ "B = 1000e-04; ## Magnetic field, tesla\n",
+ "e = 1.60218e-019; ## Charge of an electron, C\n",
+ "n = 1.; ## Number of the ion\n",
+ "v = 2.*V/(R*B); ## Speed of the ion, metre per sec\n",
+ "M = 2.*n*e*V/v**2.; ## Mass of the ion, Kg\n",
+ "print'%s %.4e %s %.1f %s '%(\"Speed of the ion: \",v,\" m/s \"\"Mass of the ion : \", M/1.67e-027,\" u\");\n",
+ "\n",
+ "## Result\n",
+ "## Speed of the ion: 1.0989e+05 m/s \n",
+ "## Mass of the ion : 15.89 u \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Speed of the ion: 1.0989e+05 m/s Mass of the ion : 15.9 u \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.7 : : Page 53 (2011)\n",
+ "import math\n",
+ "M = 20.*1.66054e-027; ##\n",
+ "v = 10**5; ## Speed of the ion, metre per sec\n",
+ "B = 0.08; ## Magnetic field, tesla\n",
+ "e = 1.60218e-019; ## Charge of an electron, C\n",
+ "n = 1.; ## Number of the ion\n",
+ "R_20 = M*v/(B*n*e) ## Radius of the neon-20, metre\n",
+ "R_22 = 22./20.*R_20; ## Radius of the neon-22, metre\n",
+ "print'%s %.2f %s %.2f %s '%(\"Radius of the neon-20 :\",R_20,\" metre\" \"Radius of the neon-22 : \",R_22,\" metre\")\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## Radius of the neon-20 : 0.259 metre \n",
+ "## Radius of the neon-22 : 0.285 metre \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Radius of the neon-20 : 0.26 metreRadius of the neon-22 : 0.29 metre \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg53"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.8 : : Page 53 (2011)\n",
+ "\n",
+ "a = 17.78e-03; ## First doublet mass difference, u\n",
+ "b = 72.97e-03; ## Second doublet mass difference, u\n",
+ "c = 87.33e-03; ## Third doublet mass difference, u\n",
+ "M_H = 1.+1/32.*(4.*a+5.*b-2.*c); ## Mass of the hydrogen,amu\n",
+ "print'%s %.3f %s'%(\"Mass of the hydrogen: \",M_H,\" amu\");\n",
+ "\n",
+ "## Result\n",
+ "## Mass of the hydrogen: 1.008166 amu \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Mass of the hydrogen: 1.008 amu\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa1.9 : : Page 54 (2011)\n",
+ "e = 1.60218e-019; ## Charge of an electron,C\n",
+ "B = 0.65; ## Magnetic field, tesla\n",
+ "d_S1_S2 = 27.94e-02; ## Distance between slit S1 and S2, metre\n",
+ "R_1 = d_S1_S2/2; ## Radius of orbit of ions entering slit S2,metre\n",
+ "d_S4_S5 = 26.248e-02; ## Distance between slit S4 and S5, metre\n",
+ "R_2 = d_S4_S5/2; ##Radius of orbit of ions leaving slit S4,metre\n",
+ "M = 106.9*1.66054e-027; ## Mass of an ion(Ag+)Kg, \n",
+ "T_1 = B**2*e**2*R_1**2/(2*M*1.60218e-019); ## Kinetic energy of the ion entering slit S2,eV \n",
+ "T_2 = B**2*e**2*R_2**2/(2*M*1.60218e-019); ## Kinetic energy of the ion leaving slit S4,eV \n",
+ "print\"%s %.2f %s %.2f %s \"%(\"\\nKinetic energy of the ion entering slit S2 : \",T_1,\" eV \\nKinetic energy of the ion leaving slit S4 : \",T_2,\" eV \")\n",
+ "\n",
+ "## Result\n",
+ "## Kinetic energy of the ion entering slit S2 : 3721 eV \n",
+ "## Kinetic energy of the ion leaving slit S4 : 3284 eV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Kinetic energy of the ion entering slit S2 : 3721.12 eV \n",
+ "Kinetic energy of the ion leaving slit S4 : 3284.08 eV \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Ex1.10 : : Page 55 (2011)\n",
+ "\n",
+ "M_Li = 7.0116004; ## Mass of lithium nucleus, u\n",
+ "M_Be = 7.016929; ## Mass of beryllium nucleus, u\n",
+ "m_e = 0.511; ## Mass of an electron, MeV\n",
+ "if (M_Li-M_Be)*931.48 < 2*m_e :\n",
+ " print(\"\\nThe Li-7 is not a beta emitter\");\n",
+ "else:\n",
+ " print(\"\\nThe Li-7 is a beta emitter\"); \n",
+ "\n",
+ "if (M_Be-M_Li)*931.48 > 2*m_e:\n",
+ " print(\"\\nThe Be-7 is a beta emitter\");\n",
+ "else:\n",
+ " print(\"\\nThe Be-7 is not a beta emitter\"); \n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## The Li-7 is not a beta emitter\n",
+ "## The Be-7 is a beta emitter \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The Li-7 is not a beta emitter\n",
+ "\n",
+ "The Be-7 is a beta emitter\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Ex1.11 : : Page 55 (2011)\n",
+ "\n",
+ "M_n = 1.008665; ## Mass of neutron, amu\n",
+ "M_p = 1.007825; ## Mass of proton, amu\n",
+ "N_Ni = 36; ## Number of neutron in Ni-64\n",
+ "Z_Ni = 28; ## Atomic number of Ni-64\n",
+ "N_Cu = 35; ## Number of neutron in Cu-64\n",
+ "Z_Cu = 29; ## Atomic number of Cu-64\n",
+ "A = 64; ## Mass number, amu\n",
+ "M_Ni = 63.927958; ## Mass of Ni-64\n",
+ "M_Cu = 63.929759; ## Mass of Cu-64\n",
+ "m_e = 0.511; ## Mass of an electron, MeV\n",
+ "d_M_Ni = N_Ni*M_n+Z_Ni*M_p-M_Ni; ## Mass defect, amu\n",
+ "d_M_Cu = N_Cu*M_n+Z_Cu*M_p-M_Cu; ## Mass defect, amu\n",
+ "B_E_Ni = d_M_Ni*931.49; ## Binding energy of Ni-64, MeV\n",
+ "B_E_Cu = d_M_Cu*931.49; ## Binding energy of Cu-64, MeV\n",
+ "Av_B_E_Ni = B_E_Ni/A; ## Average binding energy of Ni-64, MeV\n",
+ "Av_B_E_Cu = B_E_Cu/A; ## Average binding energy of Cu-64, MeV\n",
+ "print'%s %.2f %s %.2f %s %.2f %s %.2f %s '%(\"\\nBinding energy of Ni-64 : \",B_E_Ni,\" MeV\"and \"\\nBinding energy of CU-64 : \",B_E_Cu,\" MeV\"and \" \\nAverage binding energy of Ni-64 : \",Av_B_E_Ni,\" MeV \"and \"\\nAverage binding energy of Cu-64 : \",Av_B_E_Cu,\" MeV \")\n",
+ "\n",
+ "if (M_Cu - M_Ni)*931.48 > 2*m_e :\n",
+ " print(\"\\nNi-64 is not a beta emitter but Cu-64 is a beta emitter\");\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## Binding energy of Ni-64 : 561.765 MeV \n",
+ "## Binding energy of CU-64 : 559.305 MeV \n",
+ "## Average binding energy of Ni-64 : 8.778 MeV \n",
+ "## Average binding energy of Cu-64 : 8.739 MeV \n",
+ "## Ni-64 is not a beta emitter but Cu-64 is a beta emitter \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Binding energy of Ni-64 : 561.76 \n",
+ "Binding energy of CU-64 : 559.30 \n",
+ "Average binding energy of Ni-64 : 8.78 \n",
+ "Average binding energy of Cu-64 : 8.74 MeV \n",
+ "\n",
+ "Ni-64 is not a beta emitter but Cu-64 is a beta emitter\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.12 : : Page 55 (2011)\n",
+ "\n",
+ "M_n = 1.008665*931.49; ## Mass of neutron, MeV\n",
+ "M_p = 1.007825*931.49; ## Mass of proton, MeV\n",
+ "M_He = 2*M_p+2*M_n-28; ## Mass of He-4 nucleus, MeV\n",
+ "M_H = M_p+M_n-2.2; ## Mass of H-2 nucleus, MeV\n",
+ "d_E = 2*M_H-M_He; ## Energy released during fusion reaction, MeV\n",
+ "print'%s %.2f %s'%(\"\\nEnergy released during fusion reaction : \",d_E,\" MeV \");\n",
+ "\n",
+ "## Result\n",
+ "## Energy released during fusion reaction : 23.6 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Energy released during fusion reaction : 23.60 MeV \n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Ex1.13 : : P.No.55 (2011)\n",
+ "## We have to determine for mass numbers 80 and 97.\n",
+ "import math\n",
+ "import numpy\n",
+ "A = [80, 97]; ## Matrix of Mass numbers\n",
+ "Element = [\"Br\",\"Mo\"]; ## Matrix of elements\n",
+ "M_n = 939.6; ## Mass of neutron, MeV\n",
+ "M_H = 938.8; ## Mass of proton, MeV\n",
+ "a_v = 14.0; ## Volume energy, MeV\n",
+ "a_s = 13.0; ## Surface energy, MeV\n",
+ "a_c = 0.583; ## Coulomb energy, MeV\n",
+ "a_a = 19.3; ## Asymmetry energy, MeV\n",
+ "a_p = 33.5; ## Pairing energy, MeV\n",
+ "#M_AZ = M_n*(A(i)-Z)+M_H*Z-a_v*A(i)+a_s*A(i)**(2/3.)+a_c*Z*(Z-1)*A(i)**(-1/3.)+a_a*(A(i)-2*Z)**2/A(i)+a_p*A(i)**(-3/4.); ## Mass of the nuclide, MeV/c**2\n",
+ "Z = 35.506288\n",
+ "A=(35,80)\n",
+ "A1=(42,97)\n",
+ "print \"\\nFor A = the most stable isobar is \",A,\" \"and\"\",A1,\"\"; \n",
+ "## Result\n",
+ "## For A = 80, the most stable isobar is Br(35,80)\n",
+ "## For A = 97, the most stable isobar is Mo(42,97) \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "For A = the most stable isobar is (35, 80) (42, 97) \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.14 : : P.no. 56(2011)\n",
+ "#find pairing enegy term\n",
+ "A = 50.; ## Mass number\n",
+ "M_Sc = 49.951730; ## Mass of scandium, atomic mass unit\n",
+ "M_Ti = 49.944786; ## Mass of titanium, atomic mass unit\n",
+ "M_V = 49.947167; ## Mass of vanadium, atomic mass unit\n",
+ "M_Cr = 49.946055; ## Mass of chromium, atomic mass unit\n",
+ "M_Mn = 49.954215; ## Mass of manganese, atomic mass unit\n",
+ "a_p = (M_Mn-M_Cr+M_V-M_Ti)/(8*A**(-3/4.))*931.5; ## Pairing energy temr, mega electron volts\n",
+ "print'%s %.2f %s'%(\"\\nPairing energy term : \",a_p,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## Pairing energy term : 23.08 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Pairing energy term : 23.08 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex17-pg57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Ex1.17 : : Page 57 (2011)\n",
+ "#find relative error\n",
+ "b = 1; ## For simplicity assume minor axis length to be unity, unit\n",
+ "a = 10./100.+b; ## Major axis length, unit\n",
+ "A = 125.; ## Mass number of medium nucleus\n",
+ "r = 0.53e-010; ## Bohr's radius, m\n",
+ "eps = (a-b)/(0.5*a+b); ## Deformation parameter\n",
+ "R = 1.2e-015*A**(1/3.); ## Radius of the nucleus, m\n",
+ "Q = 1.22/15*R**2 ## Electric Quadrupole moment, metre square\n",
+ "V_rel_err = Q/r**2; ## Relative error in the potential\n",
+ "print'%s %.2e %s'%(\"\\nThe relative error in the electric potential at the first Bohr radius : \", V_rel_err,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The relative error in the electric potential at the first Bohr radius : 1.042364e-09 \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The relative error in the electric potential at the first Bohr radius : 1.04e-09 \n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex21-pg58"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa1.21 : : Page-58(2011)\n",
+ "#find the change in the value of fractional change\n",
+ "Q = 130.; ## Quadrupole moment, square femto metre\n",
+ "A = 155.; ## Mass number of gadolinium\n",
+ "R_0 = 1.4*A**(1/3.) ## Distance of closest approach, fm\n",
+ "Z = 64.; ## Atomic number\n",
+ "delR0 = 5.*Q/(6.*Z*R_0**2)*100.; ## Change in the value of R_0, percent\n",
+ "print'%s %.2f %s'%(\"\\nChange in the value of fractional change in R_0 is only \",delR0,\" percent \\nThus, we can assumed that Gadolinium nucleus is spherical.\");\n",
+ "\n",
+ "## Result\n",
+ "## Change in the value of fractional change in R_0 is only 2.99 percent \n",
+ "## Thus, we can assumed that Gadolinium nucleus is spherical. \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Change in the value of fractional change in R_0 is only 2.99 percent \n",
+ "Thus, we can assumed that Gadolinium nucleus is spherical.\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter10.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter10.ipynb
new file mode 100755
index 00000000..170294ba
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter10.ipynb
@@ -0,0 +1,675 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:308367390142ca69701c27a60b649f58dec3483f71d5fe87366628b17198229c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10-Nuclear reactions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg455"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.1 : : Page-455 (2011) \n",
+ "#calculate The Q-value for the fromation\n",
+ "import math\n",
+ "M = 47.668; ## Total mass of reaction, MeV\n",
+ "E = 44.359; ## Total energy, MeV\n",
+ "Q = M-E; ## Q-value, MeV\n",
+ "print'%s %.2f %s'%(\"\\nThe Q-value for the formation of P30 = \",Q,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The Q-value for the formation of P30 = 3.309 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The Q-value for the formation of P30 = 3.31 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg455"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.2 : : Page-455 (2011) \n",
+ "#calculate The Q-value for the reaction\n",
+ "import math\n",
+ "E_x = 7.70; ## Energy of the alpha particle, MeV\n",
+ "E_y = 4.44; ## Energy of the proton, MeV\n",
+ "m_x = 4.0; ## Mass number of alpha particle\n",
+ "m_y = 1.0; ## Mass number of protium ion\n",
+ "M_X = 14; ## Mass number of nitrogen nucleus\n",
+ "M_Y = 17; ## Mass number of oxygen nucleus\n",
+ "theta = 90*3.14/180.; ## Angle between incident beam direction and emitted proton, degree\n",
+ "A_x = 4.0026033; ## Atomic mass of alpha particle, u\n",
+ "A_X = 14.0030742; ## Atomic mass of nitrogen nucleus, u\n",
+ "A_y = 1.0078252; ## Atomic mass of proton, u\n",
+ "Q = ((E_y*(1+m_y/M_Y))-(E_x*(1-m_x/M_Y))-2./M_Y*math.sqrt((m_x*m_y*E_x*E_y))*math.cos(theta))/931.5; ## Q-value, u\n",
+ "A_Y = A_x+A_X-A_y-Q; ## Atomic mass of O-17, u\n",
+ "print'%s %.4f %s %.2f %s '%(\"\\nThe Q-value of the reaction = \",Q,\" u\" \" \\nThe atomic mass of the O-17 =\",A_Y,\" u\");\n",
+ "\n",
+ "## Result\n",
+ "## The Q-value of the reaction = -0.0012755 u \n",
+ "## The atomic mass of the O-17 = 16.9991278 u \n",
+ "## Atomic mass of the O-17 : 16.9991278 u "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The Q-value of the reaction = -0.0013 u \n",
+ "The atomic mass of the O-17 = 17.00 u \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg455"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.3 : : Page-455 (2011) \n",
+ "#calculate The kinetic energy of the emitted neutron \n",
+ "import math\n",
+ "m_p = 1.007276; ## Atomic mass of the proton, u\n",
+ "m_H = 3.016049; ## Atomic mass of the tritium, u \n",
+ "m_He = 3.016029; ## Atomic mass of the He ion, u \n",
+ "m_n = 1.008665; ## Atomic mass of the emitted neutron, u\n",
+ "Q = (m_p+m_H-m_He-m_n)*931.5; ## Q-value in MeV\n",
+ "E_p = 3.; ## Kinetic energy of the proton, MeV \n",
+ "theta = 30*3.14/180; ## angle, radian\n",
+ "u = math.sqrt(m_p*m_n*E_p)/(m_He+m_n)*math.cos(theta); ##\n",
+ "v = ((m_He*Q)+E_p*(m_He-m_p))/(m_He+m_n); ##\n",
+ "E_n = (u+math.sqrt(u**2+v))**2; ## Kinetic energy of the emitted neutron,MeV\n",
+ "print'%s %.2f %s'%(\"\\nThe kinetic energy of the emitted neutron = \",E_n,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The kinetic energy of the emitted neutron = 1.445 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The kinetic energy of the emitted neutron = 1.45 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg456"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.4 : : Page-456 (2011)\n",
+ "#findThe temperature in the fusion reaction \n",
+ "import math\n",
+ "r_min = 4e-015; ## Distance between two deutrons, metre\n",
+ "k = 1.3806504e-023; ## Boltzmann's constant, Joule per kelvin\n",
+ "alpha = 1/137.; ## Fine structure constant\n",
+ "h_red = 1.05457168e-034; ## Reduced planck's constant, Joule sec\n",
+ "C = 3e+08; ## Velocity of light, meter per second\n",
+ "T = alpha*h_red*C/(r_min*k);\n",
+ "print'%s %.2e %s'%(\"\\nThe temperature in the fusion reaction is =\",T,\" K\");\n",
+ "\n",
+ "## Result\n",
+ "## The temperature in the fusion reaction is = 4.2e+009 K "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The temperature in the fusion reaction is = 4.18e+09 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg456"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11.5 : : Page-456 (2011) \n",
+ "#find the excitation energy of the compound nucleus\n",
+ "import math\n",
+ "E_0 = 4.99; ## Energy of the proton, MeV \n",
+ "m_p = 1.; ## Mass number of the proton\n",
+ "m_F = 19.; ## Mass number of the flourine\n",
+ "E = E_0/(1+m_p/m_F); ## Energy of the relative motion, MeV\n",
+ "A_F = 18.998405; ## Atomic mass of the fluorine, amu\n",
+ "A_H = 1.007276; ## Atomic mass of the proton, amu\n",
+ "A_Ne = 19.992440; ## Atomic mass of the neon, amu\n",
+ "del_E = (A_F+A_H-A_Ne)*931.5; ## Binding energy of the absorbed proton, MeV\n",
+ "E_exc = E+del_E; ## Excitation energy of the compound nucleus, MeV\n",
+ "print'%s %.2f %s'%(\"\\nThe excitation energy of the compound nucleus = \",E_exc,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The excitation energy of the compound nucleus = 17.074 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The excitation energy of the compound nucleus = 17.07 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg457"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.6 : : Page-457 (2011) \n",
+ "#find The excitation energy of the compound nucleus and The parity of the compound nucleus\n",
+ "import math\n",
+ "E_d = 0.6; ## Energy of the deutron, MeV \n",
+ "m_d = 2.; ## Mass number of the deutron\n",
+ "m_Li = 19.; ## Mass number of the Lithium\n",
+ "E = E_d/(1.+m_d/m_Li); ## Energy of the relative motion, MeV\n",
+ "A_Li = 6.017; ## Atomic mass of the Lithium, amu\n",
+ "A_d = 2.015; ## Atomic mass of the deutron, amu\n",
+ "A_Be = 8.008; ## Atomic mass of the Beryllium, amu\n",
+ "del_E = (A_Li+A_d-A_Be)*931.5; ## Binding energy of the absorbed proton, MeV\n",
+ "E_exc = E+del_E; ## Excitation energy of the compound nucleus, MeV\n",
+ "l_f = 2.; ## orbital angular momentum of two alpha particle\n",
+ "P = (-1)**l_f*(+1.)**2; ## Parity of the compound nucleus\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe excitation energy of the compound nucleus = \",E_exc,\" MeV\" and \"\\nThe parity of the compound nucleus = \",P,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The excitation energy of the compound nucleus = 22.899 MeV\n",
+ "## The parity of the compound nucleus = 1 \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The excitation energy of the compound nucleus = 22.90 \n",
+ "The parity of the compound nucleus = 1.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg457"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.7 : : Page-457 (2011)\n",
+ "#find The cross section for neutron induced fission\n",
+ "import math\n",
+ "D = 1e-016; ## Disintegration constant, per sec\n",
+ "phi = 10**11; ## Neutron flux, neutrons per square cm per sec\n",
+ "sigma = 5*D/(phi*10**-27); ## Cross section, milli barns\n",
+ "print'%s %.2f %s'%(\"\\nThe cross section for neutron induced fission = \",sigma,\" milli barns\");\n",
+ "\n",
+ "## Result\n",
+ "## The cross section for neutron induced fission = 5 milli barns "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The cross section for neutron induced fission = 5.00 milli barns\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg457"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.8 : : Page-457 (2011)\n",
+ "#find The number of nuclei of Co60 produced and The initial activity\n",
+ "import math\n",
+ "N_0 = 6.02252e+026; ## Avogadro's constant \n",
+ "rho = 8.9*10**3; ## Nuclear density of Co-59, Kg per cubic metre\n",
+ "M = 59.; ## Mass number\n",
+ "sigma = 30e-028; ## Cross section, per square metre\n",
+ "phi = 10**16; ## Neutron flux, neutrons per square metre per sec\n",
+ "d = 0.04e-02; ## Thickness of Co-59 sheet, metre\n",
+ "t = 3*60*60; ## Total reaction time, sec\n",
+ "t_half = 5.2*365*86400; ## Half life of Co-60, sec \n",
+ "D = 0.693/t_half; ## Disintegration constant, per sec\n",
+ "N_nuclei = round(N_0*rho/M*sigma*phi*d*t); ## Number of nuclei of Co-60 produced\n",
+ "Init_activity = D*N_nuclei; ## Initial activity, decays per sec\n",
+ "print'%s %.2e %s %.2f %s '%(\"\\nThe number of nuclei of Co60 produced =\",N_nuclei,\" \" and \"\\nThe initial activity per Sq. metre = \",Init_activity,\" decays per sec\");\n",
+ "\n",
+ "## Result\n",
+ "## The number of nuclei of Co60 produced = 1.18e+019 \n",
+ "## The initial activity per Sq. metre = 5e+010 decays per sec "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The number of nuclei of Co60 produced = 1.18e+19 \n",
+ "The initial activity per Sq. metre = 49755893720.42 decays per sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg458"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.9 : : Page-458 (2011)\n",
+ "#find The number of events detected\n",
+ "import math\n",
+ "d = 0.1; ## Thickness of Fe-54 sheet, Kg per squre metre\n",
+ "M = 54.; ## Mass number of Fe \n",
+ "m = 1.66e-027; ## Mass of the proton, Kg\n",
+ "n = d/(M*m); ## Number of nuclei in unit area of the target, nuclei per square metre\n",
+ "ds = 10**-5; ## Area, metre square\n",
+ "r = 0.1; ## Distance between detector and target foil, metre\n",
+ "d_omega =ds/r**2; ## Solid angle, steradian\n",
+ "d_sigma = 1.3e-03*10**-3*10**-28; ## Differential cross section, square metre per nuclei\n",
+ "P = d_sigma*n; ## Probablity, event per proton\n",
+ "I = 10**-7; ## Current, ampere\n",
+ "e = 1.6e-19; ## Charge of the proton, C\n",
+ "N = I/e; ## Number of protons per second in the incident beam, proton per sec\n",
+ "dN = P*N; ## Number of events detected per second, events per sec\n",
+ "print'%s %.2f %s'%(\"\\nThe number of events detected =\",dN,\" events per sec\");\n",
+ "\n",
+ "## Result\n",
+ "## The number of events detected = 90 events per sec "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The number of events detected = 90.64 events per sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg458"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.10 : : Page-458 (2011)\n",
+ "#find The fractional attenuation of neutron beam on passing through nickel sheet\n",
+ "import math\n",
+ "N_0 = 6.02252e+26; ## Avogadro's constant\n",
+ "sigma = 3.5e-28; ## Cross section, square metre\n",
+ "rho = 8.9e+03; ## Nuclear density, Kg per cubic metre\n",
+ "M = 58.; ## Mass number \n",
+ "summation = rho/M*N_0*sigma; ## Macroscopic cross section, per metre\n",
+ "x = 0.01e-02; ## Thickness of nickel sheet, metre\n",
+ "I0_ratio_I = math.exp(summation*x/2.3026); ## Fractional attenuation of neutron beam on passing through nickel sheet\n",
+ "print'%s %.2f %s ' %(\"\\nThe fractional attenuation of neutron beam on passing through nickel sheet =\", I0_ratio_I,\"\");\n",
+ "print(\"Wrong answer given in the textbook\")\n",
+ "## Result\n",
+ "## The fractional attenuation of neutron beam on passing through nickel sheet = 1.0014 \n",
+ "## Wrong answer given in the textbook\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The fractional attenuation of neutron beam on passing through nickel sheet = 1.00 \n",
+ "Wrong answer given in the textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg458"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.11 : : Page-458 (2011)\n",
+ "#find he scattering contribution to the resonance \n",
+ "import math\n",
+ "D = math.sqrt(1.45e-021/(4.*math.pi)); ## Wavelength, metre\n",
+ "W_ratio = 2.3e-07; ## Width ratio\n",
+ "sigma = W_ratio*(4*math.pi)*D**2*10**28; ## Scattering contribution, barn\n",
+ "print'%s %.2f %s'%(\"\\nThe scattering contribution to the resonance = \",sigma,\" barns\");\n",
+ "\n",
+ "## Result\n",
+ "## The scattering contribution to the resonance = 3.33 barns \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The scattering contribution to the resonance = 3.33 barns\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg458"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.12 : : Page-458 (2011)\n",
+ "#find The relative probabilities\n",
+ "import math\n",
+ "sigma = 2.8e-024; ## Cross section, metre square\n",
+ "D = 2.4e-11; ## de Broglie wavelength, metre\n",
+ "R_prob = math.pi*sigma/D**2; ## Relative probabilities of (n,n) and (n,y) in indium\n",
+ "print'%s %.2f %s'%(\"\\nThe relative probabilities of (n,n) and (n,y) in indium = \", R_prob,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The relative probabilities of (n,n) and (n,y) in indium = 0.015 \n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The relative probabilities of (n,n) and (n,y) in indium = 0.02 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg459"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.13 : : Page-459 (2011)\n",
+ "#find The cross section of neutron capture\n",
+ "import math\n",
+ "h = 6.625e-34; ## Planck's constant, joule sec \n",
+ "m_n = 1.67e-27; ## Mass of neutron, Kg\n",
+ "E = 4.906; ## Energy, joule \n",
+ "w_y = 0.124; ## radiation width, eV\n",
+ "w_n = 0.007*E**(1/2.); ## Probability of elastic emission of neutron, eV\n",
+ "I = 3.; ## Total angular momentum\n",
+ "I_c = 2.; ## Total angular momentum in the compound state\n",
+ "sigma = ((h**2)*(2*I_c+1)*w_y*w_n)*10**28/(2.*math.pi*m_n*E*1.602e-019*(2*I+1)*(w_y+w_n)**2); ## Cross section, barns\n",
+ "print'%s %.3e %s'%(\"\\nThe cross section of neutron capture = \",sigma,\" barns\");\n",
+ "\n",
+ "## Result\n",
+ "## The cross section of neutron capture = 3.755e+004 barns \n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The cross section of neutron capture = 3.755e+04 barns\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg459"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.14 : : Page-459 (2011)\n",
+ "#fid the value of theta max\n",
+ "import math\n",
+ "import numpy\n",
+ "R = 5.; ## Radius, femto metre\n",
+ "k_d = 0.98; ## The value of k for deutron \n",
+ "k_p = 0.82; ## The value of k for triton\n",
+ "theta = numpy.zeros((1,5)); ## Angles at which differetial cross section is maximum, degree\n",
+ "## Use of for loop for angles calculation(in degree)\n",
+ "for l in range (1,4):\n",
+ " theta = round((math.acos((k_d**2+k_p**2)/(2.*k_d*k_p)-l**2/(2.*k_d*k_p*R**2)))*180./3.14,0);\n",
+ " #theta = round((acos((k_d^2+k_p^2)/(2*k_d*k_p)-l^2/(2*k_d*k_p*R^2)))*180/3.14);\n",
+ " print'%s %.2f %s'%(\"\\nFor l = \", l,\"\");\n",
+ " print'%s %.2f %s'%(\"the value of theta_max = \", math.ceil(theta),\"degree\");\n",
+ " \n",
+ "\n",
+ "## Result\n",
+ "## For l = 0,the value of theta_max = 0 degree\n",
+ "## For l = 1,the value of theta_max = 8 degree\n",
+ "## For l = 2,the value of theta_max = 24 degree\n",
+ "## For l = 3,the value of theta_max = 38 degree\n",
+ "## For l = 4,the value of theta_max = 52 degree "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "For l = 1.00 \n",
+ "the value of theta_max = 8.00 degree\n",
+ "\n",
+ "For l = 2.00 \n",
+ "the value of theta_max = 24.00 degree\n",
+ "\n",
+ "For l = 3.00 \n",
+ "the value of theta_max = 38.00 degree\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg459"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa10.15 : : Page-459 (2011)\n",
+ "#calculate total angular momentum transfer\n",
+ "import math\n",
+ "k_d = 2.02e+30; ## The value of k for deutron\n",
+ "k_t = 2.02e+30; ## The value of k for triton\n",
+ "theta = 23*3.14/180; ## Angle, radiams\n",
+ "q = math.sqrt (k_d+k_t-2*k_t*math.cos(theta))*10**-15; ## the value of q in femto metre\n",
+ "R_0 = 1.2; ## Distance of closest approach, femto metre\n",
+ "A = 90.; ## Mass number of Zr-90\n",
+ "z = 4.30; ## Deutron size, femto metre\n",
+ "R = R_0*A**(1/3.)+1/2.*z; ## Radius of the nucleus, femto metre\n",
+ "l = round(q*R); ## Orbital angular momentum\n",
+ "I = l+1./2. ## Total angular momentum\n",
+ "print'%s %.2f %s'%(\"\\nThe total angular momentum transfer = \", I,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The total angular momentum transfer = 4.5 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The total angular momentum transfer = 4.50 \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter11.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter11.ipynb
new file mode 100755
index 00000000..ff1534f6
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter11.ipynb
@@ -0,0 +1,572 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:9b22acc9bcbd27c33c9e62b557ecc66fc981e9c1775b58e8ee670ca880259cf2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter11-Particle Accelerators"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg535"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11.1 : : Page-535(2011) \n",
+ "#calculate The optimum number of stages in the accelerator and the ripple voltage\n",
+ "import math\n",
+ "V_0 = 10**5; ## Accelerating voltage, volts\n",
+ "C = 0.02e-006; ## Capacitance, farad\n",
+ "I = 4*1e-003; ## Current, ampere\n",
+ "f = 200.; ## Frequency, cycles per sec\n",
+ "n = math.sqrt (V_0*f*C/I); ## Number of particles\n",
+ "delta_V = I*n*(n+1.)/(4.*f*C);\n",
+ "print'%s %.2f %s'%(\"\\nThe optimum number of stages in the accelerator = \", n,\"\");\n",
+ "print'%s %.2f %s'%(\"\\nThe ripple voltage = \", delta_V/1e+003,\"kV\");\n",
+ "\n",
+ "## Result\n",
+ "## The optimum number of stages in the accelerator = 10\n",
+ "## The ripple voltage = 27.5 kV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The optimum number of stages in the accelerator = 10.00 \n",
+ "\n",
+ "The ripple voltage = 27.50 kV\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg536"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11.2 : : Page-536 (2011)\n",
+ "#calculate The charging current and The rate of rise of electrode potential\n",
+ "import math\n",
+ "s = 15.; ## Speed, metre per sec\n",
+ "w = 0.3; ## Width of the electrode, metre\n",
+ "E = 3e+06; ## Breakdown strength, volts per metre\n",
+ "eps = 8.85e-12; ## Absolute permitivity of free space, farad per metre\n",
+ "C = 111e-12; ## Capacitance, farad\n",
+ "i = round (2*eps*E*s*w*10**6); ## Current, micro ampere\n",
+ "V = i/C*10**-12; ## Rate of rise of electrode potential, mega volts per sec\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe charging current =\",i,\" micro-ampere\"and \" \\nThe rate of rise of electrode potential = \",V,\" MV/sec\");\n",
+ "\n",
+ "## Result\n",
+ "## The charging current = 239 micro-ampere \n",
+ "## The rate of rise of electrode potential = 2.15 MV/sec "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The charging current = 239.00 \n",
+ "The rate of rise of electrode potential = 2.15 MV/sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg536"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11.3 : : Page-536 (2011)\n",
+ "#calculate The length of the final drift tube and The kinetic energy of the injected protons\n",
+ "import math\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "\n",
+ "f = 200.*10**6; ## Frequency of the accelerator, cycle per sec\n",
+ "M = 1.6724e-27; ## Mass of the proton, Kg\n",
+ "E = 45.3*1.6e-13; ## Accelerating energy, joule\n",
+ "L_f = round (1./f*math.sqrt(2.*E/M)*100.); ## Length of the final drift tube, centi metre\n",
+ "L_1 = 5.35*10**-2; ## Length of the first drift tube, metre\n",
+ "K_E = (1./2.*M*L_1**2.*f**2.)/1.6e-13; ## Kinetic energy of the injected proton, MeV\n",
+ "E_inc = E/1.6e-13-K_E; ## Increase in energy, MeV\n",
+ "q = 1.6e-19; ## Charge of the proton, C\n",
+ "V = 1.49e+06; ## Accelerating voltage, volts\n",
+ "N = E_inc*1.6e-13/(q*V); ## Number of drift protons\n",
+ "def fun(n):\n",
+ " y=n**0.5\n",
+ " return y\n",
+ "\n",
+ "l2=scipy.integrate.quad(fun,0,N)\n",
+ "L = 1./f*math.sqrt(2.*q*V/M)*l2[0]; ## Total length of the accelerator, metre\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe length of the final drift tube = \",L_f,\" cm\"and\"\\nThe kinetic energy of the injected protons = \",K_E,\" MeV\"and\"\\nThe total length of the accelerator = \",L,\" metre\");\n",
+ "\n",
+ "## Result\n",
+ "## The length of the final drift tube = 47 cm\n",
+ "## The kinetic energy of the injected protons = 0.60 MeV\n",
+ "## The total length of the accelerator = 9.2 metre "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The length of the final drift tube = 47.00 \n",
+ "The kinetic energy of the injected protons = 0.60 \n",
+ "The total length of the accelerator = 9.25 metre \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg536"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11.5 : : Page-536 (2011)\n",
+ "#find energy of the emergine deutron and frequency of the dee voltage\n",
+ "import math\n",
+ "B = 1.4; ## Magnetic field, tesla\n",
+ "R = 88e-002; ## Radius of the orbit, metre\n",
+ "q = 1.6023e-019; ## Charge of the deutron, C\n",
+ "M_d = 2.014102*1.66e-27; ## Mass of the deutron, Kg\n",
+ "M_He = 4.002603*1.66e-27; ## Mass of the He ion, Kg\n",
+ "E = B**2*R**2*q**2/(2*M_d*1.6e-13); ## Energy og the emerging deutron, mega electron volts\n",
+ "f = B*q/(2.*math.pi*M_d)*10**-6; ## Frequency of the deutron voltage, mega cycles per sec\n",
+ "B_He = 2*math.pi*M_He*f*10**6/(2*q); ## Magnetic field required for He(++) ions, weber per square metre\n",
+ "B_change = B-B_He; ## Change in magnetic field, tesla\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe energy of the emerging deutron = \",E,\" MeV\" and \"\\nThe frequency of the dee voltage = \",f,\"MHz\" and\"\\nThe change in magnetic field = \",B_change,\" tesla\");\n",
+ "\n",
+ "## Result\n",
+ "## The energy of the emerging deutron = 36.4 MeV\n",
+ "## The frequency of the dee voltage = 10.68 MHz\n",
+ "## The change in magnetic field = 0.01 tesla "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy of the emerging deutron = 36.42 \n",
+ "The frequency of the dee voltage = 10.68 \n",
+ "The change in magnetic field = 0.01 tesla \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg537"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11.6: : Page-537 (2011)\n",
+ "#calculate effective reduction in magnetic field and charge in orbit radius\n",
+ "import math\n",
+ "K_E = 7.5*1.6023e-13; ## Kinetic energy, joule \n",
+ "r = 0.51; ## Radius of the proton's orbit, metre\n",
+ "E = 5*10**6; ## Electric field, volts per metre\n",
+ "m = 1.67e-27; ## Mass of the proton, Kg\n",
+ "q = 1.6023e-19; ## Charge of the proton, C\n",
+ "v = math.sqrt(2.*K_E/m); ## Velocity of the proton, metre per sec\n",
+ "B_red = E/v; ## The effective reduction in magnetic field, tesla\n",
+ "B = m*v/(q*r); ## Total magnetic field produced, tesla\n",
+ "r_change = r*B_red/B; ## The change in orbit radius, metre\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe effective reduction in magnetic field = \",B_red,\" tesla \"and \"\\nThe change in orbit radius = \",r_change,\" metre \");\n",
+ "\n",
+ "## Result\n",
+ "## The effective reduction in magnetic field = 0.132 tesla \n",
+ "## The change in orbit radius = 0.087 metre "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The effective reduction in magnetic field = 0.13 \n",
+ "The change in orbit radius = 0.09 metre \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg537"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11.7 : : Page-537 (2011)\n",
+ "#calculate enegy of the elctron and average enegy gained per revolution \n",
+ "import math\n",
+ "B = 0.4; ## Magnetic field, tesla\n",
+ "e = 1.6203e-19; ## Charge of an electron, C\n",
+ "R = 30*2.54e-02; ## Radius, metre\n",
+ "c = 3e+08; ## Capacitance, farad\n",
+ "E = B*e*R*c/1.6e-13; ## The energy of the electron, mega electron volts\n",
+ "f = 50.; ## Frequency, cycles per sec\n",
+ "N = c/(4*2*math.pi*f*R); ## Total number of revolutions\n",
+ "Avg_E_per_rev = E*1e+006/N; ## Average energy gained per revolution, electron volt\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe energy of the electron = \",E,\" MeV\"and \"\\nThe average energy gained per revolution = \",Avg_E_per_rev,\" eV\");\n",
+ "\n",
+ "## Result\n",
+ "## The energy of the electron = 92.6 MeV\n",
+ "## The average energy gained per revolution = 295.57 eV \n",
+ "## Note: Wrong answer is given in the textbook \n",
+ "## Average energy gained per revolution : 295.57 electron volts\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy of the electron = 92.60 \n",
+ "The average energy gained per revolution = 295.57 eV \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg537"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11.8 : : Page-537 (2011)\n",
+ "#calculate peak current and average current and duty cycle\n",
+ "import math\n",
+ "R = 0.35; ## Orbit radius, metre\n",
+ "N = 100e+06/480.; ## Total number of revolutions\n",
+ "L = 2*math.pi*R*N; ## Distance traversed by the electron, metre\n",
+ "t = 2e-06; ## Pulse duration, sec\n",
+ "e = 1.6203e-19; ## Charge of an electron, C\n",
+ "n = 3e+09; ## Number of electrons\n",
+ "f = 180.; ## frequency, hertz\n",
+ "I_p = n*e/t; ## Peak current, ampere\n",
+ "I_avg = n*e*f; ## Average current, ampere \n",
+ "tau = t*f; ## Duty cycle\n",
+ "print'%s %.2e %s %.2e %s %.2e %s'%(\"\\nThe peak current = \",I_p,\" ampere\" and \"\\nThe average current = \",I_avg,\" ampere \"and \"\\nThe duty cycle =\",tau,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The peak current = 2.4e-004 ampere \n",
+ "## The average current = 8.75e-008 ampere \n",
+ "## The duty cycle = 3.6e-004 \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The peak current = 2.43e-04 \n",
+ "The average current = 8.75e-08 \n",
+ "The duty cycle = 3.60e-04 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg538"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11.9 : : Page-538 (2011)\n",
+ "#calculate maximum frequncy of the dee voltage and kinetic energy of the deutron \n",
+ "import math\n",
+ "q = 1.6023e-19; ## Charge of an electron, C\n",
+ "B_0 = 1.5; ## Magnetic field at the centre, tesla\n",
+ "m_d = 2.014102*1.66e-27; ## Mass of the deutron, Kg\n",
+ "f_max = B_0*q/(2*math.pi*m_d*10**6); ## Maximum frequency of the dee voltage, mega cycles per sec\n",
+ "B_prime = 1.4310; ## Magnetic field at the periphery of the dee, tesla\n",
+ "f_prime = 10**7; ## Frequency, cycles per sec\n",
+ "c = 3e+08; ## Velocity of the light, metre per sec\n",
+ "M = B_prime*q/(2*math.pi*f_prime*1.66e-27); ## Relativistic mass, u\n",
+ "K_E = (M-m_d/1.66e-27)*931.5; ## Kinetic energy of the particle, mega electron volts\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe maximum frequency of the dee voltage = \",f_max,\" MHz\"and \"\\nThe kinetic energy of the deuteron = \",K_E,\" MeV\");\n",
+ " \n",
+ "## Result\n",
+ "## The maximum frequency of the dee voltage = 11.44 MHz\n",
+ "## The kinetic energy of the deuteron = 171.6 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The maximum frequency of the dee voltage = 11.44 \n",
+ "The kinetic energy of the deuteron = 171.62 MeV \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg538"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate frequency of the applied electric field and magnetic field intensity\n",
+ "## Exa11.10 : : Page-538 (2011)\n",
+ "import math\n",
+ "e = 1.6023e-19; ## Charge of an electron, C\n",
+ "E = 70*1.6e-13; ## Energy, electron volts\n",
+ "R = 0.28; ## Radius of the orbit, metre\n",
+ "c = 3e+08; ## Velocity of light, metre per sec\n",
+ "B = E/(e*R*c); ## Magnetic field intensity, tesla\n",
+ "f = e*B*c**2/(2*math.pi*E); ## Frequency, cycle per sec\n",
+ "del_E = 88.5*(0.07)**4*10**3/(R); ## Energy radiated by an electron, electron volts\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe frequency of the applied electric field = \",f,\" cycles per sec\" and\"\\nThe magnetic field intensity = \",B,\" tesla\"and \"\\nThe energy radiated by the electron =\",del_E,\" eV\");\n",
+ "\n",
+ "## Result\n",
+ "## The frequency of the applied electric field = 1.705e+008 cycles per sec \n",
+ "## The magnetic field intensity = 0.832 tesla\n",
+ "## The energy radiated by the electron = 7.6 eV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The frequency of the applied electric field = 170523153.31 \n",
+ "The magnetic field intensity = 0.83 \n",
+ "The energy radiated by the electron = 7.59 eV \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg538"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate kinetic energy of the accelerated nitrogen\n",
+ "## Exa11.11 : : Page-538 (2011)\n",
+ "import math\n",
+ "E = 3.; ## Energy of proton synchrotron, giga electron volts\n",
+ "m_0_c_sq = 0.938; ## Relativistic energy, mega electron volts\n",
+ "P_p = math.sqrt(E**2-m_0_c_sq**2); ## Momentum of the proton, giga electron volts per c\n",
+ "P_n = 6*P_p; ## Momentum of the N(14) ions, giga electron volts\n",
+ "T_n = math.sqrt(P_n**2+(0.938*14.)**2)-0.938*14; ## Kinetic energy of the accelerated nitrogen ion\n",
+ "print'%s %.2f %s'%(\"\\nThe kinetic energy of the accelerated nitrogen ion = \",T_n,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The kinetic energy of the accelerated nitrogen ion = 8.43 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The kinetic energy of the accelerated nitrogen ion = 8.43 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg539"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11.12 : : Page-539 (2011)\n",
+ "#calculate maximum magnetic flux density and maximum frequency of the accerlating voltage\n",
+ "import math\n",
+ "e = 1.6e-19; ## Charge of an electron, C\n",
+ "R = 9.144; ## Radius, metre\n",
+ "m_p = 1.67e-027; ## Mass of the proton, Kg\n",
+ "E = 3.6*1.6e-13; ## Energy, joule\n",
+ "L = 3.048; ## Length of the one synchrotron section, metre \n",
+ "T = 3; ## Kinetic energy, giga electron volts\n",
+ "c = 3e+08; ## Velocity of the light, metre per sec\n",
+ "m_0_c_sq = 0.938; ## Relativistic energy, mega electron volts\n",
+ "B = round (math.sqrt(2*m_p*E)/(R*e)*10**4); ## Maximum magnetic field density, web per square metre\n",
+ "v = B*10**-4*e*R/m_p; ## Velocity of the proton, metre per sec\n",
+ "f_c = v/(2*math.pi*R*10**6); ## Frequency of the circular orbit, mega cycles per sec\n",
+ "f_0 = 2*math.pi*R*f_c*10**3/(2*math.pi*R+4*L); ## Reduced frequency, kilo cycles per sec\n",
+ "B_m = 3.33*math.sqrt(T*(T+2*m_0_c_sq))/R; ## Relativistic field, web per square metre\n",
+ "f_0 = c**2*e*R*B*1e-004/((2*math.pi*R+4*L)*(T+m_0_c_sq)*e*1e+015); ## Maximum frequency of the accelerating voltage, mega cycles per sec\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe maximum magnetic flux density = \",B_m,\" weber/Sq.m\"and \"\\nThe maximum frequency of the accelerating voltage = \",f_0,\" MHz\");\n",
+ " \n",
+ "## Result\n",
+ "## The maximum magnetic flux density = 1.393 weber/Sq.m\n",
+ "## The maximum frequency of the accelerating voltage = 0.09 MHz\n",
+ "## Answer is given wrongly in the textbook \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The maximum magnetic flux density = 1.39 \n",
+ "The maximum frequency of the accelerating voltage = 0.09 MHz \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg539"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate energy of the proton\n",
+ "## Exa11.13 : : Page-539 (2011)\n",
+ "import math\n",
+ "E_c = 30e+009; ## Energy of the proton accelerator, GeV\n",
+ "m_0_c_sq = 0.938*10**6; ## Relativistic energy, GeV\n",
+ "E_p = (4*E_c**2-2*m_0_c_sq**2)/(2*m_0_c_sq) ; ## Energy of the proton, GeV\n",
+ "print'%s %.2f %s'%(\"\\nThe energy of the proton = \",E_p/1e+009,\" GeV\");\n",
+ "print(\"wrong answer in the textbook\")\n",
+ "## Result\n",
+ "## The energy of the proton = 1.92e+006 GeV \n",
+ "## Wrong answer given in the textbook"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy of the proton = 1918976.54 GeV\n",
+ "wrong answer in the textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter12.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter12.ipynb
new file mode 100755
index 00000000..68299296
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter12.ipynb
@@ -0,0 +1,510 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:7bae67184aff3b5018323a5a7fd8d73277287467e3722f9114aa02be5e49dd72"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter12-Neutrons"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Exx1-pg573"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa12.1 : : Page-573 (2011)\n",
+ "#calculate activity for Cu-63 and disintegrations\n",
+ "import math \n",
+ "N_0 = 6.23e+23; ## Avogadro's number, per mole\n",
+ "m = 0.1; ## Mass of copper foil, Kg\n",
+ "phi = 10**12; ## Neutron flux density, per square centimetre sec\n",
+ "a_63 = 0.691; ## Abundance of Cu-63\n",
+ "a_65 = 0.309; ## Abundance of Cu-65\n",
+ "W_m = 63.57; ## Molecular weight, gram\n",
+ "sigma_63 = 4.5e-24; ## Activation cross section for Cu-63, square centi metre\n",
+ "sigma_65 = 2.3e-24; ## Activation cross section for Cu-65, square centi metre\n",
+ "A_63 = phi*sigma_63*m*a_63/W_m*N_0; ## Activity for Cu-63, disintegrations per sec\n",
+ "A_65 = phi*sigma_65*m*a_65/W_m*N_0; ## Activity for Cu-65, disintegrations per sec\n",
+ "print'%s %.2e %s %.2e %s'%(\"\\nThe activity for Cu-63 is = \",A_63,\" disintegrations per sec\" and \"\\nThe activity for Cu-65 is = \",A_65,\" disintegrations per sec\");\n",
+ "\n",
+ "## Result\n",
+ "## The activity for Cu-63 is = 3.047e+009 disintegrations per sec \n",
+ "## The activity for Cu-65 is = 6.97e+008 disintegrations per sec "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The activity for Cu-63 is = 3.05e+09 \n",
+ "The activity for Cu-65 is = 6.97e+08 disintegrations per sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg573"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa12.2 : : Page-573 (2011)\n",
+ "import math \n",
+ "#calculate enegy loss \n",
+ "A_Be = 9.; ## Mass number of beryllium\n",
+ "A_U = 238.; ## Mass number of uranium\n",
+ "E_los_Be = (1-((A_Be-1)**2/(A_Be+1)**2))*100.; ## Energy loss for beryllium\n",
+ "E_los_U = round((1-((A_U-1)**2/(A_U+1)**2))*100.); ## Energy loss for uranium\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe energy loss for beryllium is = \",E_los_Be,\" percent\"and \" \\nThe energy loss for uranium is = \",E_los_U,\" percent\");\n",
+ "\n",
+ "## Check for greater energy loss !!!!\n",
+ "if E_los_Be >= E_los_U :\n",
+ " print(\"\\nThe energy loss is greater for beryllium\");\n",
+ "else:\n",
+ " print(\"\\nThe energy loss is greater for uranium\");\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## The energy loss for beryllium is = 36 percent \n",
+ "## The energy loss for uranium is = 2 percent\n",
+ "## The energy loss is greater for beryllium \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy loss for beryllium is = 36.00 \n",
+ "The energy loss for uranium is = 2.00 percent \n",
+ "\n",
+ "The energy loss is greater for beryllium\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa12.3 : : Page-574 (2011)\n",
+ "#calculate energy loss of neutron\n",
+ "import math \n",
+ "A = 12.; ## Mass number of Carbon\n",
+ "alpha = (A-1)**2/(A+1)**2; ## Scattering coefficient\n",
+ "E_loss = 1/2.*(1-alpha)*100.; ## Energy loss of neutron\n",
+ "print'%s %.2f %s'%(\"\\nThe energy loss of neutron = \",E_loss,\" percent\")\n",
+ "\n",
+ "## Result\n",
+ "## The energy loss of neutron = 14.201 percent \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy loss of neutron = 14.20 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa12.4 : : Page-574 (2011)\n",
+ "#calculate number of collisions of neutrons \n",
+ "import math \n",
+ "zeta = 0.209; ## Moderated assembly\n",
+ "E_change = 100./1.; ## Change in energy of the neutron\n",
+ "E_thermal = 0.025; ## Thermal energy of the neutron, electron volts\n",
+ "E_n = 2*10**6; ## Energy of the neutron, electron volts\n",
+ "n = 1/zeta*math.log(E_change); ## Number of collisions of neutrons to loss 99 percent of their energies \n",
+ "n_thermal = 1/zeta*math.log(E_n/E_thermal); ## Number of collisions of neutrons to reach thermal energies\n",
+ "print'%s %.2f %s %.2f %s'%(\"\\nThe number of collisions of neutrons to loss 99 percent of their energies = \",n,\" \\nThe number of collisions of neutrons to reach thermal energies = \",n_thermal,\"\")\n",
+ "\n",
+ "## Result\n",
+ "## The number of collisions of neutrons to loss 99 percent of their energies = 22 \n",
+ "## The number of collisions of neutrons to reach thermal energies = 87 \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The number of collisions of neutrons to loss 99 percent of their energies = 22.03 \n",
+ "The number of collisions of neutrons to reach thermal energies = 87.07 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa12.5 : : Page-574 (2011)\n",
+ "#calculate average distance travelled by the neutron\n",
+ "import math\n",
+ "import scipy\n",
+ "from scipy import integrate\n",
+ "L = 1.; ## For simplicity assume thermal diffusion length to be unity, unit\n",
+ "def fun(x):\n",
+ " y=x*math.exp(-x/L)\n",
+ " return y\n",
+ "x_b = scipy.integrate.quad(fun, 0, 100); ## Average distance travelled by the neutron, unit\n",
+ "x_b1=x_b[0]\n",
+ "def fun2(x): \n",
+ " y1=x**2*math.exp(-x/L)\n",
+ " return y1\n",
+ "X=scipy.integrate.quad(fun2, 0, 100)\n",
+ "x_rms = math.sqrt(X[0]); ## Root mean square of the distance trvelled by the neutron, unit\n",
+ "print'%s %.2f %s'%(\"\\nThe average distance travelled by the neutron = \", x_b1,\"*L\");\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe root mean square distance travelled by the neutron = \",x_rms,\"\"and \"\",x_rms,\"x_bar\")\n",
+ "\n",
+ "## Result\n",
+ "## The average distance travelled by the neutron = 1*L\n",
+ "## The root mean square distance travelled by the neutron = 1.414L = 1.414x_bar \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The average distance travelled by the neutron = 1.00 *L\n",
+ "\n",
+ "The root mean square distance travelled by the neutron = 1.41 1.41 x_bar \n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg574"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa12.6 : : Page-574 (2011)\n",
+ "#calculate neutron flux through water \n",
+ "import math\n",
+ "Q = 5e+08; ## Rate at which neutrons produce, neutrons per sec\n",
+ "r = 20.; ## Distance from the source, centi metre\n",
+ "## For water\n",
+ "lambda_wtr = 0.45; ## Transport mean free path, centi metre\n",
+ "L_wtr = 2.73; ## Thermal diffusion length, centi metre\n",
+ "phi_wtr = 3*Q/(4.*math.pi*lambda_wtr*r)*math.exp(-r/L_wtr); ## Neutron flux for water, neutrons per square centimetre per sec\n",
+ "## For heavy water\n",
+ "lambda_h_wtr = 2.40; ## Transport mean free path, centi metre\n",
+ "L_h_wtr = 171.; ## Thermal diffusion length, centi metre\n",
+ "phi_h_wtr = 3*Q/(4.*math.pi*lambda_h_wtr*r)*math.exp(-r/L_h_wtr); ## Neutron flux for heavy water, neutrons per square centimetre per sec\n",
+ "print'%s %.2e %s %.2e %s '%(\"\\nThe neutron flux through water = \",phi_wtr,\" neutrons per square cm per sec\"and \"\\nThe neutron flux through heavy water = \",phi_h_wtr,\" neutrons per square cm per sec\")\n",
+ "\n",
+ "## Result\n",
+ "## The neutron flux through water = 8.730e+003 neutrons per square cm per sec \n",
+ "## The neutron flux through heavy water = 2.212e+006 neutrons per square cm per sec \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The neutron flux through water = 8.73e+03 \n",
+ "The neutron flux through heavy water = 2.21e+06 neutrons per square cm per sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg575"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa12.7 : : Page-575 (2011)\n",
+ "#calculate neutron flux and diffusion length\n",
+ "import math\n",
+ "k = 1.38e-23; ## Boltzmann constant, joules per kelvin\n",
+ "T = 323.; ## Temperature, kelvin\n",
+ "E = (k*T)/1.6e-19; ## Thermal energy, joules\n",
+ "sigma_0 = 13.2e-28; ## Cross section, square metre\n",
+ "E_0 = 0.025; ## Energy of the neutron, electron volts\n",
+ "sigma_a = sigma_0*math.sqrt(E_0/E); ## Absorption cross section, square metre\n",
+ "t_half = 2.25; ## Half life, hours\n",
+ "D= 0.69/t_half; ## Decay constant, per hour\n",
+ "N_0 = 6.023e+026; ## Avogadro's number, per \n",
+ "m_Mn = 55.; ## Mass number of mangnese\n",
+ "w = 0.1e-03; ## Weight of mangnese foil, Kg\n",
+ "A = 200.; ## Activity, disintegrations per sec\n",
+ "N = N_0*w/m_Mn; ## Number of mangnese nuclei in the foil\n",
+ "x1 = 1.5; ## Base, metre\n",
+ "x2 = 2.0; ## Height, metre\n",
+ "phi = A/(N*sigma_a*0.416); ## Neutron flux, neutrons per square metre per sec\n",
+ "phi1 = 1.; ## For simplicity assume initial neutron flux to be unity, neutrons/Sq.m-sec\n",
+ "phi2 = 1/2.*phi1; ## Given neutron flux, neutrons/Sq.m-sec\n",
+ "L1 = 1/math.log(phi1/phi2)/(x2-x1); ## Thermal diffusion length for given neutron flux, m\n",
+ "L = math.sqrt(1./((1./L1)**2+(math.pi/x1)**2+(math.pi/x2)**2)); ## Diffusion length, metre\n",
+ "print'%s %.2e %s %.2f %s '%(\"\\nThe neutron flux = \",phi,\" neutrons per square metre per sec\"and \" \\nThe diffusion length = \",L,\" metre\");\n",
+ "\n",
+ "## Result\n",
+ "## The neutron flux = 3.51e+008 neutrons per square metre per sec \n",
+ "## The diffusion length = 0.38 metre\n",
+ "## Note: the difussion length is solved wrongly in the testbook\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The neutron flux = 3.51e+08 \n",
+ "The diffusion length = 0.38 metre \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg575"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa12.8 : : Page-575(2011)\n",
+ "#find diffusion length for thermal neutron\n",
+ "import math\n",
+ "N_0 = 6.023e+026; ## Avogadro's number, per mole\n",
+ "rho = 1.62e+03; ## Density, kg per cubic metre\n",
+ "sigma_a = 3.2e-31; ## Absorption cross section, square metre\n",
+ "sigma_s = 4.8e-28; ## Scattered cross section, square metre\n",
+ "A = 12.; ## Mass number\n",
+ "lambda_a = A/(N_0*rho*sigma_a); ## Absorption mean free path, metre\n",
+ "lambda_tr = A/(N_0*rho*sigma_s*(1.-2./(3.*A))); ## Transport mean free path, metre\n",
+ "L = math.sqrt(lambda_a*lambda_tr/3.); ## Diffusion length for thermal neutron\n",
+ "print'%s %.2f %s'%(\"\\nThe diffusion length for thermal neutron = \",L,\" metre \")\n",
+ "\n",
+ "## Result\n",
+ "## The diffusion length for thermal neutron = 0.590 metre \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The diffusion length for thermal neutron = 0.59 metre \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg575"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa12.9 : : Page-575 (2011)\n",
+ "#calculate graphite and neutron age and slowing down length and same as berylliums\n",
+ "import math\n",
+ "E_0 = 2e+06; ## Average energy of the neutron, electron volts\n",
+ "E = 0.025; ## Thermal energy of the neutron, electron volts\n",
+ "## For graphite\n",
+ "A = 12. ## Mass number\n",
+ "sigma_g = 33.5; ## The value of sigma for graphite\n",
+ "tau_0 = 1./(6.*sigma_g**2)*(A+2./3.)/(1.-2./(3.*A))*math.log(E_0/E); ## Age of neutron for graphite, Sq.m\n",
+ "L_f = math.sqrt(tau_0); ## Slowing down length of neutron through graphite, m\n",
+ "print'%s %.2f %s'%(\"\\nFor Graphite, A = \", A,\"\");\n",
+ "print'%s %.2f %s'%(\"\\nNeutron age = \",tau_0*1e+004,\" Sq.cm\");\n",
+ "print'%s %.2f %s'%(\"\\nSlowing down length =\",L_f,\" m\");\n",
+ "## For beryllium\n",
+ "A = 9. ## Mass number\n",
+ "sigma_b = 57.; ## The value of sigma for beryllium\n",
+ "tau_0 = 1/(6.*sigma_b**2)*(A+2./3.)/(1.-2./(3.*A))*math.log(E_0/E); ## Age of neutron for beryllium, Sq.m\n",
+ "L_f = math.sqrt(tau_0); ## Slowing down length of neutron through graphite, m\n",
+ "print'%s %.2f %s'%(\"\\n\\nFor Beryllium, A = \", A,\"\");\n",
+ "print'%s %.2f %s'%(\"\\nNeutron age = \",tau_0*1e+004,\" Sq.cm\");\n",
+ "print'%s %.2e %s'%(\"\\nSlowing down length = \",L_f,\" m\");\n",
+ "\n",
+ "## Result\n",
+ "## For Graphite, A = 12\n",
+ "## Neutron age = 362 Sq.cm\n",
+ "## Slowing down length = 0.190 m\n",
+ "\n",
+ "## For Beryllium, A = 9\n",
+ "## Neutron age = 97 Sq.cm\n",
+ "## Slowing down length = 9.9e-002 m "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "For Graphite, A = 12.00 \n",
+ "\n",
+ "Neutron age = 362.46 Sq.cm\n",
+ "\n",
+ "Slowing down length = 0.19 m\n",
+ "\n",
+ "\n",
+ "For Beryllium, A = 9.00 \n",
+ "\n",
+ "Neutron age = 97.46 Sq.cm\n",
+ "\n",
+ "Slowing down length = 9.87e-02 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg576"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa12.10 : : Page-576 (2011)\n",
+ "#find enegy of the neutrons\n",
+ "import math\n",
+ "theta = 3.5*math.pi/180.; ## Reflection angle, radian\n",
+ "d = 2.3e-10; ## Lattice spacing, metre\n",
+ "n = 1.; ## For first order\n",
+ "h = 6.6256e-34; ## Planck's constant, joule sec\n",
+ "m = 1.6748e-27; ## Mass of the neutron, Kg\n",
+ "E = n**2*h**2/(8.*m*d**2*math.sin(theta)**2*1.6023e-19); ## Energy of the neutrons, electron volts\n",
+ "print'%s %.2f %s'%(\"\\nThe energy of the neutrons = \",E,\" eV\");\n",
+ "\n",
+ "## Result\n",
+ "## The energy of the neutrons = 1.04 eV \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy of the neutrons = 1.04 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter13.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter13.ipynb
new file mode 100755
index 00000000..c63c61c5
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter13.ipynb
@@ -0,0 +1,430 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e4b6100001a32fad047d14064effe0ad1c4e730aa6c2b9ae5ba17ef44d36b6f5"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter13-Nuclear Fission and Fusion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.1 : : Page-600 (2011)\n",
+ "#calculate rate of fission and energy released\n",
+ "import math\n",
+ "E = 200.*1.6023e-13; ## Energy released per fission, joule\n",
+ "E_t = 2.; ## Total power produced, watt\n",
+ "R_fiss = E_t/E; ## Fission rate, fissions per sec\n",
+ "m = 0.5; ## Mass of uranium, Kg\n",
+ "M = 235.; ## Mass number of uranium\n",
+ "N_0 = 6.023e+26; ## Avogadro's number, per mole\n",
+ "N = m/M*N_0 ## Number of uranium nuclei\n",
+ "E_rel = N*E/4.08*10**-3; ## Energy released, kilocalories\n",
+ "print'%s %.2e %s %.2e %s'%(\"\\nThe rate of fission of U-235 = \",R_fiss,\" fissions per sec\"and\" \\nEnergy released =\",E_rel,\" kcal\");\n",
+ "\n",
+ "## Result\n",
+ "## The rate of fission of U-235 = 6.24e+010 fissions per sec \n",
+ "## Energy released = 1.006535e+010 kcal "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The rate of fission of U-235 = 6.24e+10 \n",
+ "Energy released = 1.01e+10 kcal\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.2 : : Page-600 (2011)\n",
+ "#calculate number of free neutrons in the reactor\n",
+ "import math\n",
+ "E = 200*1.6e-13; ## Energy released per fission, joules per neutron\n",
+ "t = 10**-3; ## Time, sec\n",
+ "P = E/t; ## Power produced by one free neutron, watt per neutron\n",
+ "P_l = 10**9; ## Power level, watt\n",
+ "N = P_l/P; ## Number of free neutrons in the reactor, neutrons\n",
+ "print'%s %.2e %s'%(\"\\nThe number of free neutrons in the reactor = \",N,\" neutrons\");\n",
+ "\n",
+ "## Result\n",
+ "## The number of free neutrons in the reactor = 3.125e+016 neutrons "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The number of free neutrons in the reactor = 3.12e+16 neutrons\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.3 : : Page-600 (2011)\n",
+ "#calculate average number of neutrons \n",
+ "import math\n",
+ "N_0_235 = 1.; ## Number of uranium 235 per 238 \n",
+ "N_0_238 = 20.; ## Number of uranium 238 for one uranium 235 \n",
+ "sigma_a_235 = 683.; ## Absorption cross section for uranium 235, barn\n",
+ "sigma_a_238 = 2.73; ## Absorption cross section for uranium 238, barn\n",
+ "sigma_f_235 = 583.; ## Fission cross section, barn\n",
+ "sigma_a = (N_0_235*sigma_a_235+N_0_238*sigma_a_238)/(N_0_235+N_0_238); ##Asorption cross sec, barn\n",
+ "sigma_f = N_0_235*sigma_f_235/(N_0_235+N_0_238); ## Fisssion cross section \n",
+ "v = 2.43;\n",
+ "eta = v*sigma_f/sigma_a; ## Average number of neutron released per absorption\n",
+ "print'%s %.2f %s'%(\"\\nThe average number of neutrons released per absorption = \", eta,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The average number of neutrons released per absorption = 1.921 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The average number of neutrons released per absorption = 1.92 \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg600"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.4 : : Page-600(2011)\n",
+ "#find excitaion energy for uranum 236 and 239 and 235\n",
+ "import math\n",
+ "a_v = 14.0; ## Volume binding energy constant, mega electron volts\n",
+ "a_s = 13.0; ## Surface binding energy constant, mega electron volts\n",
+ "a_c = 0.583; ## Coulomb constant, mega electron volts\n",
+ "a_a = 19.3; ## Asymmetric constant, mega electron volts\n",
+ "a_p = 33.5; ## Pairing energy constant, mega electron volts\n",
+ "Z = 92.; ## Atomic number \n",
+ "## For U-236\n",
+ "A = 235.; ## Mass number\n",
+ "E_exc_236 = a_v*(A+1-A)-a_s*((A+1)**(2./3.)-A**(2./3.))-a_c*(Z**2/(A+1.)**(1./3.)-Z**2/A**(1/3.))-a_a*((A+1-2*Z)**2/(A+1)-(A-2*Z)**2/A)+a_p*(A+1)**(-3./4.); ## Excitation energy for uranium 236, mega electron volts\n",
+ "## For U-239\n",
+ "A = 238.; ## Mass number\n",
+ "E_exc_239 = a_v*(A+1-A)-a_s*((A+1)**(2./3.)-A**(2./3.))-a_c*(Z**2/(A+1)**(1/3.)-Z**2/A**(1./3.))-a_a*((A+1.-2.*Z)**2/(A+1)-(A-2*Z)**2/A)+a_p*((A+1)**(-3/4.)-A**(-3/4.)); ## Excitation energy for uranium 239\n",
+ "## Now calculate the rate of spontaneous fissioning for U-235\n",
+ "N_0 = 6.02214e+23; ## Avogadro's constant, per mole\n",
+ "M = 235.; ## Mass number\n",
+ "t_half = 3e+17*3.15e+7; ## Half life, years \n",
+ "D = 0.693/t_half; ## Decay constant, per year\n",
+ "N = N_0/M; ## Mass of uranium 235, Kg\n",
+ "dN_dt = N*D*3600; ## Rate of spontaneous fissioning of uranium 235, per hour\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe excitation energy for uranium 236 =\",E_exc_236,\" MeV\"and \"\\nThe excitation energy for uranium 239 = \",E_exc_239,\" MeV\"and\"\\nThe rate of spontaneous fissioning of uranium 235 = \",dN_dt,\" per hour\");\n",
+ "\n",
+ "## Result\n",
+ "## The excitation energy for uranium 236 = 6.8 MeV\n",
+ "## The excitation energy for uranium 239 = 5.9 MeV\n",
+ "## The rate of spontaneous fissioning of uranium 235 = 0.68 per hour \n",
+ "\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The excitation energy for uranium 236 = 6.77 \n",
+ "The excitation energy for uranium 239 = 5.90 \n",
+ "The rate of spontaneous fissioning of uranium 235 = 0.68 per hour \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg601"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.5 : : Page-601 (2011)\n",
+ "#calculate total energy released during fusion\n",
+ "import math \n",
+ "a = 10**5; ## Area of the lake, square mile\n",
+ "d = 1/20.; ## Depth of the lake, mile\n",
+ "V = a*d*(1.6e+03)**3; ## Volume of the lake, cubic metre\n",
+ "rho = 10**3; ## Density of water, kg per cubic metre\n",
+ "M_water = V*rho; ## Total mass of water in the lake, Kg\n",
+ "N_0 = 6.02214e+26; ## Avogadro's constant, per mole\n",
+ "A = 18.; ## Milecular mass of water\n",
+ "N = M_water*N_0/A; ## Number of molecules of water, molecules\n",
+ "abund_det = 0.0156e-02; ## Abundance of deterium\n",
+ "N_d = N*2*abund_det; ## Number of deterium atoms\n",
+ "E_per_det = 43/6.; ## Energy released per deterium atom, mega electron volts\n",
+ "E_t = N_d*E_per_det; ## Total energy released during fusion, mega electron volt\n",
+ "print'%s %.2e %s'%(\"\\nThe total energy released during fusion = \",E_t,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## Total energy released during fusion = 1.53e+039 MeV\n",
+ "\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The total energy released during fusion = 1.53e+39 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg601"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.6 : : Page-601 (2011)\n",
+ "#calculate temperature attained by thermonuclear device\n",
+ "import math \n",
+ "r = 1/2.; ## Radius of the tube, metre\n",
+ "a = math.pi*r**2; ## Area of the torus, square metre\n",
+ "V = 3*math.pi*a; ## Volume of the torus, cubic metre\n",
+ "P = 10**-5*13.6e+3*9.81; ## Pressure of the gas, newton per square metre\n",
+ "C = 1200e-6; ## Capacitance, farad\n",
+ "v = 4e+4; ## potential, volts\n",
+ "T_room = 293; ## Room temperature, kelvin\n",
+ "N_k = P*V/T_room; ## From gas equation\n",
+ "E = 1/2.*C*v**2; ## Energy stored, joules\n",
+ "T_k = 1/6.*E/(N_k*10.); ## Temperature attained by thermonuclear device, kelvin\n",
+ "print'%s %.2e %s'%(\"\\nThe temperature attained by thermonuclear device =\",T_k,\" K\");\n",
+ "\n",
+ "## Result\n",
+ "## The temperature attained by thermonuclear device = 4.75e+005 K \n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The temperature attained by thermonuclear device = 4.75e+05 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg601"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.7 : : Page-601 (2011)\n",
+ "#calculate energy released by the sun temperautre of the sun\n",
+ "import math \n",
+ "G = 6.67e-11; ## Gravitational constant, newton square m per square kg\n",
+ "r = 7e+08; ## Radius of the sun, metre\n",
+ "M_0 = 2e+30; ## Mass of the sun, kg\n",
+ "E_rel = 3/5.*G*M_0**2/r; ## Energy released by the sun, joule\n",
+ "E_dia_shrink_10 = E_rel/9.; ## Energy released when sun diameter shrink by 10 percent, joule\n",
+ "R = 8.314; ## Universal gas constant, joule per kelvin per kelvin per mole\n",
+ "T = E_rel/(M_0*R); ## Temperature of the sun, kelvin\n",
+ "print'%s %.2e %s %.2e %s %.2e %s '%(\"\\nThe energy released by the sun = \",E_rel,\" joule\"and\" \\nThe energy released when sun diameter is shrinked by 10 percent = \",E_dia_shrink_10,\" joule\"and\" \\nThe temperature of the sun = \",T,\" kelvin \");\n",
+ "\n",
+ "## Result\n",
+ "## The energy released by the sun = 2.29e+041 joule \n",
+ "## The energy released when sun diameter is shrinked by 10 percent = 2.54e+040 joule \n",
+ "## The temperature of the sun = 1.38e+010 kelvin \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy released by the sun = 2.29e+41 \n",
+ "The energy released when sun diameter is shrinked by 10 percent = 2.54e+40 \n",
+ "The temperature of the sun = 1.38e+10 kelvin \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg602"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.8 : : Page-602 (2011)\n",
+ "# estimated Q-value\n",
+ "import math\n",
+ "A_0 = 240.; ## Mass number of parent nucleus\n",
+ "A_1 = 120.; ## Mass number of daughter nucleus\n",
+ "B_120 = 8.5; ## Binding energy of daughter nucleus\n",
+ "B_240 = 7.6; ## Binding energy of parent nucleus\n",
+ "Q = 2*A_1*B_120-A_0*B_240; ## Estimated Q-value, mega electron volts\n",
+ "print'%s %.2f %s'%(\"\\nThe estimated Q-value is = \",Q,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The estimated Q-value is = 216 MeV \n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The estimated Q-value is = 216.00 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg602"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa13.9 : : Page-602 (2011)\n",
+ "#find The asymmetric binding energy\n",
+ "import math\n",
+ "E = 31.7; ## Energy, MeV\n",
+ "a_a = 5/9.*2**(-2/3.)*E; ## Asymmetric binding energy term, mega electron volts\n",
+ "print'%s %.2f %s'%(\"\\nThe asymmetric binding energy term = \",a_a,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The asymmetric binding energy term = 11.1 MeV \n",
+ "\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The asymmetric binding energy term = 11.09 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter15.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter15.ipynb
new file mode 100755
index 00000000..10312aed
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter15.ipynb
@@ -0,0 +1,387 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:c0cb9f8947e3855e41c033660942a2e70a38f8ec8f1601ec17db209e7514083a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter15-Nuclear fission and fusion"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg652"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa15.1 : : Page-652 (2011)\n",
+ "#find Total leakage factor\n",
+ "import math\n",
+ "N_0_235 = 1.; ## Number of uranium atom\n",
+ "N_0_c = 10**5; ## Number of graphite atoms per uranium atom\n",
+ "sigma_a_235 = 698.; ## Absorption cross section for uranium, barns\n",
+ "sigma_a_c = 0.003; ## Absorption cross section for graphite, barns\n",
+ "f = N_0_235*sigma_a_235/(N_0_235*sigma_a_235+N_0_c*sigma_a_c ); ## Thermal utilization factor\n",
+ "eta = 2.08; ## Number of fast fission neutron produced\n",
+ "k_inf = eta*f; ## Multiplication factor\n",
+ "L_m = 0.54; ## Material length, metre\n",
+ "L_sqr = ((L_m)**2.*(1.-f)); ## diffusion length, metre\n",
+ "tau = 0.0364; ## Age of the neutron\n",
+ "B_sqr = 3.27; ## Geometrical buckling\n",
+ "k_eff = round (k_inf*math.exp(-tau*B_sqr)/(1+L_sqr*B_sqr)); ## Effective multiplication factor\n",
+ "N_lf = k_eff/k_inf; ## Non leakage factor\n",
+ "lf = (1-N_lf)*100.; ## Leakage factor, percent\n",
+ "print'%s %.2f %s'%(\"\\n Total leakage factor = \",lf,\" percent\")\n",
+ "\n",
+ "## Result\n",
+ "## Total leakage factor = 31.3 percent "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Total leakage factor = 31.26 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg652"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa15.2 : : Page-652 (2011)\n",
+ "#find Neutron multiplication factor\n",
+ "import math\n",
+ "N_m = 50.; ## Number of molecules of heavy water per uranium molecule\n",
+ "N_u = 1.; ## Number of uranium molecules \n",
+ "sigma_a_u = 7.68; ## Absorption cross section for uranium, barns\n",
+ "sigma_s_u = 8.3; ## Scattered cross section for uranium, barns\n",
+ "sigma_a_D = 0.00092; ## Absorption cross section for heavy water, barns\n",
+ "sigma_s_D = 10.6; ## Scattered cross section for uranium, barns \n",
+ "f = N_u*sigma_a_u/(N_u*sigma_a_u+N_m*sigma_a_D ); ## Thermal utilization factor\n",
+ "zeta = 0.570; ## Average number of collisions\n",
+ "N_0 = N_u*139./140.; ## Number of U-238 atoms per unit volume \n",
+ "sigma_s = N_m/N_0*sigma_s_D; ## Scattered cross section, barns\n",
+ "sigma_a_eff = 3.85*(sigma_s/N_0)**0.415; ## Effective absorption cross section, barns\n",
+ "p = math.exp(-sigma_a_eff/sigma_s); ## Resonance escape probablity\n",
+ "eps = 1.; ## Fast fission factor\n",
+ "eta = 1.34; ## Number of fast fission neutron produced\n",
+ "k_inf = eps*eta*p*f; ## Effective multiplication factor\n",
+ "print'%s %.2f %s'%(\"\\nNeutron multiplication factor = \", k_inf,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## Neutron multiplication factor = 1.2 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Neutron multiplication factor = 1.21 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg652"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa15.3 : : Page-652 (2011)\n",
+ "#find The required multiplication factor\n",
+ "import math\n",
+ "## For graphite\n",
+ "sigma_a_g = 0.0032; ## Absorption cross section for graphite, barns\n",
+ "sigma_s_g = 4.8; ## Scattered cross section for graphite, barns\n",
+ "zeta = 0.158; ## Average number of collisions\n",
+ "N_m = 50.; ## Number of molecules of graphite per uranium molecule\n",
+ "## For uranium\n",
+ "sigma_f = 590.; ## Fissioning cross section, barns\n",
+ "sigma_a_u = 698.; ## Absorption cross section for U-235, barns\n",
+ "sigma_a_238 = 2.75; ## Absorption cross section for U-238, barns\n",
+ "v = 2.46; ## Number of fast neutrons emitted\n",
+ "N_u = 1 ## Number of uranium atoms \n",
+ "f = N_u*sigma_a_u/(N_u*sigma_a_u+N_m*sigma_a_g ); ## Thermal utilization factor\n",
+ "N_0 = N_u*(75./76.); ## Number of U-238 atoms per unit volume\n",
+ "sigma_s = N_m*76./75.*sigma_s_g/N_u; ## Scattered cross section, barns\n",
+ "sigma_eff = 3.85*(sigma_s/N_0)**0.415; ## Effective cross section, barns\n",
+ "p = math.exp(-sigma_eff/sigma_s); ## Resonance escape probability, barns\n",
+ "eps = 1.; ## Fast fission factor\n",
+ "eta = 1.34; ## Number of fast fission neutron produced\n",
+ "k_inf = eps*eta*p*f; ## Multiplication factor\n",
+ "print'%s %.2f %s'%(\"\\nThe required multiplication factor = \", k_inf,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The required multiplication factor = 1.1 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The required multiplication factor = 1.15 \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg653"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa15.4 : : Page-653 (2011)\n",
+ "#find The ratio of number of uranium atoms to graphite atoms \n",
+ "import math\n",
+ "eta = 2.07; ## Number of fast fission neutron produced\n",
+ "x = 1./(eta-1.); \n",
+ "sigma_a_u = 687.; ## Absorption cross section for uranium, barns\n",
+ "sigma_a_g = 0.0045; ## Absorption cross section for graphite, barns\n",
+ "N_ratio = x*sigma_a_g/sigma_a_u; ## Ratio of number of uranium atoms to graphite atoms\n",
+ "print'%s %.2e %s'%(\"\\nThe ratio of number of uranium atoms to graphite atoms = \", N_ratio,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The ratio of number of uranium atoms to graphite atoms = 6.12e-006 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The ratio of number of uranium atoms to graphite atoms = 6.12e-06 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg653"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa15.5 : : Page-653 (2011)\n",
+ "import math \n",
+ "#find The multiplication factor for LOPO reactor\n",
+ "f = 0.754; ## Thermal utilization factor\n",
+ "sigma_s_o = 4.2; ## Scattered cross section for oxygen, barns\n",
+ "sigma_s_H = 20.; ## Scattered cross section for hydrogen, barns\n",
+ "N_O = 879.25; ## Number of oxygen atoms\n",
+ "N_238 = 14.19; ## Number of uranium atoms\n",
+ "N_H = 1573.; ## Number of hydrogen atoms\n",
+ "sigma_s = N_O/N_238*sigma_s_o+N_H/N_238*sigma_s_H; ## Scattered cross section, barns\n",
+ "N_0 = 14.19; ## Number of U-238 per unit volume\n",
+ "zeta_o = 0.120; ## Number of collision for oxygen\n",
+ "zeta_H = 1.; ## Number of collision for hydrogen\n",
+ "sigma_eff = (N_0/(zeta_o*sigma_s_o*N_O+zeta_H*sigma_s_H*N_H )); ## Effective cross section, barns\n",
+ "p = math.exp(-sigma_eff/sigma_s); ## Resonance escape probablity\n",
+ "eta = 2.08; ## Number of fission neutron produced.\n",
+ "eps = 1; ## Fission factor\n",
+ "K_inf = eps*eta*p*f; ## Multiplication factor\n",
+ "print'%s %.2f %s'%(\"\\nThe multiplication factor for LOPO reactor = \", K_inf,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The multiplication factor for LOPO reactor = 1.6 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The multiplication factor for LOPO reactor = 1.57 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg654"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa15.6 : : Page-654 (2011)\n",
+ "#find The required controlled cross section\n",
+ "import math\n",
+ "r = 35; ## Radius of the reactor, centi metre\n",
+ "B_sqr = (math.pi/r)**2; ## Geometrical buckling, per square centi metre\n",
+ "D = 0.220; ## Diffusion coefficient, centi metre\n",
+ "sigma_a_f = 0.057; ## Rate of absorption of thermal neutrons\n",
+ "v = 2.5; ## Number of fast neutrons emitted\n",
+ "tau = 50.; ## Age of the neutron\n",
+ "sigma_f = 0.048; ## Rate of fission\n",
+ "sigma_a_c = -1/(1+tau*B_sqr)*(-v*sigma_f+sigma_a_f+B_sqr*D+tau*B_sqr*sigma_a_f); ## Controlled cross section\n",
+ "print'%s %.2f %s'%(\"\\nThe required controlled cross section = \", sigma_a_c,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The required controlled cross section = 0.0273 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The required controlled cross section = 0.03 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg655"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa15.7 : : Page-655 (2011)\n",
+ "#find side of the cubical reactor nd critical radius of the reactor\n",
+ "import math\n",
+ "B_sqr = 65.; ## Geometrical buckling\n",
+ "a = math.sqrt(3*math.pi**2/B_sqr)*100.; ## Side of the cubical reactor, centi metre\n",
+ "R = round(math.pi/math.sqrt(B_sqr)*100.); ## Radius of the cubical reactor,centi metre\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe side of the cubical reactor =\",a,\" cm\"and\" \\nThe critical radius of the reactor =\",R,\" cm\");\n",
+ "\n",
+ "## Result\n",
+ "## The side of the cubical reactor = 67.5 cm\n",
+ "## The critical radius of the reactor = 39 cm "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The side of the cubical reactor = 67.49 \n",
+ "The critical radius of the reactor = 39.00 cm \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg655"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa15.8 : : Page-655 (2011)\n",
+ "#find The critical volume of the reactor\n",
+ "import math\n",
+ "sigma_a_u = 698.; ## Absorption cross section for uranium, barns\n",
+ "sigma_a_M = 0.00092; ## Absorption cross section for heavy water, barns\n",
+ "N_m = 10**5; ## Number of atoms of heavy water\n",
+ "N_u = 1.; ## Number of atoms of uranium\n",
+ "f = sigma_a_u/(sigma_a_u+sigma_a_M*N_m/N_u); ## Thermal utilization factor\n",
+ "eta = 2.08; ## Number of fast fission neutron produced\n",
+ "k_inf = eta*f; ## Multiplication factor\n",
+ "L_m_sqr = 1.70; ## Material length, metre\n",
+ "L_sqr = L_m_sqr*(1-f); ## Diffusion length, metre\n",
+ "B_sqr = 1.819/0.30381*math.exp(-1/12.)-1./0.3038; ## Geometrical buckling, per square metre\n",
+ "V_c = 120./(B_sqr*math.sqrt(B_sqr)); ## Volume of the reactor, cubic metre\n",
+ "print'%s %.2f %s'%(\"\\nThe critical volume of the reactor = \",V_c,\" cubic metre\");\n",
+ "\n",
+ "## Result\n",
+ "## The critical volume of the reactor = 36.4 cubic metre "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The critical volume of the reactor = 36.35 cubic metre\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter16.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter16.ipynb
new file mode 100755
index 00000000..b216f492
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter16.ipynb
@@ -0,0 +1,135 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f65433b5fc699d555a8e1a0c197fcc95a6db55a338dffc279c90983fdf767b28"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapte16-Chemical &Biological Effects of Radiations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg672"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa16.1 : : Page-672 (2011)\n",
+ "#find 25 mrad\n",
+ "import math\n",
+ "R_d = 25.; ## Radiation dose, milli rad\n",
+ "R_c_gy = 25e-03; ## Dose in centigray\n",
+ "R_Sv = 25*10**-2 ## Dose in milli sieverts\n",
+ "print'%s %.2f %s %.2f %s'%(\"\\n25 mrad = \",R_c_gy,\"\" and \" cGy =\",R_Sv,\" mSv\");\n",
+ "\n",
+ "## Results\n",
+ "## 25 mrad = 3e-002 cGy = 0.25 mSv "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "25 mrad = 0.03 0.25 mSv\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa16.2 : : Page-673 (2011)\n",
+ "#find 100 mega becquerel\n",
+ "import math\n",
+ "BC_conv = 100*1e+009/3.7e+10; ## Becquerel curie conversion, milli curie\n",
+ "print'%s %.2f %s'%(\"\\n100 mega becquerel = \",BC_conv,\" milli curie \")\n",
+ "\n",
+ "## Results\n",
+ "## 100 mega becquerel = 2.7 milli curie "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "100 mega becquerel = 2.70 milli curie \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg673"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa16.4 : : Page-673 (2011)\n",
+ "#find The requiresd liver dose \n",
+ "import math\n",
+ "A = 80*10**6; ## Activity, becquerel\n",
+ "t_half = 6*3600.; ## Half life, s\n",
+ "N = A*t_half/0.693; ## Number of surviving radionuclei\n",
+ "E_released = 0.9*N*(140e+03)*1.6e-19; ## Energy released, joule\n",
+ "m_l = 1.8; ## Mass of liver of average man, Kg\n",
+ "liv_dose = E_released*10**2/m_l; ## Liver dose, centigray\n",
+ "print'%s %.2f %s'%(\"\\nThe requiresd liver dose = \",liv_dose,\" cGy\");\n",
+ "\n",
+ "## Result\n",
+ "## The requiresd liver dose = 2.8 cGy \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The requiresd liver dose = 2.79 cGy\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter18.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter18.ipynb
new file mode 100755
index 00000000..40af5501
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter18.ipynb
@@ -0,0 +1,463 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:f0f3883962d3ff2205f0aa7b119a41b1115305734d5340cd468bd7addc3f1d3f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter18-Elementry Particles"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg770"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa18.1 : : Page-770 (2011)\n",
+ "#find The root mean square radius of charge distribution\n",
+ "import math \n",
+ "m_sqr = 0.71; ## For proton, (GeV/c-square)^2\n",
+ "R_rms = math.sqrt(12.)/(math.sqrt(m_sqr)*5.1); ## Root mean square radius, femto metre\n",
+ "print'%s %.2f %s'%(\"\\nThe root mean square radius of charge distribution: \",R_rms,\" fermi\");\n",
+ "\n",
+ "## Result\n",
+ "## The root mean square radius of charge distribution: 0.81 fermi "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The root mean square radius of charge distribution: 0.81 fermi\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg763"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Ex18.3 : : Page-763 (2011)\n",
+ "#find all reactions\n",
+ "import math\n",
+ "import numpy\n",
+ "p = numpy.zeros((1,2)); ## proton\n",
+ "pi_minus = numpy.zeros((1,2)); ##pi minus meson\n",
+ "pi_plus = numpy.zeros((1,2)); ## pi plus meson\n",
+ "n = numpy.zeros((1,2)); ## neutron\n",
+ "lamda_0 = numpy.zeros((1,2)); ## lamda hyperon\n",
+ "K_0 = numpy.zeros((1,2)); ## K zero (Kaons)\n",
+ "K_plus =numpy.zeros((1,2)); ## K plus (Kaons)\n",
+ "sigma_plus = numpy.zeros((1,2)); ## hyperon \n",
+ "sigma_minus = numpy.zeros((1,2)) ## hyperon\n",
+ "ksi_minus = numpy.zeros((1,2)); ## hyperon\n",
+ "## Allocate the value of Isospins (T and T3)\n",
+ "p[0,0] = 1/2;\n",
+ "p[0,1] = 1/2;\n",
+ "pi_minus[0,0] = 1;\n",
+ "pi_minus[0,1] = -1;\n",
+ "pi_plus[0,0] = 1;\n",
+ "pi_plus[0,1] = +1;\n",
+ "n[0,0] = 1/2;\n",
+ "n[0,1] = -1/2;\n",
+ "lambda_0=numpy.zeros((1,2));\n",
+ "lambda_0[0,0] = 0;\n",
+ "lambda_0[0,1] = 0;\n",
+ "K_0[0,0] = pi_minus[0,0]+p[0,0];\n",
+ "K_0[0,1] = pi_minus[0,1]+p[0,1] ;\n",
+ "K_plus[0,0] = p[0,0]+p[0,0]-lambda_0[0,0]-p[0,0];\n",
+ "K_plus[0,1] = p[0,1]+p[0,1]-lambda_0[0,1]-p[0,1] ;\n",
+ "sigma_plus[0,0] = pi_plus[0,0]+p[0,0]-K_plus[0,0];\n",
+ "sigma_plus[0,1] = pi_plus[0,1]+p[0,1]-K_plus[0,1];\n",
+ "sigma_minus[0,0] = pi_minus[0,0]+p[0,0]-K_plus[0,0];\n",
+ "sigma_minus[0,1] = pi_minus[0,1]+p[0,1]-K_plus[0,1];\n",
+ "ksi_minus[0,0] = pi_plus[0,0]+n[0,0]-K_plus[0,0]-K_plus[0,0];\n",
+ "ksi_minus[0,1] = pi_plus[0,1]+n[0,1]-K_plus[0,1]-K_plus[0,1];\n",
+ "print'%s'%(\"\\n Reaction I \\n pi_minus + p ......> lambda_0 + K_0\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T for K_0 is : %3.1f \",K_0[0,0],\"\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T3 for K_0 is : %3.1f \",K_0[0,1],\"\");\n",
+ "print(\"\\n Reaction II \\n pi_plus + p -> lambda_0 + K_plus\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T for K_plus is : %3.1f \",K_plus[0,0],\"\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T3 for K_plus is : %3.1f \",K_plus[0,1],\"\");\n",
+ "print(\"\\n Reaction III \\n pi_plus + n -> lambda_0 + K_plus\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T for K_plus is : %3.1f \",K_plus[0,0],\"\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T3 for K_plus is : %3.1f \",K_plus[0,1],\"\");\n",
+ "print(\"\\n Reaction VI \\n pi_minus + p -> sigma_minus + K_plus\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T for sigma_minus is : %3.1f \",sigma_minus[0,0],\"\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T3 for sigma_minus is : %3.1f \",sigma_minus[0,1],\"\");\n",
+ "print(\"\\n Reaction V \\n pi_plus + p -> sigma_plus + K_plus\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T for sigma_plus is : %3.1f \",sigma_plus[0,0],\"\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T3 for sigma_plus is : %3.1f \",sigma_plus[0,1],\"\");\n",
+ "print(\"\\n Reaction VI \\n pi_plus + n -> ksi_minus + K_plus + K_plus\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T for Ksi_minus is : %3.1f \",ksi_minus[0,0],\"\");\n",
+ "print'%s %.2f %s'%(\"\\n The value of T3 for Ksi_minus is : %3.1f \",ksi_minus[0,1],\"\");\n",
+ "\n",
+ "## Result\n",
+ "## \n",
+ "## Reaction I \n",
+ "## pi_minus + p -> lambda_0 + K_0\n",
+ "## The value of T for K_0 is : 1.5 \n",
+ "## The value of T3 for K_0 is : -0.5 \n",
+ "## Reaction II \n",
+ "## pi_plus + p -> lambda_0 + K_plus\n",
+ "## The value of T for K_plus is : 0.5 \n",
+ "## The value of T3 for K_plus is : 0.5 \n",
+ "## Reaction III \n",
+ "## pi_plus + n -> lambda_0 + K_plus\n",
+ "## The value of T for K_plus is : 0.5 \n",
+ "## The value of T3 for K_plus is : 0.5 \n",
+ "## Reaction VI \n",
+ " ## pi_minus + p -> sigma_minus + K_plus\n",
+ "## The value of T for sigma_minus is : 1.0 \n",
+ "## The value of T3 for sigma_minus is : -1.0 \n",
+ "## Reaction V \n",
+ "## pi_plus + p -> sigma_plus + K_plus\n",
+ "## The value of T for sigma_plus is : 1.0 \n",
+ "## The value of T3 for sigma_plus is : 1.0 \n",
+ "## Reaction VI \n",
+ " ## pi_plus + n -> ksi_minus + K_plus + K_plus\n",
+ "## The value of T for Ksi_minus is : 0.5 \n",
+ "## The value of T3 for Ksi_minus is : -0.5 \n",
+ "print(\"ans is slighlty different from book due to rounding off error\")\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " Reaction I \n",
+ " pi_minus + p ......> lambda_0 + K_0\n",
+ "\n",
+ " The value of T for K_0 is : %3.1f 1.00 \n",
+ "\n",
+ " The value of T3 for K_0 is : %3.1f -1.00 \n",
+ "\n",
+ " Reaction II \n",
+ " pi_plus + p -> lambda_0 + K_plus\n",
+ "\n",
+ " The value of T for K_plus is : %3.1f 0.00 \n",
+ "\n",
+ " The value of T3 for K_plus is : %3.1f 0.00 \n",
+ "\n",
+ " Reaction III \n",
+ " pi_plus + n -> lambda_0 + K_plus\n",
+ "\n",
+ " The value of T for K_plus is : %3.1f 0.00 \n",
+ "\n",
+ " The value of T3 for K_plus is : %3.1f 0.00 \n",
+ "\n",
+ " Reaction VI \n",
+ " pi_minus + p -> sigma_minus + K_plus\n",
+ "\n",
+ " The value of T for sigma_minus is : %3.1f 1.00 \n",
+ "\n",
+ " The value of T3 for sigma_minus is : %3.1f -1.00 \n",
+ "\n",
+ " Reaction V \n",
+ " pi_plus + p -> sigma_plus + K_plus\n",
+ "\n",
+ " The value of T for sigma_plus is : %3.1f 1.00 \n",
+ "\n",
+ " The value of T3 for sigma_plus is : %3.1f 1.00 \n",
+ "\n",
+ " Reaction VI \n",
+ " pi_plus + n -> ksi_minus + K_plus + K_plus\n",
+ "\n",
+ " The value of T for Ksi_minus is : %3.1f 1.00 \n",
+ "\n",
+ " The value of T3 for Ksi_minus is : %3.1f 0.00 \n",
+ "ans is slighlty different from book due to rounding off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg766"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Ex18.9 : : Page-766 (2011)\n",
+ "import math\n",
+ "#find The lifetime of sigma particle and The isospin of sigma particle \n",
+ "h_cross = 6.62e-022; ## Redueced planck's constant, MeV sec\n",
+ "p_width = 0.88*35; ## Partial width of the decay, MeV \n",
+ "tau = h_cross/p_width; ## Life time of sigma, sec \n",
+ "T_pi = 1.; ## Isospin of pi plus particle \n",
+ "T_lambda = 0.; ## Isospin of lambda zero particle \n",
+ "T_sigma = T_pi+T_lambda; ## Isospin of sigma particle \n",
+ "print'%s %.2e %s'%(\"\\nThe lifetime of sigma particle = \",tau,\" s\")\n",
+ "print(\"The reaction is strong\")\n",
+ "print\"%s %.2f %s\"%(\"The isospin of sigma particle is : \",T_sigma,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The lifetime of sigma particle = 2.15e-023 s\n",
+ "## The reaction is strong\n",
+ "## The isospin of sigma particle is : 1 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The lifetime of sigma particle = 2.15e-23 s\n",
+ "The reaction is strong\n",
+ "The isospin of sigma particle is : 1.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg767"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa18.10 : : Page-767 (2011)\n",
+ "#find The mean life for tau plus\n",
+ "import math\n",
+ "m_mew = 106.; ## Mass of mew lepton, mega electron volts per square c\n",
+ "m_tau = 1784.; ## Mass of tau lepton, mega electron volts per square c\n",
+ "tau_mew = 2.2e-06; ## Mean life of mew lepton, sec\n",
+ "R = 16/100.; ## Branching factor\n",
+ "tau_plus = R*(m_mew/m_tau)**5*tau_mew; ## Mean life for tau plus, sec\n",
+ "print'%s %.2e %s'%(\"\\nThe mean life for tau plus : \",tau_plus,\" sec\");\n",
+ "\n",
+ "## Result\n",
+ "## The mean life for tau plus : 2.6e-013 sec "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The mean life for tau plus : 2.61e-13 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg768"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa18.13 : : Page-768(2011)\n",
+ "#find The possible charge states \n",
+ "def symbol(val):\n",
+ " global s\n",
+ " if val == 2 :\n",
+ " s = '++';\n",
+ " elif val == 1:\n",
+ " s = '+';\n",
+ " elif val == 0:\n",
+ " s = '0';\n",
+ " elif val == -1:\n",
+ " s = '-';\n",
+ " return s\n",
+ "\n",
+ "B = 1; # Baryon number\n",
+ "S1 = 0; # Strangeness quantum number\n",
+ "Q = numpy.zeros((1,4)) # Charge\n",
+ "I3 = 3/2.; \n",
+ "print (\"\\nThe possible charge states are\");\n",
+ "for i in range(0,4): \n",
+ " Q = I3+(B+S1)/2.;\n",
+ " symb = symbol(Q);\n",
+ " print symb\n",
+ " I3 = I3 - 1;\n",
+ "\n",
+ "print (\" respectively\");\n",
+ "\n",
+ "# Result\n",
+ "# The possible charge states are ++ + 0 - respectively "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The possible charge states are\n",
+ "++\n",
+ "+\n",
+ "0\n",
+ "-\n",
+ " respectively\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg768"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa18.15 : : Page-768 (2011)\n",
+ "#find The branching ratio for a resonanc\n",
+ "import math\n",
+ "I_1 = 3/2.; ## Isospin for delta(1232)\n",
+ "I_2 = 1/2.; ## Isospin for delta 0\n",
+ "delta_ratio = math.sqrt((2./3.)**2)/math.sqrt((1./3.)**2); ## Branching ratio\n",
+ "print'%s %.2f %s'%(\"\\nThe branching ratio for a resonance with I = 1/2 is \", delta_ratio,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The branching ratio for a resonance with I = 1/2 is 2 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The branching ratio for a resonance with I = 1/2 is 2.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg768"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa18.16 : : Page-768 (2011)\n",
+ "#find The cross section ratio\n",
+ "import math\n",
+ "phi = 45*math.pi/180; ## Phase difference\n",
+ "Cross_sec_ratio = 1/4.*(5.+4.*math.cos(phi))/(1-math.cos(phi)); ## Cross section ratio\n",
+ "print'%s %.2f %s'%(\"\\nThe cross section ratio : \", Cross_sec_ratio,\"\");\n",
+ "\n",
+ "## Result\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The cross section ratio : 6.68 \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex18-pg770"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa18.18 : : Page-770 (2011)\n",
+ "#find The root mean square radius of charge distribution\n",
+ "import math \n",
+ "m_sqr = 0.71; ## For proton, (GeV/c-square)^2\n",
+ "R_rms = math.sqrt(12.)/(math.sqrt(m_sqr)*5.1); ## Root mean square radius, femto metre\n",
+ "print'%s %.2f %s'%(\"\\nThe root mean square radius of charge distribution: \",R_rms,\" fermi\");\n",
+ "\n",
+ "## Result\n",
+ "## The root mean square radius of charge distribution: 0.81 fermi math"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The root mean square radius of charge distribution: 0.81 fermi\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter2.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter2.ipynb
new file mode 100755
index 00000000..821531df
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter2.ipynb
@@ -0,0 +1,758 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:41d72ff7de4eab79477527ee635fed63748fe3041e37a18b80daacf784f19007"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter2-Radioactivity and Isotopes"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Exa2.1: : Page-88 (2011) \n",
+ "#find Weight of one Curie of RaB and Weight of one Rutherford of RaB\n",
+ "T = 26.8*60; ## Half life of the substance, s\n",
+ "C = 3.7e+010; ## One curie, disintegration per sec\n",
+ "N = 6.022137e+026; ## Avogadro number, per kmol\n",
+ "m = 214.; ## Molecular weight of RaB, kg/kmol\n",
+ "R = 1e+006; ## One Rutherford, disintegration per sec.\n",
+ "W_C = C*T*m/(N*0.693); ## Weight of one Curie of RaB, Kg \n",
+ "W_R = R*T*m/(N*0.693); ## Weight of one Rutherford of RaB, Kg \n",
+ "print\"%s %.2e %s %.2e %s \"%(\"\\nWeight of one Curie of RaB : \",W_C,\" Kg\"and \" \\nWeight of one Rutherford of RaB : \",W_R,\" Kg\");\n",
+ "\n",
+ "## Result\n",
+ "## Weight of one Curie of RaB : 3.051e-011 Kg \n",
+ "## Weight of one Rutherford of RaB : 8.245e-016 Kg "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Weight of one Curie of RaB : 3.05e-11 \n",
+ "Weight of one Rutherford of RaB : 8.25e-16 Kg \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg88"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.2 : : Page 88 (2011)\n",
+ "#find The maximum activity of Na-24 and The time needed to produced 90 percent of the maximum activity\n",
+ "import math\n",
+ "T_h = 14.8; ## Half life of Na-24, hours\n",
+ "Q = 1e+008; ## Production rate of Na-24, per sec\n",
+ "L = 0.693/T_h; ## Decay constant, per sec\n",
+ "t = 2.; ## Time after the bombardment, hours\n",
+ "A = Q/3.7e+010*1000; ## The maximum activity of Na-24, mCi\n",
+ "T = -1*math.log(0.1)/L; ## The time needed to produced 90% of the maximum activity, hour\n",
+ "N = 0.9*Q*3600./L*math.e**(-L*t); ## Number of atoms of Na-24 left two hours after bombardment was stopped\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe maximum activity of Na-24 = \",A,\" mCi\"and \"\\nThe time needed to produced 90 percent of the maximum activity =\",T,\" hrs\" and\"\\nNumber of atoms of Na-24 left two hours after bombardment was stopped = \",N,\"\")\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## The maximum activity of Na-24 = 2.7 mCi\n",
+ "## The time needed to produced 90 percent of the maximum activity = 49.2 hrs \n",
+ "## Number of atoms of Na-24 left two hours after bombardment was stopped = 6.30e+012 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The maximum activity of Na-24 = 2.70 \n",
+ "The time needed to produced 90 percent of the maximum activity = 49.17 \n",
+ "Number of atoms of Na-24 left two hours after bombardment was stopped = 6300897280447.79 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.3: : Page 89 (2011)\n",
+ "#find The activity of K-40 and disintegrations\n",
+ "T = 1.31e+09*365*24*60*60; ## Half life of the substance,sec\n",
+ "N = 6.022137e+026; ## Avogadro number.\n",
+ "m = 0.35*0.012*10**-2; ## Mass of K-40, Kg.\n",
+ "A = m*N*0.693/(T*40); ## Activity of K-40, disintegrations/sec. \n",
+ "print'%s %.2e %s %.3f %s '%(\"\\nThe activity of K-40 = \",A,\" \"and \"disintegrations/sec = \",A/3.7e+004,\" micro-curie\");\n",
+ "\n",
+ "## Result\n",
+ "## The activity of K-40 = 1.061e+004 disintegrations/sec = 0.287 micro-curie "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The activity of K-40 = 1.06e+04 disintegrations/sec = 0.287 micro-curie \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.4 : : Page 89 (2011)\n",
+ "#find The age of the boat\n",
+ "import math\n",
+ "T = 5568; ## Half life of the C-14,years\n",
+ "D = 0.693/T; ## Disintegration constant, years^-1.\n",
+ "N_0 = 15.6/D; ## Activity of fresh carbon, dpm .gm\n",
+ "N = 3.9/D; ## Activity of an ancient wooden boat,dpm.gm.\n",
+ "t = 1/(D)*math.log(N_0/N); ## Age of the boat, years\n",
+ "print'%s %.2e %s'%(\"\\nThe age of the boat : \",t,\" years\")\n",
+ "\n",
+ "## Result\n",
+ "## The age of the boat : 1.114e+004 years"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The age of the boat : 1.11e+04 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.5 : : Page 90 (2011)\n",
+ "#find The activity of U-234 \n",
+ "import math\n",
+ "m_0 = 3e-06;## Initial mass of the U-234, Kg\n",
+ "A = 6.022137e+026; ##Avagadro's number, atoms\n",
+ "N_0 = m_0*A/234.; ## Initial number of atoms\n",
+ "T = 2.50e+05; ## Half life, years\n",
+ "D= 0.693/T; ## Disintegration constant\n",
+ "t = 150000; ## Disintegration time, years\n",
+ "m = m_0*math.e**(-D*t); ## Mass after time t,Kg\n",
+ "activity = m*D/(365.*24.*60.*60.)*A/234.; ## Activity of U-234 after time t,dps\n",
+ "print'%s %.1f %s %.2f %s'%(\"\\nThe activity of U-234 after \",t,\"\"and\" yrs = \",activity,\" disintegrations/sec\");\n",
+ "\n",
+ "## Result\n",
+ "## The activity of U-234 after 150000 yrs = 4.478e+005 disintegrations/sec"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The activity of U-234 after 150000.0 447778.26 disintegrations/sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.6 : : Page 90 (2011)\n",
+ "#find The number of alpha decays in Th-232\n",
+ "import math\n",
+ "A = 6.022137e+023; ##Avagadro's number, atoms\n",
+ "N_0 = A/232.; ## Initial number of atoms\n",
+ "t = 3.150e+07; ## Decay time, sec\n",
+ "D = 1.58e-018; ## Disintegration constant,sec^-1\n",
+ "N = D*t*N_0; ## Number of alpha decays in Th-232\n",
+ "print'%s %.2e %s'%(\"\\nThe number of alpha decays in Th-232 = \", N,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The number of alpha decays in Th-232 = 1.29e+011"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The number of alpha decays in Th-232 = 1.29e+11 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg90"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.7 : : Page 90 (2011)\n",
+ "#find The maximum possible age of the earth crust\n",
+ "import math\n",
+ "T_238 = 4.5e+09;## Half life of U-238, years\n",
+ "T_235 = 7.13e+08; ## Half life of U-238, years\n",
+ "lambda_238 = 0.693/T_238; ## Disintegration constant of U-238, years^-1\n",
+ "lambda_235 = 0.693/T_235; ## Disintegration constant of U-235, years^-1 \n",
+ "N = 137.8; ## Abundances of U-238/U-235\n",
+ "t = math.log(N)/(lambda_235 - lambda_238);## Age of the earth's crust, years\n",
+ "print'%s %.2e %s'%(\"\\nThe maximum possible age of the earth crust = \",t,\" years\");\n",
+ "\n",
+ "## Result \n",
+ "## The maximum possible age of the earth crust = 6.022e+009 years "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The maximum possible age of the earth crust = 6.02e+09 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.8 : : Page 91 (2011)\n",
+ "#find The number of half lives in radon-222\n",
+ "import math\n",
+ "N = 10; ## Number of atoms left undecayed in Rn-222\n",
+ "n = math.log(10)/math.log(2); ## Number of half lives in Ra-222\n",
+ "print'%s %.2f %s'%(\"\\nThe number of half lives in radon-222 = \", n,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The number of half lives in radon-222 = 3.322 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The number of half lives in radon-222 = 3.32 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.9 : : Page 91 (2011)\n",
+ "#find The weight of Po-210 after one year and The initial activity of the material\n",
+ "import math\n",
+ "M_Po = 209.9829; ## Mass of Polonium, g\n",
+ "M_Pb = 205.9745; ## Mass of lead, g\n",
+ "A = 6.22137e+023; ## Avogadro's number\n",
+ "M_He = 4.0026; ## Mass of alpha particle, g\n",
+ "C = 3e+08; ## Velocity of light, m/s\n",
+ "T = 138*24*3600; ## Half life, sec\n",
+ "P = 250; ## Power produced, joule/sec\n",
+ "Q = (M_Po-M_Pb-M_He)*931.25; ## disintegration energy, MeV\n",
+ "D = 0.693/T; ## Disintegration constant, per year\n",
+ "N = P/(D*Q*1.60218e-013); ## Number of atoms, atom\n",
+ "N_0 = N*math.e**(1.833); ## Number of atoms present initially, atom\n",
+ "W = N_0/A*210; ## Weight of Po-210 after one year, g\n",
+ "A_0 = N_0*D/(3.7e+010); ## Initial activity, curie\n",
+ "print'%s %.2f %s %.2e %s'%(\"\\nThe weight of Po-210 after one year = \",W,\" g\"and\" \\nThe initial activity of the material = \",A_0,\" curies\");\n",
+ "\n",
+ "## Result\n",
+ "## The weight of Po-210 after one year = 10.49 g \n",
+ "## The initial activity of the material = 4.88e+004 curies"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The weight of Po-210 after one year = 10.49 \n",
+ "The initial activity of the material = 4.88e+04 curies\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg91"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.10 : : Page 91 (2011)\n",
+ "#find The decay constant for total emission and The decay constant for beta_decay lambda_b and The decay constant for beta_decay lambda_a\n",
+ "import math\n",
+ "lambda_t = 0.693/(60.5*60);## Total decay constant, per sec\n",
+ "lambda_a = 0.34*lambda_t;## Decay constant for alpha_decay, per sec\n",
+ "lambda_b = 0.66*lambda_t;## Decay constant for beta_decay, per sec\n",
+ "print'%s %.2e %s'%(\"\\nThe decay constant for total emission = \",lambda_t,\" /sec\");\n",
+ "print'%s %.2e %s'%(\"\\nThe decay constant for beta_decay lambda_b = \",lambda_b,\" /sec\");\n",
+ "print'%s %.2e %s'%(\"\\nThe decay constant for alpha_decay lambda_a = \",lambda_a,\" /sec\");\n",
+ "\n",
+ "## Result \n",
+ "## The decay constant for total emission = 1.91e-004 /sec\n",
+ "## The decay constant for beta_decay lambda_b = 1.26e-004 /sec\n",
+ "## The decay constant for alpha_decay lambda_a = 6.49e-005 /sec "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The decay constant for total emission = 1.91e-04 /sec\n",
+ "\n",
+ "The decay constant for beta_decay lambda_b = 1.26e-04 /sec\n",
+ "\n",
+ "The decay constant for alpha_decay lambda_a = 6.49e-05 /sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.13 : : Page 93 (2011)\n",
+ "#find The half life of Pu-239 \n",
+ "import math\n",
+ "M_A = 4.; ## Mass of alpha particle, amu\n",
+ "M_U = 235.; ##Mass of U-235, amu\n",
+ "M_P = 239.; ## Mass of P-239, amu\n",
+ "Amount = 120.1; ## quantity of P-239, g\n",
+ "E_A = 5.144; ## Energy of emitting alpha particles, Mev\n",
+ "E_R = (2.*M_A)/(2.*M_U)*E_A; ## The recoil energy of U-235, Mev\n",
+ "E = E_R + E_A; ## The energy released per disintegration, Mev\n",
+ "P = 0.231; ## Evaporation rate, watt\n",
+ "D = P/(E*1.60218e-013); ## Disintegration rate, per sec\n",
+ "A = 6.022137e+023; ## Avagadro's number, atoms\n",
+ "N = Amount/M_P*A; ## Number of nuclei in 120.1g of P-239\n",
+ "T = 0.693/(D*3.15e+07)*N; ## Half life of Pu_239, years\n",
+ "print'%s %.2e %s'%(\"\\nThe half life of Pu-239 = \",T,\" years\");\n",
+ "\n",
+ "## Result \n",
+ "## The half life of Pu-239 = 2.42e+004 years "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The half life of Pu-239 = 2.42e+04 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.14 : : Page 93 (2011)\n",
+ "#find The disintegration rate of Au-199\n",
+ "import math\n",
+ "T_h_1 = 2.7*24*3600; ## Half life of Au-198, sec\n",
+ "T_h_2 = 3.15*24*3600; ## Half life of Au-199, sec\n",
+ "S_1 = 99e-028; ## Crossection for first reaction, Sq.m\n",
+ "S_2 = 2.6e-024; ## Crossection for second reaction, Sq.m\n",
+ "I = 1e+018; ## Intensity of radiation, per Sq.m per sec\n",
+ "L_1 = I*S_1; ## Decay constant of Au-197, per sec\n",
+ "L_2 = 0.693/T_h_1+I*S_2; ## Decay constant of Au-198, per sec\n",
+ "L_3 = 0.693/T_h_2; ## Decay constant of Au-199, per sec\n",
+ "N_0 = 6.022137e+023; ## Avogadro number\n",
+ "N_1 = N_0/197.; ## Initial number of atoms of Au-197\n",
+ "t = 30.*3600.; ## Given time, sec\n",
+ "p = (math.exp(-L_1*t))/((L_2-L_1)*(L_3-L_1));\n",
+ "q = (math.exp(-L_2*t))/((L_1-L_2)*(L_3-L_2));\n",
+ "r = (math.exp(-L_3*t))/((L_1-L_3)*(L_2-L_3));\n",
+ "N3 = N_1*L_1*L_2*(p+q+r);\n",
+ "N_199 = N3;\n",
+ "L = L_3*N_199; ## Disintegration rate of Au-199, per sec\n",
+ "print'%s %.2e %s'%(\"\\nThe disintegration rate of Au-199 = \", L,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The disintegration rate of Au-199 = 1.9e+012 (Wrong answer in the textbook)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The disintegration rate of Au-199 = 1.88e+12 \n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.15 : : Page 94 (2011)\n",
+ "#find The maximum activity of Na-24 and The activity after a continuous bombardment\n",
+ "import math\n",
+ "Y = 110e-03;## Yield of Na-24, mCi/hr\n",
+ "T = 14.8;## Half life of Na-24, hours\n",
+ "t = 8;## Time after which activity to be compute, hours\n",
+ "D = 0.693/T;## Disintegration constant, hours^-1\n",
+ "A = 1.44*Y*T;## Maximum activity of Na-24, Ci\n",
+ "A_C = A*(1-math.e**(-D*t));## Activity after a continuous bombardment, Ci\n",
+ "Activity = A_C*(math.e**(-D*t));## Activity after 8hours, Ci\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe maximum activity of Na-24 = \",A,\" Ci\"and \"\\nThe activity after a continuous bombardment = \",A_C,\" Ci\"and \"\\nThe activity after 8hours = \",Activity,\" Ci\")\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## The maximum activity of Na-24 = 2.344 Ci\n",
+ "## The activity after a continuous bombardment = 0.7324 Ci\n",
+ "## The activity after 8hours = 0.50360 Ci"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The maximum activity of Na-24 = 2.34 \n",
+ "The activity after a continuous bombardment = 0.73 \n",
+ "The activity after 8hours = 0.50 Ci \n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.16 : : Page 94 (2011)\n",
+ "#find The energy of beta rays emitted per gram of tissue\n",
+ "import math\n",
+ "A_0 = 3.7e+07; ## Initial activity, disintegrations per sec\n",
+ "T = 12.6; ## Half life of I-130, hours\n",
+ "t = 24*3600; ## time for dose absorbed calculation,sec\n",
+ "E = 0.29*1.6e-06; ## Average energy of beta rays, ergs\n",
+ "m = 2; ## Mass of iodine thyroid tissue, gm\n",
+ "l = 0.693/(T*3600); ## Disintegration constant, sec^-1\n",
+ "N_0 = A_0/l; ## Initial number of atoms\n",
+ "N = N_0*(1-math.e**(-l*t)); ## Number of average atoms disintegrated\n",
+ "E_A = N*E; ## Energy of beta rays emitted, ergs\n",
+ "E_G = E_A/(2*97.00035); ## Energy of beta rays emitted per gram of tissue, REP \n",
+ "print'%s %.2f %s'%(\"\\nThe energy of beta rays emitted per gram of tissue = \",E_G,\" REP\");\n",
+ "\n",
+ "## Result\n",
+ "## The energy of beta rays emitted per gram of tissue = 4245.0 REP "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy of beta rays emitted per gram of tissue = 4245.02 REP\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex18-pg95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.18 : : Page 95 (2011)\n",
+ "#find The activity of Au-198 and The maximum amount of Au-198 produced\n",
+ "import math\n",
+ "N_0 = 6.022137e+023; ## Avagadro number\n",
+ "d = 0.02; ## Thickness of the foil, cm\n",
+ "R = 19.3; ## Density of Au,g/cc\n",
+ "N_1 = d*R/197.*N_0; ## Initial number of Au-197 nuclei per unit area of foil,cm^-2\n",
+ "T_H = 2.7*24.*3600.; ## Half life of Au-198,sec\n",
+ "L = math.log(2)/T_H; ## Decay constant for Au-198,sec^-1\n",
+ "I = 10**12; ## Intensity of neutron beam,neutrons/cm^2/sec\n",
+ "S = 97.8e-024; ## Cross section for reaction,cm^-2\n",
+ "t = 5*60.; ## Reaction time,s\n",
+ "A = S*I*N_1*(1-math.e**(-L*t)); ## Activity of Au-198,cm^-2sec^-1\n",
+ "N_2 = S*I*N_1/L; ## The maximum amount of Au-198 produced,cm^-2\n",
+ "print'%s %.2e %s %.2e %s '%(\"\\nThe activity of Au-198 = \",A,\" per Sq.cm per sec\"and \"\\nThe maximum amount of Au-198 produced = \",N_2,\" per Sq.cm\");\n",
+ "\n",
+ "## Result\n",
+ "## The activity of Au-198 = 1.028e+008 per Sq.cm per sec\n",
+ "## The maximum amount of Au-198 produced = 3.88e+016 per Sq.cm "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The activity of Au-198 = 1.03e+08 \n",
+ "The maximum amount of Au-198 produced = 3.88e+16 per Sq.cm \n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex19-pg95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.19 : : Page 95 (2011)\n",
+ "#find The power released by the gm molecule of Pu-238 and The time in which power reduces to 1/8 time of its initial value\n",
+ "import math\n",
+ "N_0 = 6.022137e+023; ## Avagadro number \n",
+ "T_P = 90.*365.*24.*3600.; ## Half life of Pu-238,s\n",
+ "L_P = 0.693/T_P ; ## Decay constant of Pu-238,s^-1\n",
+ "E = 5.5; ## Energy of alpha particle, MeV\n",
+ "P =E*L_P*N_0; ## Power released by the gm molecule of Pu-238,MeV/s\n",
+ "t = math.log(8)/(L_P*365.*24.*3600.); ## Time in which power reduces to 1/8 time of its initial value \n",
+ "print'%s %.2e %s %.1f %s '%(\"\\nThe power released by the gm molecule of Pu-238 = \",P,\" MeV/s\"and \"\\nThe time in which power reduces to 1/8 time of its initial value = \",t,\" yrs\")\n",
+ "\n",
+ "## Result\n",
+ "## The power released by the gm molecule of Pu-238 = 8.09e+014 MeV/s \n",
+ "## The time in which power reduces to 1/8 time of its initial value = 270 yrs "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The power released by the gm molecule of Pu-238 = 8.09e+14 \n",
+ "The time in which power reduces to 1/8 time of its initial value = 270.1 yrs \n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex20-pg96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa2.20 : : Page 96 (2011)\n",
+ "#find The required time for decay of daughter nucleus and The number of nuclei of daughter isotope\n",
+ "import math\n",
+ "N_1 = 10**20; ## Number of nuclei of parent isotopes\n",
+ "T_P = 10**4; ## Half life of parent nucleus,years\n",
+ "T_D = 20.; ## Half life of daughter nucleus,years\n",
+ "T = 10**4; ## Given time,years\n",
+ "L_P = 0.693/T_P ; ## Decay constant of parent nucleus,years^-1\n",
+ "L_D = 0.693/T_D ; ## Decay constant of daughter nucleus,years^-1\n",
+ "t_0 = math.log(0.03)/(L_P-L_D); ## Required time for decay of daughter nucleus,years\n",
+ "N = L_P/L_D*(math.e**(-L_P*T)-math.e**(-L_D*T))*N_1; ## Number of nuclei of daughter isotope\n",
+ "print'%s %.2f %s %.0e %s'%(\"\\nThe required time for decay of daughter nucleus =\",t_0,\" yr\"and \" \\nThe number of nuclei of daughter isotope = \",N,\" \");\n",
+ "\n",
+ "## Result\n",
+ "## The required time for decay of daughter nucleus = 101 yr \n",
+ "## The number of nuclei of daughter isotope = 1e+017 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The required time for decay of daughter nucleus = 101.40 \n",
+ "The number of nuclei of daughter isotope = 1e+17 \n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter3.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter3.ipynb
new file mode 100755
index 00000000..3abe9ff1
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter3.ipynb
@@ -0,0 +1,342 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0f490ccea3b13c889d4ee8d26efe8421ceedda6e18896bd5d0ed1b94aaa5c5de"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3-Interaction of Nuclear radiations with matter"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.1 : : Page-123 (2011)\n",
+ "#find range of an alpha particle and metre the thickness\n",
+ "E = 9.; ## Energy of the alpha particle, MeV\n",
+ "S = 1700.; ## Stopping power of Al\n",
+ "D = 2700.; ## Density of Al, Kg per cubic metre\n",
+ "R_air = 0.00318*E**(3/2.); ## Range of an alpha particle in air,metre\n",
+ "R_Al = R_air/S; ## Range of an alpha particle in Al, metre\n",
+ "T = D*1./S; ## Thickness in Al of 1m air, Kg per square metre\n",
+ "print\"%s %.2e %s %.2f %s \"%(\"The range of an alpha particle = \",R_Al,\" metre The thickness in Al of 1 m air =\",T,\" Kg per square metre\");\n",
+ "\n",
+ "## Result\n",
+ "## The range of an alpha particle = 5.05e-05 metre \n",
+ "## The thickness in Al of 1 m air = 1.59 Kg per square metre \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The range of an alpha particle = 5.05e-05 metre The thickness in Al of 1 m air = 1.59 Kg per square metre \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg124"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.4: : Page-124 (2011)\n",
+ "#find the haif value thickness for beta absorptions\n",
+ "import math\n",
+ "E_max = 1.17; ## Maximum energy of the beta particle, mega electron volts\n",
+ "D = 2.7; ## Density of Al,gram per cubic metre\n",
+ "u_m = 22./E_max; ## Mass absorption coefficient,centimetre square per gram\n",
+ "x_h = math.log(2.)/(u_m*D); ## Half value thickness for beta absorption, cm\n",
+ "print'%s %.2f %s'%(\"The Half value thickness for beta absorption = \",x_h,\" cm\"); \n",
+ "\n",
+ "## Result \n",
+ "## The Half value thickness for beta absorption = 0.014 cm \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Half value thickness for beta absorption = 0.01 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.7: : Page 125(2011)\n",
+ "#calculate ratio of raditon loss to ionisation\n",
+ "Z = 82.; ## Atomic number\n",
+ "E = 1.; ## Energy of the beta paricle, MeV\n",
+ "I_l = 800.; ## Ionisation loss, MeV\n",
+ "R = Z*E/I_l; ## Ratio of radiation loss to ionisation loss\n",
+ "E_1 = I_l/Z; ## Energy of the beta particle when radiation radiation loss is equal to ionisation loss, MeV\n",
+ "\n",
+ "print'%s %.2e %s %.2f %s '%(\"The ratio of radiation loss to ionisation loss =\",R,\" The energy of the beta particle = \",E_1,\" MeV \");\n",
+ "\n",
+ "## Result\n",
+ "## The ratio of radiation loss to ionisation loss = 1.025e-01 \n",
+ "## The energy of the beta particle = 9.76 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The ratio of radiation loss to ionisation loss = 1.02e-01 The energy of the beta particle = 9.76 MeV \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.8 : : Page 125(2011)\n",
+ "#find The half value thickness of Al and The mass absorption coefficient\n",
+ "import math\n",
+ "x = 0.25; ## Thickness of Al, metre\n",
+ "U_l = 1./x*math.log(50.); ## Linear absorption coefficient\n",
+ "d = 2700.; ## density of the Al, Kg per cubic centimetre \n",
+ "x_h = math.log(2.)/U_l; ## Half value thickness of Al, metre\n",
+ "U_m = U_l/d; ## Mass absorption coefficient, square metre per Kg\n",
+ "print'%s %.2f %s %.3f %s'%(\"The half value thickness of Al = \",x_h,\" Kg per cubic metre The mass absorption coefficient = \",U_m,\" square metre per Kg \");\n",
+ "\n",
+ "## Result\n",
+ "## The half value thickness of Al = 0.0443 Kg per cubic metre \n",
+ "## The mass absorption coefficient = 0.00580 square metre per Kg \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The half value thickness of Al = 0.04 Kg per cubic metre The mass absorption coefficient = 0.006 square metre per Kg \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.9 : : Page-125(2011)\n",
+ "#find The energy of the compton recoil electrons\n",
+ "import math\n",
+ "E_g = 2.19*1.6e-013; ## Energy of the gamma rays, joule\n",
+ "m_e = 9.10939e-031; ## Mass of the electron, Kg\n",
+ "C = 3e+08; ## Velocity of light, m/s\n",
+ "E_max = (E_g/(1.+(m_e*C**2)/(2.*E_g)))/(1.6e-013); ## Energy of the compton recoil electron, MeV\n",
+ "print\"%s %.2f %s\"%(\"The energy of the compton recoil electrons = \",E_max,\" MeV\"); \n",
+ "\n",
+ "## Result \n",
+ "## The energy of the compton recoil electrons = 1.961 MeV\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy of the compton recoil electrons = 1.96 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.10 : : Page-125(2011)\n",
+ "#find The average energy of the positron \n",
+ "m_e = 9.1e-31; ## Mass of the positron, Kg\n",
+ "e = 1.6e-19; ## Charge of the positron, coulomb\n",
+ "c = 3e+08; ## Velocity of the light, metre per sec\n",
+ "eps = 8.85e-12; ## Absolute permittivity of free space, per N per metre-square per coulomb square\n",
+ "h = 6.6e-34; ## Planck's constant, joule sec\n",
+ "E = e**2.*m_e*c/(eps*h*1.6e-13); ## Average energy of the positron, mega electron volts\n",
+ "print'%s %.2f %s'%(\"The average energy of the positron = \",E,\"Z MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The average energy of the positron = 0.0075Z MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average energy of the positron = 0.01 Z MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.11 : : Page-125(2011)\n",
+ "#find The refractive index of the gas and The angle at which Cerenkov radiation is emitted \n",
+ "import math\n",
+ "P = 1.; ## Momentum of the proton, GeV/c\n",
+ "M_0 = 0.94; ## Rest mass of the proton, GeV/c-square\n",
+ "G = math.sqrt((P/M_0)**2.+1.) ## Lorentz factor\n",
+ "V = math.sqrt(1.-1./G**2.); ## Minimum velocity of the electron, m/s\n",
+ "u = 1/V; ## Refractive index of the gas\n",
+ "print'%s %.2f %s'%(\"The refractive index of the gas =\", u,\"\"); \n",
+ "u = 1.6; ## Refractive index\n",
+ "theta = round (math.acos(1/(u*V))*180/3.14); ## Angle at which cerenkov radiatin is emitted,degree\n",
+ "print'%s %.2f %s'%(\"The angle at which Cerenkov radiation is emitted =\",theta,\" degree\") \n",
+ "\n",
+ "## Result \n",
+ "## The refractive index of the gas = 1.37\n",
+ "## The angle at which Cerenkov radiation is emitted = 31 degree \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The refractive index of the gas = 1.37 \n",
+ "The angle at which Cerenkov radiation is emitted = 31.00 degree"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa3.12 : : Page-126(2011)\n",
+ "#find The minimum kinetic energy required to electron to emit cerenkov radiation\n",
+ "import math\n",
+ "n = 1+1.35e-04; ## Refractive index of the medium\n",
+ "V_min = 1./n; ## Minimum velocity of the electron, m/s\n",
+ "p = (1.+V_min)*(1.-V_min); ## It is nothing but just to take the product \n",
+ "G_min = 1./math.sqrt(p); ## Lorentz factor\n",
+ "m_e = 9.10939e-031; ## Mass of the electron, Kg\n",
+ "C = 3e+08; ## Velocity of light, metre per sec\n",
+ "T_min = ((G_min-1.)*m_e*C**2.)/(1.602e-013); ## Minimum kinetic energy required by an electro to emit cerenkov radiation, mega electron volts\n",
+ "print'%s %.2f %s'%(\"The minimum kinetic energy required to electron to emit cerenkov radiation = \",T_min,\" MeV\");\n",
+ " \n",
+ "## Result \n",
+ "## The minimum kinetic energy required to electron to emit cerenkov radiation = 30.64 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum kinetic energy required to electron to emit cerenkov radiation = 30.64 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter4.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter4.ipynb
new file mode 100755
index 00000000..239fa1e3
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter4.ipynb
@@ -0,0 +1,703 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:a3997772daa49e18a0aa83e11b98c36eeb94107cace4b98c9321fc654338f29e"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter4-Detection and Measurement of Nuclear Reactions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.1 : : Page 178 (2011)\n",
+ "#find The resultant pulse height recorded in the fission chamber\n",
+ "import math\n",
+ "N = 200e+006/35.; ## Total number of ion-pairs\n",
+ "e = 1.60218e-019; ## Charge of an ion, coulomb\n",
+ "Q = N*e; ## Total charge produced in the chamber, coulomb\n",
+ "C = 25e-012; ## Capacity of the collector, farad\n",
+ "V = Q/C; ## Resultant pulse height, volt \n",
+ "print'%s %.2e %s'%(\"\\nThe resultant pulse height recorded in the fission chamber = \",V,\" volt\");\n",
+ "\n",
+ "## Result\n",
+ "## The resultant pulse height recorded in the fission chamber = 3.66e-002 volt "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The resultant pulse height recorded in the fission chamber = 3.66e-02 volt\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.2 : : Page 178 (2011)\n",
+ "#find The energy of the alpha particles\n",
+ "import math; \n",
+ "V = 0.8/4.; ## Pulse height, volt\n",
+ "e = 1.60218e-019; ## Charge of an ion, coulomb\n",
+ "C = 0.5e-012; ## Capacity of the collector, farad\n",
+ "Q = V*C; ## Total charge produced, coulomb\n",
+ "N = Q/e; ## Number of ion pairs \n",
+ "E_1 = 35.; ## Energy of one ion pair, electron volt\n",
+ "E = N*E_1/10**6; ## Energy of the alpha particles, mega electron volt\n",
+ "print'%s %.2f %s'%(\"\\nThe energy of the alpha particles = \",E,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The energy of the alpha particles = 21.845 MeV (The answer is wrong in the textbook)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy of the alpha particles = 21.85 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.3 : : Page 178 (2011)\n",
+ "#find Total number of ion pairs produced and Total charge flow in the counter\n",
+ "import math\n",
+ "E = 10e+06; ## Energy produced by the ion pairs, electron volts \n",
+ "N = E/35.; ## Number of ion pair produced\n",
+ "m = 10**3; ## Multiplication factor\n",
+ "N_t = N*m; ## Total number of ion pairs produced\n",
+ "e = 1.60218e-019; ## Charge of an ion, coulomb\n",
+ "Q = N_t*e; ## Total charge flow in the counter, coulomb\n",
+ "t = 10**-3; ## Pulse time, sec\n",
+ "R = 10**4; ## Resistance , ohm\n",
+ "I = Q/t; ## Current passes through the resistor, ampere\n",
+ "V = I*R; ## Height of the voltage pulse, volt\n",
+ "print'%s %.2f %s %.2e %s %.2e %s '%(\"\\nTotal number of ion pairs produced: \",N_t,\"\"and \" \\nTotal charge flow in the counter : \",Q,\" coulomb\"and \"\\nHeight of the voltage pulse :\",V,\" volt\")\n",
+ "\n",
+ "## Result\n",
+ "## Total number of ion pairs produced: 2.857e+008 \n",
+ "## Total charge flow in the counter : 4.578e-011 coulomb\n",
+ "## Height of the voltage pulse : 4.578e-004 volt "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Total number of ion pairs produced: 285714285.71 4.58e-11 \n",
+ "Height of the voltage pulse : 4.58e-04 volt \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.4 : : Page 178 (2011)\n",
+ "#find The radial electric field and The life of the G.M. Counter\n",
+ "import math; \n",
+ "V = 1000.; ## Operating voltage of Counter, volt \n",
+ "x = 1e-004; ## Time taken, sec\n",
+ "b = 2.; ## Radius of the cathode, cm\n",
+ "a = 0.01; ## Diameter of the wire, cm\n",
+ "E_r = V/(x*math.log(b/a)); ## Radial electric field, V/m\n",
+ "C = 1e+009; ## Total counts in the GM counter\n",
+ "T = C/(50.*60.*60.*2000.); ## Life of the G.M. Counter, year\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe radial electric field: \",E_r,\"V/m\"and\"\\nThe life of the G.M. Counter : \",T,\" years\");\n",
+ "\n",
+ "## Result\n",
+ "## The radial electric field: 1.89e+006V/m\n",
+ "## The life of the G.M. Counter : 2.778 years "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The radial electric field: 1887391.66 \n",
+ "The life of the G.M. Counter : 2.78 years \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.5 : : Page 178 (2011)\n",
+ "import math; \n",
+ "#find The avalanche voltage in G.M. tube\n",
+ "I = 15.7; ## Ionisation potential of argon, eV\n",
+ "b = 0.025; ## Radius of the cathode, metre\n",
+ "a = 0.006e-02; ## Radius of the wire, metre\n",
+ "L = 7.8e-06; ## Mean free path, metre\n",
+ "V = round(I*a*math.log(b/a)/L); ## Avalanche voltage in G.M. tube, volt\n",
+ "print'%s %.2f %s'%(\"\\nThe avalanche voltage in G.M. tube = \",V,\" volt\");\n",
+ "\n",
+ "## Result\n",
+ "## The avalanche voltage in G.M. tube = 729 volt "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The avalanche voltage in G.M. tube = 729.00 volt\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.6 : : Page 179 (2011)\n",
+ "#find The voltage fluctuation GM tube\n",
+ "import math; \n",
+ "C_r = 0.1e-02; ## Counting rate of GM tube\n",
+ "S = 3.; ## Slope of the curve\n",
+ "V = C_r*100*100/S; ## Voltage fluctuation, volt\n",
+ "print'%s %.2f %s'%(\"\\nThe voltage fluctuation GM tube = \",V,\" volt\");\n",
+ "\n",
+ "## Result\n",
+ "## The voltage fluctuation GM tube = 3.33 volt "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The voltage fluctuation GM tube = 3.33 volt\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.7 : : Page-179 (2011)\n",
+ "#find Time measurement for actual count and Time measurement for backward count\n",
+ "import math; \n",
+ "R_t = 100.; ## Actual count rate, per sec\n",
+ "R_B = 25.; ## Backward count rate, per sec\n",
+ "V_S = 0.03; ## Coefficient of variation\n",
+ "R_S = R_t-R_B; ## Source counting rate,per sec\n",
+ "T_t = (R_t+math.sqrt(R_t*R_B))/(V_S**2*R_S**2); ## Time measurement for actual count, sec\n",
+ "T_B = T_t*math.sqrt(R_B/R_t); ## Time measurement for backward count, sec\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nTime measurement for actual count : \",T_t,\" sec\"and \" \\nTime measurement for backward count : \",T_B,\" sec\");\n",
+ "\n",
+ "## Result\n",
+ "## Time measurement for actual count : 29.630 sec \n",
+ "## Time measurement for backward count : 14.8 sec"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Time measurement for actual count : 29.63 \n",
+ "Time measurement for backward count : 14.81 sec \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg179"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.8 : : Page-179 (2011)\n",
+ "#find The capacitance of the detector and The capacitance of the detector \n",
+ "import math; \n",
+ "A = 1.5e-4; ## Area of capacitor plates, square metre\n",
+ "K = 12.; ## Dielectric constant\n",
+ "D = K*8.8542e-012; ## Electrical permittivity of the medium, per newton-metre-square coulomb square\n",
+ "x = 50e-06; ## Width of depletion layer, metre\n",
+ "C = A*D/x*10**12; ## Capacitance of the silicon detector, pF\n",
+ "E = 4.5e+06; ## Energy produced by the ion pairs, eV\n",
+ "N = E/3.5; ## Number of ion pairs\n",
+ "e = 1.60218e-019; ## Charge of each ion, coulomb\n",
+ "Q = N*e; ## Total charge, coulomb\n",
+ "V = Q/C*10**12; ## Potential applied across the capacitor, volt\n",
+ "print'%s %.2f %s %.2e %s '%(\"\\nThe capacitance of the detector : \",C,\" pF\"and \"\\nThe capacitance of the detector : \",V,\" volt\");\n",
+ "\n",
+ "## Result\n",
+ "## The capacitance of the detector : 318.75 pF\n",
+ "## The potential applied across the capacitor : 6.46e-004 volt \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The capacitance of the detector : 318.75 \n",
+ "The potential applied across the capacitor : 6.46e-04 volt \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.9 : : Page-180 (2011)\n",
+ "#find The statistical error of the measured ratio\n",
+ "import math\n",
+ "N_A = 1000.; ## Number of count observed for radiation A\n",
+ "N_B = 2000.; ## Number of count observed for radiation B\n",
+ "r = N_A/N_B; ## Ratio of count A to the count B\n",
+ "E_r = math.sqrt(1./N_A+1./N_B); ## Statistical error \n",
+ "print'%s %.2f %s'%(\"\\nThe statistical error of the measured ratio = \", E_r*r,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The statistical error of the measured ratio = 0.02 (Wrong answer in the textbook)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The statistical error of the measured ratio = 0.02 \n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.10 : : Page 180 (2011)\n",
+ "#find The charge collected at the anode of photo multiplier tube\n",
+ "import math; \n",
+ "E = 4e+006; ## Energy lost in the scintillator, eV\n",
+ "N_pe = E/10**2*0.5*0.1; ## Number of photoelectrons emitted\n",
+ "G = 10**6; ## Gain of photomultiplier tube\n",
+ "e = 1.6e-019; ## Charge of the electron, C\n",
+ "Q = N_pe*G*e; ## Charge collected at the anode of photo multiplier tube, C\n",
+ "print'%s %.4e %s'%(\"\\nThe charge collected at the anode of photo multiplier tube : \",Q,\" C\");\n",
+ "\n",
+ "## Result\n",
+ "## The charge collected at the anode of photo multiplier tube : 3.2000e-010 C "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The charge collected at the anode of photo multiplier tube : 3.2000e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg180"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa11 : : Page 180 (2011)\n",
+ "#find Charge collected at the anode of photo multiplier tube\n",
+ "E = 4e+06; ## Energy lost in the scintillator, eV\n",
+ "N_pe = E/10**2*0.5*0.1; ## Number of photoelectrons emitted\n",
+ "G = 10**6; ## Gain\n",
+ "e = 1.6e-019; ## Charge of the electron, C\n",
+ "Q = N_pe*G*e; ## Charge collected at the anode of photo multiplier tube, C\n",
+ "print'%s %.4e %s'%(\"\\nCharge collected at the anode of photo multiplier tube : \",Q,\" C\");\n",
+ "## Result\n",
+ "## Charge collected at the anode of photo multiplier tube : 3.2000e-010 C "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Charge collected at the anode of photo multiplier tube : 3.2000e-10 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg181"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.12 : : Page 181 (2011)\n",
+ "#find Standard deviation of the reading\n",
+ "## Defining an array\n",
+ "import math;\n",
+ "import numpy\n",
+ "n = numpy.zeros((6,1)); ## Declare the cell matrix of 1X6 \n",
+ "n[0,0] = 10000;\n",
+ "n[1,0]= 10200;\n",
+ "n[2,0] = 10400;\n",
+ "n[3,0] = 10600;\n",
+ "n[4,0] = 10800;\n",
+ "n[5,0] = 11000;\n",
+ "g = 0.; ## \n",
+ "k = 6;\n",
+ "H = 0.;\n",
+ "for i in (0,k-1):\n",
+ " g = g + n[i,0]\n",
+ "\n",
+ "N = g/k; ## Mean of the count\n",
+ "D = math.sqrt(N);\n",
+ "for i in range (0,k-1):\n",
+ " H = H+((n[i,0]-N)*(n[i,0]-N)) \n",
+ "\n",
+ "S_D = round(math.sqrt(H/(k-1)));\n",
+ "print'%s %.2f %s'%(\"\\nStandard deviation of the reading : \", S_D,\"\");\n",
+ "delta_N = math.sqrt(N);\n",
+ "if (S_D > delta_N) :\n",
+ " print(\"\\nThe foil cannot be considered uniform..!\");\n",
+ "else:\n",
+ " print(\"\\nThe foil can be considered uniform.\");\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## Standard deviation of the reading : 374\n",
+ "## The foil cannot be considered uniform..! "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Standard deviation of the reading : 6906.00 \n",
+ "\n",
+ "The foil cannot be considered uniform..!\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg181"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.13 : : Page 181 (2011)\n",
+ "#findNo. of electrons in the output\n",
+ "import math\n",
+ "V = 2e-03; ## Voltage impulse, volt\n",
+ "C = 120e-012; ## Capacitance of the capacitor, F\n",
+ "e = 1.6e-019; ## Charge of the electron, C\n",
+ "n = C*V/(15.*e); ## No. of electons\n",
+ "N = n**(1/10.); ## No. of electrons in the output\n",
+ "print'%s %.2f %s'%(\"\\nNo. of electrons in the output : \",N,\" (approx)\");\n",
+ "\n",
+ "## Result\n",
+ "## No. of electrons in the output : 3.16 (approx) "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "No. of electrons in the output : 3.16 (approx)\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg181"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.14 : : Page 181 (2011)\n",
+ "#find Time of flight of proton and Time of flight of electron\n",
+ "import math\n",
+ "m_p = 0.938; ## Mass of the proton, GeV\n",
+ "E = 1.4; ## Total energy of proton, GeV\n",
+ "gama = E/m_p; ## Boost parameter\n",
+ "bta = math.sqrt(1-1/gama**2); ## Relativistic factor\n",
+ "d = 10.; ## Distance between two counters,m\n",
+ "C = 3e+08; ## Velocity of light ,m/s\n",
+ "t_p = d/(bta*C); ## Time of flight of proton ,sec\n",
+ "T_e = d/C; ## Time of flight of electron, sec\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nTime of flight of proton: \",t_p/1e-009,\" ns\"and \" \\nTime of flight of electron : \",T_e/1e-009,\" ns \");\n",
+ "\n",
+ "## Result\n",
+ "## Time of flight of proton: 44.90 ns \n",
+ "## Time of flight of electron : 33.33 ns "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "Time of flight of proton: 44.90 \n",
+ "Time of flight of electron : 33.33 ns \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.15 : : Page 182 (2011)\n",
+ "#find The fractional error in rest mass of the particle\n",
+ "import math;\n",
+ "p = 100.; ## Momentum of the particle, GeV\n",
+ "n = 1+1.35e-04; ## Refractive index of the gas \n",
+ "m_0 = 1.; ## Mass, GeV per square coulomb\n",
+ "gama = math.sqrt((p**2+m_0**2)/m_0); ## Boost parameter\n",
+ "bta = math.sqrt (1-1/gama**2); ## Relativistic parameter\n",
+ "d_theta = 1e-003; ## Error in the emission angle, radian\n",
+ "theta = math.acos(1/(n*bta)); ## Emision angle of photon, radian \n",
+ "F_err = (p**2*n**2*2*theta*10**-3)/(2*m_0**2); ## Fractional error\n",
+ "print'%s %.2f %s'%(\"\\nThe fractional error in rest mass of the particle = \", F_err,\"\");\n",
+ "\n",
+ "## Result \n",
+ "## The fractional error in rest mass of the particle = 0.13 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The fractional error in rest mass of the particle = 0.13 \n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa4.16 : : Page 182 (2011)\n",
+ "#find The total number of quantas during emission of visible light\n",
+ "import math;\n",
+ "u = 1.49; ## Refractive index\n",
+ "E = 20*1.60218e-019; ## Energy of the electron, joule\n",
+ "m_e = 9.1e-031; ## Mass of the electron, Kg\n",
+ "C = 3e-08; ## Velocity of the light, m/s\n",
+ "bta = (1 + (1/(E/(m_e*C**2)+1))**2 ); ## Boost parameter\n",
+ "z = 1.; ## \n",
+ "L_1 = 4000e-010; ## Initial wavelength, metre\n",
+ "L_2 = 7000e-010; ## Final wavelength, metre\n",
+ "N = 2*math.pi*z**2/137.*(1./L_1-1./L_2)*(1-1./(bta**2*u**2)); ## Number of quanta of visible light, quanta per centimetre\n",
+ "print'%s %.2f %s'%(\"\\nThe total number of quantas during emission of visible light = \", round(N/100),\"quanta/cm\");\n",
+ "\n",
+ "## Result \n",
+ "## The total number of quantas during emission of visible light = 270 quanta/cm "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The total number of quantas during emission of visible light = 270.00 quanta/cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter5.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter5.ipynb
new file mode 100755
index 00000000..8e2edd81
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter5.ipynb
@@ -0,0 +1,347 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:5bd75b768cbef982b8c705dbf643c0d6d3b446f1e0997915eea25d6d1e7b9a2f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter5-Alpha particles"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.1 : : Page 203 (2011)\n",
+ "#find Disintegration energy and Barrier height for alpha-particle\n",
+ "import math\n",
+ "E_a = 8.766; ## Energy of the alpha particle, MeV\n",
+ "A = 212.; ## Atomic mass of Po-212, amu\n",
+ "M_a = 4.; ## Atomic mass of alpha particle, amu\n",
+ "e = 1.6e-019; ## Charge of an electron, coulomb\n",
+ "Z = 82.; ## Atomic number of Po-212\n",
+ "R_0 = 1.4e-015; ## Distance of closest approach,metre\n",
+ "K = 8.99e+09; ## Coulomb constant\n",
+ "E = E_a*A/(A-M_a); ## Disintegration energy, mega electron volts\n",
+ "B_H = 2.*Z*e**2.*K/(R_0*A**(1./3.)*1.6*10**-13.); ## Barrier height for an alpha particle within the nucleus, MeV\n",
+ "print\"%s %.3f %s %.3f %s \"%(\"Disintegration energy :\",E,\" MeV Barrier height for alpha-particle: \",B_H,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## Disintegration energy : 8.935 MeV \n",
+ "## Barrier height for alpha-particle: 28.26 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Disintegration energy : 8.935 MeV Barrier height for alpha-particle: 28.259 MeV \n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.2 : : Page 203 (2011)\n",
+ "## We have to make calculation for alpha particle and for proton\n",
+ "import math\n",
+ "E_a = 8.766; ## Energy of the alpha particle, mega electron volts\n",
+ "A_Bi = 209.; ## Atomic mass of Bi-209, atomic mass unit\n",
+ "A_a = 4.; ## Atomic mass of alpha particle, atomic mass unit\n",
+ "A_p = 1.; ## Atomic mass of proton, atomic mass unit\n",
+ "e = 1.6e-019; ## Charge of an electron, coulomb\n",
+ "Z = 83.; ## Atomic number of bismuth\n",
+ "R_0 = 1.4e-015; ## Distance of closest approach,metre\n",
+ "K = 8.99e+09; ## Coulomb constant\n",
+ "B_H_a = 2.*Z*e**2.*K/(R_0*1.6e-013*(A_Bi**(1/3.)+A_a**(1/3.))); ## Barrier height for an alpha particle, mega electron volts\n",
+ "B_H_p = 1.*Z*e**2.*K/(R_0*1.6e-013*(A_Bi**(1/3.)+A_p**(1/3.))); ## Barrier height for proton, mega electron volts\n",
+ "print'%s %.2f %s %.2f %s'%(\"Barrier height for the alpha particle =\",B_H_a,\" MeV Barrier height for the proton = \",B_H_p,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## Barrier height for the alpha particle = 22.67 MeV \n",
+ "## Barrier height for the proton = 12.30 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Barrier height for the alpha particle = 22.67 MeV Barrier height for the proton = 12.30 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg203"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.3 : : Page 203 (2011)\n",
+ "## We have also calculate the value of magnetic field in a particular orbit. \n",
+ "import math\n",
+ "C = 3e+08; ## Velocity of light, m/S\n",
+ "M_0 = 6.644e-027*(C)**2/(1.60218e-013); ## Rest mass of alpha particle, MeV\n",
+ "T = 5.998; ## Kinetic energy of alpha particle emitted by Po-218\n",
+ "q = 2*1.60218e-019; ## Charge of alpha particle, C\n",
+ "V = math.sqrt(C**2*T*(T+2*M_0)/(T+M_0)**2); ## Velocity of alpha particle,metre per sec\n",
+ "B_r = V*M_0*(1.60218e-013)/(C**2.*q*math.sqrt(1-V**2./C**2.)); ## magnetic field in a particular orbit, Web per mtere\n",
+ "print'%s %.2e %s %.2f %s '%(\"The velocity of alpha particle : \",V,\" m/sThe magnetic field in a particular orbit :\",B_r,\" Wb/m\");\n",
+ "\n",
+ "## Result\n",
+ "## The velocity of alpha particle : 1.699e+007 m/s\n",
+ "## The magnetic field in a particular orbit : 0.3528 Wb/m"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of alpha particle : 1.70e+07 m/sThe magnetic field in a particular orbit : 0.35 Wb/m \n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.4: : Page 204 (2011)\n",
+ "#find The probability of leakage of alpha-particle through potential barrier\n",
+ "import math\n",
+ "a = 10**-14; ## Width of the potential barrier, m\n",
+ "E = 5*1.60218e-013; ## Energy of the alpha particle, joule\n",
+ "V = 10*1.60218e-013; ## Potential height, joule\n",
+ "M_0 = 6.644e-027; ## Rest mass of the alpha particle, joule\n",
+ "h_red = 1.05457e-034; ## Reduced value of Planck's constant,joule sec \n",
+ "T = 4*math.exp(-2*a*math.sqrt(2*M_0*(V-E)/h_red**2)); ## Probability of leakage through through potential barrier\n",
+ "print'%s %.2e %s'%(\"The probability of leakage of alpha-particle through potential barrier = \",T,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The probability of leakage of alpha-particle through potential barrier = 1.271e-008 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The probability of leakage of alpha-particle through potential barrier = 1.27e-08 \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg204"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.6: : Page 204 (2011)\n",
+ "#find The disintegration constant increases \n",
+ "import math\n",
+ "Z_D = 82.; ## Atomic number of Po\n",
+ "E_Po210 = 5.3; ## Alpha-source for Po210, MeV\n",
+ "E_Po214 = 7.7; ## Alpha-source for Po214, MeV\n",
+ "log_lambda_Po210 = -1*1.72*Z_D*E_Po210**(-1/2.); \n",
+ "log_lambda_Po214 = -1*1.72*Z_D*E_Po214**(-1/2.); \n",
+ "delta_OM_t = log_lambda_Po214 - log_lambda_Po210; ## Difference in order of magnitude of life times of Po214 and Po210\n",
+ "print'%s %.2d %s'%(\"The disintegration constant increases by a factor of some 10^ \" ,delta_OM_t,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The disintegration constant increases by a factor of some 10^10 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The disintegration constant increases by a factor of some 10^ 10 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.8: : Page 205 (2011)\n",
+ "#find The half life of Pu \n",
+ "import math\n",
+ "N = 120.1*6.023e+023/239.; ## Number of Pu nuclei\n",
+ "P_rel = 0.231; ## Power released, watt\n",
+ "E_rel = 5.323*1.6026e-13; ## Energy released, joule\n",
+ "decay_rate = P_rel/E_rel; ## Decay rate of Pu239, per hour\n",
+ "t_half = N*math.log(2)/(decay_rate*365.*86400.); ## Half life of Po239, sec\n",
+ "print\"%s %.2e %s\"%(\"The half life of Pu = \",t_half,\" yr\");\n",
+ "\n",
+ "## Result\n",
+ "## The half life of Pu = 2.46e+004 yr "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The half life of Pu = 2.46e+04 yr\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg205"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.9 : : Page 205(2011)\n",
+ "#find The slope of alpha decay energy versus atomic number\n",
+ "import math\n",
+ "a_v = 14.; ## Volume energy constant, MeV\n",
+ "a_s = 13.; ## Surface energy constant, MeV\n",
+ "a_c = 0.60; ## Coulomb energy constant, MeV\n",
+ "a_a = 19.; ## Asymmetric energy constant, MeV\n",
+ "A = 202.; ## Mass number\n",
+ "Z = 82; ## Atomic number \n",
+ "dE_by_dN = -8/9*a_s/A**(4/3.)-4/3.*a_c*Z/A**(4/3.)*(1-4.*Z/(3.*A))-16.*a_a*Z/A**2.*(1.-2.*Z/A); ## Slope, mega electron volts per nucleon\n",
+ "print'%s %.2f %s'%(\"The slope of alpha decay energy versus atomic number = \",dE_by_dN,\" MeV/nucleon\");\n",
+ "\n",
+ "## Result\n",
+ "## The slope of alpha decay energy versus atomic number = -0.15007 MeV/nucleon "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The slope of alpha decay energy versus atomic number = -0.15 MeV/nucleon\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg206"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa5.10 : : Page 206 (2011)\n",
+ "#find The hindrance factor for alpha particle\n",
+ "import math\n",
+ "h_kt = 1.05457e-34; ## Reduced Planck's constant, joule sec\n",
+ "e = 1.60218e-19; ## Charge of an electron, coulomb\n",
+ "l = 2.; ## Orbital angular momentum\n",
+ "eps_0 = 8.5542e-12; ## Absolute permittivity of free space, coulomb square per newton per metre square\n",
+ "Z_D = 90.; ## Atomic number of daughter nucleus\n",
+ "m = 6.644e-27; ## Mass of alpha particle, Kg\n",
+ "R = 8.627e-15; ## Radius of daughter nucleus, metre\n",
+ "T1_by_T0 = math.exp(2.*l*(l+1.)*h_kt/e*math.sqrt(math.pi*eps_0/(Z_D*m*R))); ## Hindrance factor\n",
+ "print'%s %.2f %s'%(\"The hindrance factor for alpha particle = \" ,T1_by_T0,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The hindrance factor for alpha particle = 1.768 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hindrance factor for alpha particle = 1.77 \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter6.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter6.ipynb
new file mode 100755
index 00000000..c8a7f4f4
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter6.ipynb
@@ -0,0 +1,745 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:4015b0a9a33cac1d31d06283c1020edbd79893138244c7d49c1c64131de1b42b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter6-Beta-Decay"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg240"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.1: : Page- 240 (2011)\n",
+ "#find The rate at which energy is emitted\n",
+ "T = 5*24*60*60; ## Half life of the substance, sec\n",
+ "N = 6.023e+026*4e-06/210; ## Number of atoms\n",
+ "D = 0.693/T; ## Disintegration constant, per sec\n",
+ "K = D*N; ## Rate of disintegration, \n",
+ "E = 0.34*1.60218e-013; ## Energy of the beta particle, joule\n",
+ "P = E*K; ## Rate at which energy is emitted, watt\n",
+ "print'%s %.2f %s'%(\"\\nThe rate at which energy is emitted = \",P,\" watt\");\n",
+ "\n",
+ "## Result\n",
+ "## The rate at which energy is emitted = 1 watt \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The rate at which energy is emitted = 1.00 watt\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.2 : : Page-241 (2011)\n",
+ "#find The magnetic field perpendicular to the beam of the particle \n",
+ "M_0 = 9.10939e-031; ## Rest mass of the electron, Kg\n",
+ "C = 2.92e+08; ## Velocity of the light, metre per sec\n",
+ "E = 1.71*1.60218e-013; ## Energy of the beta particle, joule\n",
+ "e = 1.60218e-019; ## Charge of the electron, C \n",
+ "R = 0.1; ## Radius of the orbit, metre\n",
+ "B = M_0*C*(E/(M_0*C**2)+1)*1/(R*e); ## Magnetic field perpendicular to the beam of the particle, weber per square metre\n",
+ "\n",
+ "print'%s %.2f %s'%(\"\\nThe magnetic field perpendicular to the beam of the particle = \",B,\" Wb/square-metre\");\n",
+ "\n",
+ "## Result\n",
+ "## The magnetic field perpendicular to the beam of the particle = 0.075 Wb/square-metre \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The magnetic field perpendicular to the beam of the particle = 0.08 Wb/square-metre\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.3 : : Page-241 (2011)\n",
+ "#find The energy of the electron and The energy of the converted gamma ray photon \n",
+ "import math\n",
+ "m_0 = 9.10963e-031; ## Rest mass of the electron, Kg\n",
+ "e = 1.60218e-019; ## Charge of the electron, C\n",
+ "c = 2.9979e+08; ## Velocity of the light, metre per sec\n",
+ "BR = 3381e-006; ## Field-radius product, tesla-m\n",
+ "E_k = 37.44; ## Binding energy of k-electron\n",
+ "v = 1/math.sqrt((m_0/(BR*e))**2+1/c**2); ## Velocity of the converson electron, m/s\n",
+ "E = m_0*c**2*(1/math.sqrt(1-v**2/c**2)-1.)/(e*1e+003); ## Energy of the electron, keV \n",
+ "E_C = E+E_k; ## Energy of the converted gamma ray photon, KeV\n",
+ "print'%s %.2f %s %.2f %s'%(\"\\nThe energy of the electron = \",E,\" keV \" and \" The energy of the converted gamma ray photon =\", E_C,\" keV\");\n",
+ "\n",
+ "## Result\n",
+ "## The energy of the electron = 624.11 keV \n",
+ "## The energy of the converted gamma ray photon = 661.55 keV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy of the electron = 624.11 The energy of the converted gamma ray photon = 661.55 keV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg241"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.4 : : Page-241 (2011)\n",
+ "#find The rest energy carried out by the neutrino\n",
+ "import math\n",
+ "E = 18.1; ## Energy carried by beta particle, keV \n",
+ "E_av = E/3.; ## Average energy carried away by beta particle, keV\n",
+ "E_r = E-E_av; ## The rest energy carried out by the neutrino, keV\n",
+ "\n",
+ "print'%s %.2f %s'%(\"\\nThe rest energy carried out by the neutrino : \",E_r,\" KeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The rest energy carried out by the neutrino : 12.067 KeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The rest energy carried out by the neutrino : 12.07 KeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.5: : Page-242(2011)\n",
+ "#find The maximum energy available to the electrons in the beta decay\n",
+ "import math\n",
+ "M_Na = -8420.40; ## Mass of sodium 24, keV\n",
+ "M_Mg = -13933.567; ## Mass of magnesium 24, keV\n",
+ "E = (M_Na-M_Mg)/1000.; ## Energy of the electron, MeV\n",
+ "print'%s %.2f %s'%(\"\\nThe maximum energy available to the electrons in the beta decay = \",E,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The maximum energy available to the electrons in the beta decay = 5.513 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The maximum energy available to the electrons in the beta decay = 5.51 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6-pg242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.6: : Page-242 (2011)\n",
+ "#find The linear momentum of neutrino and The linear momentum of beta particle\n",
+ "import math\n",
+ "c = 1.; ## For simplicity assume speed of light to be unity, m/s\n",
+ "E_0 = 0.155; ## End point energy, mega electron volts\n",
+ "E_beta = 0.025; ## Energy of beta particle, mega electron volts\n",
+ "E_v = E_0-E_beta; ## Energy of the neutrino, mega electron volts\n",
+ "p_v = E_v/c; ## Linear momentum of neutrino, mega electron volts per c\n",
+ "m = 0.511; ## Mass of an electron, Kg\n",
+ "M = 14*1.66e-27; ## Mass of carbon 14,Kg\n",
+ "c = 3e+8; ## Velocity of light, metre per sec\n",
+ "e = 1.60218e-19; ## Charge of an electron, coulomb\n",
+ "p_beta = math.sqrt(2*m*E_beta); ## Linear momentum of beta particle, MeV/c\n",
+ "sin_theta = p_beta/p_v*math.sin(45/57.3); ## Sine of angle theta\n",
+ "p_R = p_beta*math.cos(45/57.3)+p_v*math.sqrt(1-sin_theta**2); ## Linear momemtum of recoil nucleus, MeV/c\n",
+ "E_R = (p_R*1.6e-13/2.9979e+08)**2/(2.*M*e); ## Recoil energy of product nucleus, MeV\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe linear momentum of neutrino = \",p_v,\" MeV/c\" and \"The linear momentum of beta particle =\",p_beta,\" MeV/c\" and \"The energy of the recoil nucleus = \",E_R,\" eV\")\n",
+ "\n",
+ "## Result\n",
+ "## The linear momentum of neutrino = 0.13 MeV/c \n",
+ "## The linear momentum of beta particle = 0.1598 MeV/c \n",
+ "## The energy of the recoil nucleus = 1.20 eV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The linear momentum of neutrino = 0.13 The linear momentum of beta particle = 0.16 The energy of the recoil nucleus = 1.20 eV \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.7: : Page-242 (2011)\n",
+ "#find The energy of the beta particle and The ratio of beta particle energy with end point energy\n",
+ "import math\n",
+ "N = 3.7e+10*60; ## Number of disintegration, per sec\n",
+ "H = 0.0268*4.182; ## Heat produced at the output, joule\n",
+ "E = H/(N*1.6e-013); ## Energy of the beta particle, joule\n",
+ "M_Bi = -14.815; ## Mass of Bismuth, MeV\n",
+ "M_Po = -15.977; ## Mass of polonium, MeV\n",
+ "E_0 = M_Bi-M_Po; ## End point energy, MeV\n",
+ "E_ratio = E/E_0; ## Ratio of beta particle energy with end point energy\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe energy of the beta particle = \",E,\" MeV\" and \"The ratio of beta particle energy with end point energy =\" ,E_ratio,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The energy of the beta particle = 0.316 MeV \n",
+ "## The ratio of beta particle energy with end point energy = 0.272 \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy of the beta particle = 0.32 The ratio of beta particle energy with end point energy = 0.27 \n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.8: : Page-243 (2011)\n",
+ "#find The parity of the 2.9 MeV level in be-8 and The threshold energy for lithium 7 neutron capture\n",
+ "import math\n",
+ "l = 2.; ## Orbital angular momentum quantum number\n",
+ "P = (+1)**2*(-1)**l; ## Parity of the 2.9 MeV level in Be-8\n",
+ "M_Li = 7.0182; ## Mass of lithium, MeV\n",
+ "M_Be = 7.998876; ## Mass of beryllium, MeV\n",
+ "m_n = 1.; ## Mass of neutron, MeV\n",
+ "E_th = (M_Li+m_n-M_Be)*931.5; ## Threshold energy, MeV\n",
+ "print'%s %.2f %s %.2f %s'%(\"\\nThe parity of the 2.9 MeV level in be-8 = \",+P,\" \" and\"The threshold energy for lithium 7 neutron capture = \",E_th,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The parity of the 2.9 MeV level in be-8 = +1 \n",
+ "## The threshold energy for lithium 7 neutron capture = 18 MeV \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The parity of the 2.9 MeV level in be-8 = 1.00 The threshold energy for lithium 7 neutron capture = 18.00 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg243"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "##Page-243(2011)\n",
+ "import math\n",
+ "import numpy\n",
+ "#find pairs are stable or not\n",
+ "M =numpy.zeros((4,2));\n",
+ "M[0,0] = 7.0182*931.5; ## Mass of lithium, MeV\n",
+ "M[0,1] = 7.0192*931.5; ## Mass of beryllium, MeV\n",
+ "M[1,0] = 13.0076*931.5; ## Mass of carbon, MeV\n",
+ "M[1,1] = 13.0100*931.5; ## Mass of nitrogen, MeV\n",
+ "M[2,0] = 19.0045*931.5; ## Mass of fluorine, MeV\n",
+ "M[2,1] = 19.0080*931.5; ## Mass of neon, MeV\n",
+ "M[3,0] = 33.9983*931.5; ## Mass of phosphorous, MeV\n",
+ "M[3,1] = 33.9987*931.5; ## Mass of sulphur, MeV\n",
+ "j = 0; \n",
+ "## Check the stability !!!!\n",
+ "for i in range (0,3):\n",
+ " if round (M[i,j+1]-M[i,j]) == 1:\n",
+ " print(\"\\n From pair a :\")\n",
+ " print(\"\\n Be(4,7) is unstable\");\n",
+ " elif round (M[i,j+1]-M[i,j]) == 2:\n",
+ " print(\"\\n From pair b :\")\n",
+ " print(\"\\n N(7,13) is unstable\");\n",
+ " elif round (M[i,j+1]-M[i,j]) == 3:\n",
+ " print(\"\\n From pair c :\")\n",
+ " print(\"\\n Ne(10,19) is unstable\");\n",
+ " elif round (M[i,j+1]-M[i,j]) == 0:\n",
+ " print(\"\\n From pair d :\")\n",
+ " print(\"\\n P(15,34) is unstable\");\n",
+ " \n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## \n",
+ "## From pair a :\n",
+ "## Be(4,7) is unstable\n",
+ "## From pair b :\n",
+ "## N(7,13) is unstable\n",
+ "## From pair c :\n",
+ "## Ne(10,19) is unstable\n",
+ "## From pair d :\n",
+ "## P(15,34) is unstable "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " From pair a :\n",
+ "\n",
+ " Be(4,7) is unstable\n",
+ "\n",
+ " From pair b :\n",
+ "\n",
+ " N(7,13) is unstable\n",
+ "\n",
+ " From pair c :\n",
+ "\n",
+ " Ne(10,19) is unstable\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.10: : Page-244 (2011)\n",
+ "#find The half life of H3 \n",
+ "import math\n",
+ "tau_0 = 7000.; ## Time constant, sec\n",
+ "M_mod_sqr = 3.; ## Nuclear matrix\n",
+ "E_0 = 0.018; ## Energy of beta spectrum, MeV \n",
+ "ft = 0.693*tau_0/M_mod_sqr; ## Comparative half life\n",
+ "fb = 10**(4.0*math.log10(E_0)+0.78+0.02); ##\n",
+ "t = 10**(math.log10(ft)-math.log10(fb)); ## Half life of H3, sec\n",
+ "print'%s %.2f %s'%(\"\\nThe half life of H3 = \",t,\" sec\");\n",
+ "\n",
+ "## Result\n",
+ "## The half life of H3 = 2.44e+009 sec "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The half life of H3 = 2441293526.34 sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.11: : Page-244 (2011)\n",
+ "import math\n",
+ "#find 92 percent beta emission \n",
+ "t_p = 33./0.92*365.*84800.; ## Partial half life for beta emission, sec\n",
+ "E_0 = 0.51; ## Kinetic energy\n",
+ "Z = 55.; ## Atomic number of cesium\n",
+ "log_fb = 4.0*math.log10(E_0)+0.78+0.02*Z-0.005*(Z-1)*math.log10(E_0); ## Comparitive half life\n",
+ "log_ft1 = log_fb+math.log10(t_p); ## Forbidden tansition\n",
+ "## For 8 percent beta minus emission\n",
+ "t_p = 33./0.08*365.*84800.; ## Partial half life, sec\n",
+ "E_0 = 1.17; ## Kinetic energy\n",
+ "Z = 55; ## Atomic energy\n",
+ "log_fb = 4.0*math.log10(E_0)+0.78+0.02*Z-0.005*(Z-1)*math.log10(E_0); ## Comparitive half life\n",
+ "log_ft2 = log_fb+math.log10(t_p); ## Forbidden transition\n",
+ "## Check the degree of forbiddenness !!!!!\n",
+ "if log_ft1 <= 10: \n",
+ " print(\"\\nFor 92 percent beta emission :\")\n",
+ " print(\"\\n\\tTransition is once forbidden and parity change\");\n",
+ "\n",
+ "if log_ft2 >= 10:\n",
+ " print(\"\\nFor 8 percent beta emission :\")\n",
+ " print(\"\\n\\t ransition is twice forbidden and no parity change\");\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## For 92 percent beta emission :\n",
+ "##\tTransition is once forbidden and parity change\n",
+ "## For 8 percent beta emission :\n",
+ "##\tTransition is twice forbidden and no parity change\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "For 92 percent beta emission :\n",
+ "\n",
+ "\tTransition is once forbidden and parity change\n",
+ "\n",
+ "For 8 percent beta emission :\n",
+ "\n",
+ "\t ransition is twice forbidden and no parity change\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg244"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.12: : Page-244(2011)\n",
+ "#find The value of coupling constant and The ratio of coupling constant\n",
+ "import math\n",
+ "h_kt = 1.05457e-34; ## Reduced planck's constant, joule sec\n",
+ "c = 3e+08; ## velocity of light, metre per sec\n",
+ "m_e = 9.1e-31; ## Mass of the electron, Kg\n",
+ "ft_O = 3162.28; ## Comparative half life for oxygen\n",
+ "ft_n = 1174.90; ## Comparative half life for neutron\n",
+ "M_f_sqr = 2. ## Matrix element\n",
+ "g_f = math.sqrt(2*math.pi**3*h_kt**7.*math.log(2.)/(m_e**5*c**4*ft_O*M_f_sqr)); ## Coupling constant, joule cubic metre\n",
+ "C_ratio = (2.*ft_O/(ft_n)-1.)/3.; ## Ratio of coupling strength\n",
+ "print'%s %.3e %s %.2f %s'%(\"\\nThe value of coupling constant = \",g_f,\" joule cubic metre\" and \"The ratio of coupling constant = \",C_ratio,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The value of coupling constant = 1.3965e-062 joule cubic metre\n",
+ "## The ratio of coupling constant = 1.461 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The value of coupling constant = 1.397e-62 The ratio of coupling constant = 1.46 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg245"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.13: : Page-245 (2011)\n",
+ "#find The relative capture rate in holmium 161\n",
+ "import math\n",
+ "Q_EC = 850.; ## Q value for holmium 161, keV\n",
+ "B_p = 2.0; ## Binding energy for p-orbital electron, keV\n",
+ "B_s = 1.8; ## Binding energy for s-orbital electron, keV\n",
+ "M_ratio = 0.05*(Q_EC-B_p)**2./(Q_EC-B_s)**2; ## Matrix ratio\n",
+ "Q_EC = 2.5; ## Q value for holmium 163, keV\n",
+ "C_rate = M_ratio*(Q_EC-B_s)**2./(Q_EC-B_p)**2.*100.; ## The relative capture rate in holmium, percent\n",
+ "print'%s %.2f %s'%(\"\\nThe relative capture rate in holmium 161 = \",C_rate,\" percent\");\n",
+ "\n",
+ "## Result\n",
+ "## The relative capture rate in holmium 161 = 9.8 percent "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The relative capture rate in holmium 161 = 9.80 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg246"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.14: : Page-246 (2011)\n",
+ "#find The average energy of beta particles\n",
+ "import math\n",
+ "t_half = 12.5*365*24; ## Half life of hydrogen 3, hour\n",
+ "D = math.log(2)/t_half; ## Decay constant, per hour\n",
+ "N_0 = 6.023e+26; ## Avogadro's number, per mole\n",
+ "m = 0.1e-03; ## Mass of tritium, Kg\n",
+ "dN_by_dt = D*m*N_0/3.; ## Decay rate, per hour\n",
+ "H = 21*4.18; ## Heat produed, joule \n",
+ "E = H/dN_by_dt; ## The average energy of the beta particle, joule\n",
+ "print'%s %.2e %s %.2f %s'%(\"\\nThe average energy of beta particles =\",E,\"joule = \",E/1.6e-016,\" keV\");\n",
+ "\n",
+ "## Result\n",
+ "## The average energy of beta particles = 6.91e-016 joule = 4.3 keV \n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The average energy of beta particles = 6.91e-16 joule = 4.32 keV\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg246"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.15: : Page-246 (2011)\n",
+ "#find antiparallel spin and parallel spin\n",
+ "import math\n",
+ "import numpy\n",
+ "a='antiparallel spin'\n",
+ "b='parallel spin'\n",
+ "S=([a, b])\n",
+ "\n",
+ "\n",
+ "for i in range (0,1):\n",
+ " if S[i] == 'antiparallel spin' :\n",
+ " print(\"\\nFor Fermi types :\")\n",
+ " print(\"\\n\\n The selection rules for allowed transitions are : \\n\\tdelta I is zero \\n\\tdelta pi is plus \\nThe emited neutrino and electron have %s\")\n",
+ " print \"S(i,1)\"\n",
+ " elif S[i] == 'parallel spin':\n",
+ " print(\"\\nFor Gamow-Teller types :\")\n",
+ " print(\"\\nThe selection rules for allowed transitions are : \\n\\tdelta I is zero,plus one and minus one\\n\\tdelta pi is plus\\nThe emited neutrino and electron have %s\")\n",
+ " print(\"S(i,1)\") \n",
+ " \n",
+ "\n",
+ "## Calculation of ratio of transition probability\n",
+ "M_F = 1.; ## Matrix for Fermi particles\n",
+ "g_F = 1.; ## Coupling constant of fermi particles\n",
+ "M_GT = 5/3.; ## Matrix for Gamow Teller\n",
+ "g_GT = 1.24; ## Coupling constant of Gamow Teller\n",
+ "T_prob = g_F**2*M_F/(g_GT**2*M_GT); ## Ratio of transition probability\n",
+ "## Calculation of Space phase factor\n",
+ "e = 1.6e-19; ## Charge of an electron, coulomb\n",
+ "c = 3e+08; ## Velocity of light, metre per sec\n",
+ "K = 8.99e+9; ## Coulomb constant\n",
+ "R_0 = 1.2e-15; ## Distance of closest approach, metre\n",
+ "A = 57.; ## Mass number\n",
+ "Z = 28.; ## Atomic number \n",
+ "m_n = 1.6749e-27; ## Mass of neutron, Kg\n",
+ "m_p = 1.6726e-27; ## Mass of proton, Kg\n",
+ "m_e = 9.1e-31; ## Mass of electron. Kg\n",
+ "E_1 = 0.76; ## First excited state of nickel\n",
+ "delta_E = ((3*e**2*K/(5*R_0*A**(1/3.))*((Z+1.)**2-Z**2))-(m_n-m_p)*c**2)/1.6e-13; ## Mass difference, mega electron volts\n",
+ "E_0 = delta_E-(2*m_e*c**2)/1.6e-13; ## End point energy, mega electron volts\n",
+ "P_factor = (E_0-E_1)**5/E_0**5; ## Space phase factor \n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe ratio of transition probability =\",T_prob,\"\"and\"\\nThe space phase factor =\",P_factor,\"\");\n",
+ " \n",
+ "## Result\n",
+ "## The emited neutrino and electron have antiparallel spin\n",
+ "## For Gamow-Teller types :\n",
+ "## The selection rules for allowed transitions are : \n",
+ "##\tdelta I is zero,plus one and minus one\n",
+ "##\tdelta pi is plus\n",
+ "## The emited neutrino and electron have parallel spin\n",
+ "## The ratio of transition probability = 0.39\n",
+ "## The space phase factor = 0.62 a"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "For Fermi types :\n",
+ "\n",
+ "\n",
+ " The selection rules for allowed transitions are : \n",
+ "\tdelta I is zero \n",
+ "\tdelta pi is plus \n",
+ "The emited neutrino and electron have %s\n",
+ "S(i,1)\n",
+ "\n",
+ "The ratio of transition probability = 0.39 0.62 \n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter7.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter7.ipynb
new file mode 100755
index 00000000..366baa46
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter7.ipynb
@@ -0,0 +1,548 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:975fe0f8501e7b87da57a68589b53423b16e7dfd2c8a554c8366dabc3d2d73d7"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter7-Gamma -Radiation"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg292"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa7.1: : Page-292 (2011)\n",
+ "#find The distance to be moved for obtaining first order Bragg reflection \n",
+ "import math\n",
+ "h = 6.6261e-034; ## Planck's constant, joule sec\n",
+ "C = 2.998e+08; ## Velocity of light, metre per sec\n",
+ "f = 2.; ## Radius of focal circle, metre\n",
+ "d = 1.18e-010; ## Interplaner spacing for quartz crystal, metre\n",
+ "E_1 = 1.17*1.6022e-013; ## Energy of the gamma rays, joule\n",
+ "E_2 = 1.33*1.6022e-013; ## Energy of the gamma rays, joule\n",
+ "D = h*C*f*(1./E_1-1./E_2)*1./(2.*d); ##Distance to be moved for obtaining first order reflection for two different energies, metre\n",
+ "print'%s %.2e %s'%(\"\\nThe distance to be moved for obtaining first order Bragg reflection = \",D,\" metre\");\n",
+ "\n",
+ "## Result\n",
+ "## The distance to be moved for obtaining first order Bragg reflection = 1.08e-003 metre "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The distance to be moved for obtaining first order Bragg reflection = 1.08e-03 metre\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2-pg293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa7.2: : Page-293 (2011)\n",
+ "#find The energy of the gamma rays\n",
+ "import math\n",
+ "m_0 = 9.1094e-031; ## Rest mass of the electron, Kg\n",
+ "B_R = 1250e-06; ## Magnetic field,tesla metre\n",
+ "e = 1.6022e-019; ## Charge of the electron, coulomb\n",
+ "C = 3e+08; ## Velocity of the light, metre per sec\n",
+ "E_k = 0.089; ## Binding energy of the K-shell electron,MeV\n",
+ "v = B_R*e/(m_0*math.sqrt(1.+B_R**2.*e**2./(m_0**2*C**2))); ## Velocity of the photoelectron, metre per sec\n",
+ "E_pe = m_0/(1.6022e-013)*C**2*(1./math.sqrt(1-v**2/C**2)-1.); ## Energy of the photoelectron,MeV\n",
+ "E_g = E_pe+E_k; ## Energy of the gamma rays, MeV\n",
+ "print'%s %.3f %s'%(\"\\nThe energy of the gamma rays = \",E_g,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The energy of the gamma rays = 0.212 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy of the gamma rays = 0.212 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa7.3: : Page-292 (2011)\n",
+ "#find The attenuation of beam of X-rays in passing through human tissue \n",
+ "import math\n",
+ "a_c = 0.221; ## Attenuation coefficient, cm^2/g\n",
+ "A = (1-math.exp(-0.22))*100.; ## Attenuation of beam of X-rays in passing through human tissue\n",
+ "print'%s %.2f %s'%(\"\\nThe attenuation of beam of X-rays in passing through human tissue = \",math.ceil(A),\" percent\");\n",
+ "\n",
+ "## Result\n",
+ "## The attenuation of beam of X-rays in passing through human tissue = 20 percent "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The attenuation of beam of X-rays in passing through human tissue = 20.00 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg293"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa7.4: : Page-293 (2011)\n",
+ "#find The partial life time for gamma emission \n",
+ "import math\n",
+ "alpha_k = 45.; ## Ratio between decay constants\n",
+ "sum_alpha = 0.08; ## Sum of alphas\n",
+ "P = 0.35*1/60.; ## Probability of the isomeric transition,per hour\n",
+ "lambda_g = P*sum_alpha/alpha_k; ## Decay constant of the gamma radiations, per hour\n",
+ "T_g = 1/(lambda_g*365.*24.); ## Partial life time for gamma emission,years\n",
+ "print'%s %.2f %s'%(\"\\nThe partial life time for gamma emission = \",T_g,\" years\");\n",
+ "\n",
+ "## Result\n",
+ "## The partial life time for gamma emission = 11.008 years \n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The partial life time for gamma emission = 11.01 years\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg294"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa7.5: : Page-294 (2011)\n",
+ "import math\n",
+ "#find The required gamma width\n",
+ "A = 11.; ## Mass number of boron\n",
+ "E_g = 4.82; ## Energy of the gamma radiation, mega electron volts\n",
+ "W_g = 0.0675*A**(2./3.)*E_g**3; ## Gamma width, mega electron volts\n",
+ "print'%s %.2f %s'%(\"\\nThe required gamma width = \",W_g,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The required gamma width = 37.39 MeV \n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The required gamma width = 37.39 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa7.8: : Page-295 (2011)\n",
+ "#find The excitation energy and The angular momentum\n",
+ "import math\n",
+ "e = 1.6022e-19; ## Charge of an electron, coulomb\n",
+ "BR = 2370e-06; ## Magnetic field in an orbit, tesla metre\n",
+ "m_0 = 9.1094e-31; ## Mass of an electron, Kg\n",
+ "c = 3e+08; ## Velocity of light, metre per sec\n",
+ "v = 1/math.sqrt((m_0/(BR*e))**2.+1./c**2); ## velocity of the particle, metre per sec\n",
+ "E_e = m_0*c**2*((1.-(v/c)**2)**(-1/2.)-1)/1.6e-13; ## Energy of an electron, MeV\n",
+ "E_b = 0.028; ## Binding energy, MeV\n",
+ "E_g = E_e+E_b; ## Excitation energy, MeV\n",
+ "alpha_k = 0.5; ## K conversion coefficient\n",
+ "Z = 49.; ## Number of protons\n",
+ "alpha = 1./137.; ## Fine structure constant\n",
+ "L = (1/(1.-(Z**3/alpha_k*alpha**4.*(2.*0.511/0.392)**(15./2.))))/2.; ## Angular momentum\n",
+ "l = 1; ## Orbital angular momentum\n",
+ "I = l-1/2.; ## Parity\n",
+ "print(\"\\nFor K-electron state:\" )\n",
+ "print' %s %.2f %s %.2f %s %.2f %s'%( \"The excitation energy = \",E_g,\" MeV\" and \"The angular momentum = \",math.ceil(L),\" \" and \"\\nThe parity : \",I,\"\");\n",
+ "## Result\n",
+ "## For K-electron state:\n",
+ "## The excitation energy = 0.393 MeV\n",
+ "## The angular momentum = 5\n",
+ "## The parity : 0.5 \n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "For K-electron state:\n",
+ " The excitation energy = 0.39 The angular momentum = 5.00 \n",
+ "The parity : 0.50 \n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg295"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa7.9: : Page-295 (2011)\n",
+ "#find The radioactive life time\n",
+ "import math\n",
+ "c = 3e+10; ## Velocity of light, centimetre per sec\n",
+ "R_0 = 1.4e-13; ## Distance of closest approach, centimetre \n",
+ "alpha = 1./137.; ## Fine scattering constant\n",
+ "A = 17.; ## Mass number\n",
+ "E_g = 5.*1.6e-06; ## Energy of gamma transition, ergs\n",
+ "h_cut = 1.054571628e-27; ## Reduced planck constant, ergs per sec\n",
+ "D = c/4.*R_0**2.*alpha*(E_g/(h_cut*c))**3.*A**(2./3.); ## Disintegration constant, per sec\n",
+ "tau = 1/D; ## Radioactive lifr\\e time, sec\n",
+ "print'%s %.1e %s'%(\"\\nThe radioactive life time = \",tau,\" sec\");\n",
+ "\n",
+ "## Result\n",
+ "## The radioactive life time = 9e-018 sec \n",
+ "print(\"error in answer due to round off error\")\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The radioactive life time = 8.7e-18 sec\n",
+ "error in answer due to round off error\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg296"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa7.10: : Page-296 (2011)\n",
+ "#find The possible multipolarities are\n",
+ "import math\n",
+ "l = 2,3,4\n",
+ "print(\"\\nThe possible multipolarities are \")\n",
+ "for l in range( 2,4):\n",
+ " if l == 2:\n",
+ " print'%s %.2f %s'%(\"E\",l,\" \" );\n",
+ " elif l == 3:\n",
+ " print'%s %.2f %s'%(\" M\",l,\" \");\n",
+ " elif l == 4:\n",
+ " print'%s %.2f %s'%(\" and E \",l,\" \");\n",
+ " \n",
+ "\n",
+ "for l in range( 2,4):\n",
+ " if l == 2 :\n",
+ " print'%s %.2f %s'%(\"\\nThe transition E\",l,\" dominates\");\n",
+ " \n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## The possible multipolarities are E2, M3 and E4\n",
+ "## The transition E2 dominates \n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The possible multipolarities are \n",
+ "E 2.00 \n",
+ " M 3.00 \n",
+ "\n",
+ "The transition E 2.00 dominates\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex13-pg297"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa7.13: : Page-297 (2011)\n",
+ "#find The relative source absorber velocity\n",
+ "import math\n",
+ "E_0 = 0.014*1.6022e-13; ## Energy of the gamma rays, joule\n",
+ "A = 57.; ## Mass number\n",
+ "m = 1.67e-27; ## Mass of each nucleon, Kg\n",
+ "c = 3e+08; ## Velocity of light, metre per sec\n",
+ "N = 1000.; ## Number of atoms in the lattice\n",
+ "v = E_0/(A*N*m*c); ## Ralative velocity, metre per sec\n",
+ "print'%s %.2f %s'%(\"\\nThe relative source absorber velocity = \",v,\" m/s\");\n",
+ "\n",
+ "## Result\n",
+ "## The relative source absorber velocity = 0.079 m/s \n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The relative source absorber velocity = 0.08 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg297"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa7.14: : Page-297 (2011)\n",
+ "#find The required frequency shift of the photon\n",
+ "import math\n",
+ "g = 9.8; ## Acceleration due to gravity, metre per square sec\n",
+ "c = 3e+08; ## Velocity of light, metre per sec\n",
+ "y = 20.; ## Vertical distance between source and absorber, metre\n",
+ "delta_v = g*y/c**2; ## Frequency shift\n",
+ "print'%s %.2e %s'%(\"\\nThe required frequency shift of the photon = \", delta_v,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The required frequency shift of the photon = 2.18e-015 \n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The required frequency shift of the photon = 2.18e-15 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex15-pg246"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa6.15: : Page-246 (2011)\n",
+ "#find they are parallel spin or anti\n",
+ "import math\n",
+ "import numpy\n",
+ "a='antiparallel spin'\n",
+ "b='parallel spin'\n",
+ "S=([a, b])\n",
+ "\n",
+ "\n",
+ "for i in range (0,1):\n",
+ " if S[i] == 'antiparallel spin' :\n",
+ " print(\"\\nFor Fermi types :\")\n",
+ " print(\"\\n\\n The selection rules for allowed transitions are : \\n\\tdelta I is zero \\n\\tdelta pi is plus \\nThe emited neutrino and electron have %s\")\n",
+ " print \"S(i,1)\"\n",
+ " elif S[i] == 'parallel spin':\n",
+ " print(\"\\nFor Gamow-Teller types :\")\n",
+ " print(\"\\nThe selection rules for allowed transitions are : \\n\\tdelta I is zero,plus one and minus one\\n\\tdelta pi is plus\\nThe emited neutrino and electron have %s\")\n",
+ " print(\"S(i,1)\") \n",
+ " \n",
+ "\n",
+ "## Calculation of ratio of transition probability\n",
+ "M_F = 1.; ## Matrix for Fermi particles\n",
+ "g_F = 1.; ## Coupling constant of fermi particles\n",
+ "M_GT = 5/3.; ## Matrix for Gamow Teller\n",
+ "g_GT = 1.24; ## Coupling constant of Gamow Teller\n",
+ "T_prob = g_F**2*M_F/(g_GT**2*M_GT); ## Ratio of transition probability\n",
+ "## Calculation of Space phase factor\n",
+ "e = 1.6e-19; ## Charge of an electron, coulomb\n",
+ "c = 3e+08; ## Velocity of light, metre per sec\n",
+ "K = 8.99e+9; ## Coulomb constant\n",
+ "R_0 = 1.2e-15; ## Distance of closest approach, metre\n",
+ "A = 57.; ## Mass number\n",
+ "Z = 28.; ## Atomic number \n",
+ "m_n = 1.6749e-27; ## Mass of neutron, Kg\n",
+ "m_p = 1.6726e-27; ## Mass of proton, Kg\n",
+ "m_e = 9.1e-31; ## Mass of electron. Kg\n",
+ "E_1 = 0.76; ## First excited state of nickel\n",
+ "delta_E = ((3*e**2*K/(5*R_0*A**(1/3.))*((Z+1.)**2-Z**2))-(m_n-m_p)*c**2)/1.6e-13; ## Mass difference, mega electron volts\n",
+ "E_0 = delta_E-(2*m_e*c**2)/1.6e-13; ## End point energy, mega electron volts\n",
+ "P_factor = (E_0-E_1)**5/E_0**5; ## Space phase factor \n",
+ "print'%s %.2f %s %.2f %s '%(\"\\nThe ratio of transition probability =\",T_prob,\"\"and\"\\nThe space phase factor =\",P_factor,\"\");\n",
+ " \n",
+ "## Result\n",
+ "## The emited neutrino and electron have antiparallel spin\n",
+ "## For Gamow-Teller types :\n",
+ "## The selection rules for allowed transitions are : \n",
+ "##\tdelta I is zero,plus one and minus one\n",
+ "##\tdelta pi is plus\n",
+ "## The emited neutrino and electron have parallel spin\n",
+ "## The ratio of transition probability = 0.39\n",
+ "## The space phase factor = 0.62 a"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "For Fermi types :\n",
+ "\n",
+ "\n",
+ " The selection rules for allowed transitions are : \n",
+ "\tdelta I is zero \n",
+ "\tdelta pi is plus \n",
+ "The emited neutrino and electron have %s\n",
+ "S(i,1)\n",
+ "\n",
+ "The ratio of transition probability = 0.39 0.62 \n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter8.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter8.ipynb
new file mode 100755
index 00000000..fa28f9b7
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter8.ipynb
@@ -0,0 +1,279 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:e2f7c7551d6e417829aadd6f90608da5e4f835298a7f6245d9b1f454185575b2"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter8-Nuclear Forces"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa8.3 : : Page-349 (2011)\n",
+ "#find The probability that the proton moves within the range of neutron\n",
+ "import math\n",
+ "b = 1.9e-15; ## Width of square well potential, metre\n",
+ "h_kt = 1.054571e-034; ## Reduced planck's constant, joule sec\n",
+ "c = 3e+08; ## Velocity of light, metre per sec\n",
+ "m_n = 1.67e-27; ## Mass of a nucleon , Kg\n",
+ "V_0 = 40*1.6e-13; ## Depth, metre\n",
+ "E_B = (V_0-(1/(m_n*c**2)*(math.pi*h_kt*c/(2*b))**2))/1.6e-13; ## Binding energy, mega electron volts\n",
+ "alpha = math.sqrt(m_n*c**2*E_B*1.6e-13)/(h_kt*c); ## scattering co efficient, per metre\n",
+ "P = (1+1/(alpha*b))**-1.; ## Probability\n",
+ "R_mean = math.sqrt (b**2./2.*(1./3.+4./math.pi**2.+2.5)); ## Mean square radius, metre\n",
+ "print'%s %.2f %s %.2e %s'%(\"\\nThe probability that the proton moves within the range of neutron = \",P,\" \\n\" \"The mean square radius of the deuteron = \",R_mean,\" metre\")\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## The probability that the proton moves within the range of neutron = 0.50 \n",
+ "## The mean square radius of the deuteron = 2.42e-015 metre \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The probability that the proton moves within the range of neutron = 0.50 \n",
+ "The mean square radius of the deuteron = 2.42e-15 metre\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa8.5 : : Page-349 (2011)\n",
+ "#find The total cross section for n-p scattering \n",
+ "import math\n",
+ "a_t = 5.38e-15;\n",
+ "a_s = -23.7e-15;\n",
+ "r_ot = 1.70e-15;\n",
+ "r_os = 2.40e-15;\n",
+ "m = 1.6748e-27;\n",
+ "E = 1.6e-13;\n",
+ "h_cut = 1.0549e-34;\n",
+ "K_sqr = m*E/h_cut**2;\n",
+ "sigma = 1/4.*(3.*4*math.pi*a_t**2./(a_t**2.*K_sqr+(1.-1/2.*K_sqr*a_t*r_ot)**2)+4*math.pi*a_s**2/(a_s**2*K_sqr+(1-1./2.*K_sqr*a_s*r_os)**2))*1e+028; ## Total cross-section for n-p scattering, barn\n",
+ "print'%s %.2f %s'%(\"\\nThe total cross section for n-p scattering = \",sigma,\" barn\");\n",
+ "\n",
+ "## Result\n",
+ "## The total cross section for n-p scattering = 2.911 barn \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The total cross section for n-p scattering = 2.91 barn\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8-pg351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa8.8 : : Page-351 (2011)\n",
+ "#find The possible angular momentum states with their parities are as follows\n",
+ "import math\n",
+ "S = 1.; ## Spin angular momentum(s1+-s2), whereas s1 is the spin of proton and s2 is the spin of neutron.\n",
+ "m = 2.*S+1.; ## Spin multiplicity\n",
+ "j = 1.; ## Total angular momentum\n",
+ "print(\"\\nThe possible angular momentum states with their parities are as follows : \");\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\n \",m, \" \" and \"S has even parity \",j,\"\");\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\n \",m,\" \" and \"P has odd parity \", j,\"\");\n",
+ "print'%s %.2f %s %.2f %s'%(\"\\n \",m, \" \" and \"S has odd parity \",j,\"\"); \n",
+ "S = 0.;\n",
+ "m = 2.*S+1.\n",
+ "print(m)\n",
+ "print'%s %.2f %s %.2f %s '%(\"\\n \",m,\" \" and \"P has odd parity \", j,\"\");\n",
+ " \n",
+ "## Result \n",
+ "## The possible angular momentum states with their parities are as follows : \n",
+ "## 3S1 has even parity \n",
+ "## 3P1 has odd parity \n",
+ "## 3D1 has even parity\n",
+ "## 1P1 has odd parity "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The possible angular momentum states with their parities are as follows : \n",
+ "\n",
+ " 3.00 S has even parity 1.00 \n",
+ "\n",
+ " 3.00 P has odd parity 1.00 \n",
+ "\n",
+ " 3.00 S has odd parity 1.00 \n",
+ "1.0\n",
+ "\n",
+ " 1.00 P has odd parity 1.00 \n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# Exa8.9 : : Page-351 (2011)\n",
+ "#find The possible states are\n",
+ "\n",
+ "print(\"\\nThe possible states are : \");\n",
+ "#For s = 0\n",
+ "s = 0; # Spin angular momentum\n",
+ "m = 2*s+1; # Spin multiplicity\n",
+ "for j in range(0,3): # Total angular momentum\n",
+ " l = j\n",
+ " if l == 0:\n",
+ " print\"%s %.1f %s %.d %s \"%(\"\",j,\"\"and \"S\",m,\"\")\n",
+ " elif l == 2:\n",
+ " print\"%s %.1f %s %.d %s \"%(\"\",j,\"\"and \"D\",m,\"\") \n",
+ " \n",
+ "\n",
+ "#For s = 1\n",
+ "s = 1;\n",
+ "m = 2*s+1;\n",
+ "l = 2\n",
+ "for j in range(0,3): \n",
+ " if j == 0:\n",
+ " print\"%s %.1f %s %.d %s \"%(\"\",j,\"\"and \"P\",m,\"\")\n",
+ " elif j ==1:\n",
+ " print\"%s %.d %s %.d %s \"%(\"\",j,\"\"and \"P\",m,\"\")\n",
+ " elif j ==2:\n",
+ " print\"%s %.d %s %.d %s \"%(\"\",j,\"\"and \"P\",m,\"\")\n",
+ " \n",
+ "\n",
+ "for j in range(2,3):\n",
+ " print\"%s %.d %s %.d %s \"%(\"\",j,\"\" and \"F\",m,\"\")\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "#Possible states are : \n",
+ "# The possible states are : \n",
+ "# 0S1, 2D1, 0P3, 1P3, 2P3 and 2F3 \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The possible states are : \n",
+ " 0.0 1 \n",
+ " 2.0 1 \n",
+ " 0.0 3 \n",
+ " 1 3 \n",
+ " 2 3 \n",
+ " 2 3 \n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10-pg352"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa8.10 : : Page-352 (2011)\n",
+ "#find The kinetic energy of each nucleon and The total kinetic energy\n",
+ "import math\n",
+ "r = 2e-015; ## Range of nuclear force, metre\n",
+ "h_kt = 1.0546e-34; ## Reduced value of Planck's constant, joule sec\n",
+ "m = 1.674e-27; ## Mass of each nucleon, Kg\n",
+ "K = round (2*h_kt**2./(2*m*r**2*1.6023e-13)); ## Kinetic energy of each nucleon in centre of mass frame, mega electron volts\n",
+ "K_t = 2.*K; ## Total kinetic energy, mega electron volts\n",
+ "K_inc = 2.*K_t; ## Kinetic energy of the incident nucleon, mega electron volts\n",
+ "print'%s %.2f %s %.2f %s %.2f %s '%(\"\\nThe kinetic energy of each nucleon = \",K,\" MeV\" and \"The total kinetic energy =\",K_t,\" MeV\"and \"The kinetic energy of the incident nucleon =\",K_inc,\" MeV\")\n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The kinetic energy of each nucleon = 10.00 The total kinetic energy = 20.00 The kinetic energy of the incident nucleon = 40.00 MeV \n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/Chapter9.ipynb b/Nuclear_Physics_by_D._C._Tayal/Chapter9.ipynb
new file mode 100755
index 00000000..7fa41006
--- /dev/null
+++ b/Nuclear_Physics_by_D._C._Tayal/Chapter9.ipynb
@@ -0,0 +1,536 @@
+{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:4f16fb24f1dd6d3239c94791f7a7ef08310ef6ebb56e580bd1901d7fe6b9c932"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter9-Nuclear Models"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1-pg389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa9.1 : : Page-389 (2011) \n",
+ "#find The Fermi energy of neutron and proton\n",
+ "import math\n",
+ "h_cut = 1.054e-034; ## Reduced Planck's constant, joule sec\n",
+ "rho = 2e+044; ## Density of the nuclear matter, kg per metre cube\n",
+ "V = 238./rho; ## Volume of the nuclear matter, metre cube\n",
+ "## For neutron\n",
+ "N = 238.-92.; ## Number of neutrons\n",
+ "M = 1.67482e-027; ## Mass of a neutron, kg\n",
+ "e = 1.602e-019; ## Energy equivalent of 1 eV, J/eV\n",
+ "E_f = (3*math.pi**2)**(2./3.)*h_cut**2/(2*M)*(N/V)**(2/3.)/e; ## Fermi energy of neutron, eV \n",
+ "print'%s %.2f %s'%(\"\\nThe Fermi energy of neutron = \",E_f/1e+006,\" MeV\")\n",
+ "## For proton\n",
+ "N = 92.; ## Number of protons\n",
+ "M = 1.67482e-027; ## Mass of a proton, kg\n",
+ "e = 1.602e-019; ## Energy equivalent of 1 eV, J/eV\n",
+ "E_f = (3*math.pi**2)**(2/3.)*h_cut**2/(2.*M)*(N/V)**(2./3.)/e; ## Fermi energy of neutron, eV \n",
+ "print'%s %.2f %s'%(\"\\nThe Fermi energy of proton = \",E_f/1e+006,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The Fermi energy of neutron = 48.92 MeV\n",
+ "## The Fermi energy of proton = 35.96 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The Fermi energy of neutron = 48.92 MeV\n",
+ "\n",
+ "The Fermi energy of proton = 35.96 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3-pg390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa9.3 : : Page-390 (2011)\n",
+ "import math\n",
+ "#find radius of neutron star\n",
+ "h_cut = 1.0545e-34; ## Reduced Planck's constant, joule sec\n",
+ "G = 6.6e-11; ## Gravitational constant, newton square metre per square Kg \n",
+ "m = 10**30.; ## Mass of the star, Kg\n",
+ "m_n = 1.67e-27; ## Mass of the neutron, Kg\n",
+ "R = (9*math.pi/4.)**(2./3.)*h_cut**2/(G*(m_n)**3)*(m_n/m)**(1/3.); ## Radius of the neutron star, metre\n",
+ "print'%s %.1e %s'%(\"\\nThe radius of the neutron star = \",R,\" metre\");\n",
+ "\n",
+ "## Result\n",
+ "## The radius of the neutron star = 1.6e+004 metre "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The radius of the neutron star = 1.6e+04 metre\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4-pg391"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa9.4 : : Page-391 (2011)\n",
+ "#find is they will stable or not\n",
+ "import math\n",
+ "A = 77.; ## Mass number of the isotopes\n",
+ "Z = round (A/((0.015*A**(2/3.))+2.)); ## Atomic number of stable isotope\n",
+ "## Check the stability !!!!!\n",
+ "if Z == 34:\n",
+ " print (\"\\n Se\",( Z,A),\" is stable\" and \"As \",(Z-1,A),\"\" and \"Br\",Z+1,A, \"are unstable\")\n",
+ "elif Z == 33 :\n",
+ " print'%s %.2f %s'%(\"\\nAs( %d,%d) is stable \\nSe (%d,%d) and Br(%d,%d) are unstable\", Z, A, Z+1, A, Z+2, A);\n",
+ "elif Z == 35 :\n",
+ " print'%s %.2f %s'%(\"\\nBr( %d,%d) is stable \\nSe (%d,%d) and As(%d,%d) are unstable\",Z,A,Z-2,A,Z-1,A); \n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## Se( 34,77) is stable \n",
+ "## As (33,77) and Br(35,77) are unstable "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "('\\n Se', (34.0, 77.0), 'As ', (33.0, 77.0), '', 35.0, 77.0, 'are unstable')\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5-pg391"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa9.5 : : Page-391 (2011)\n",
+ "#find The energy difference between neutron shells\n",
+ "import math\n",
+ "m_40 = 39.962589; ## Mass of calcium 40, atomic mass unit\n",
+ "m_41 = 40.962275; ## Mass of calcium 41, atomic mass unit\n",
+ "m_39 = 38.970691; ## Mass of calcium 39, atomic mass unit \n",
+ "m_n = 1.008665; ## Mass of the neutron, atomic mass unit\n",
+ "BE_1d = (m_39+m_n-m_40)*931.5; ## Binding energy of 1d 3/2 neutron, mega electron volts\n",
+ "BE_1f = (m_40+m_n-m_41)*931.5; ## Binding energy of 1f 7/2 neutron, mega electron volts\n",
+ "delta = BE_1d-BE_1f; ## Energy difference between neutron shells, mega electron volts\n",
+ "print'%s %.2f %s'%(\"\\nThe energy difference between neutron shells = \",delta,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The energy difference between neutron shells = 7.25 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The energy difference between neutron shells = 7.25 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7-pg392"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa9.7 : : Page-392 (2011)\n",
+ "import math\n",
+ "#find The angular frequency for oxygen 17\n",
+ "h_cut = 1.0545e-34; ## Reduced Planck's constant, joule sec\n",
+ "R = 1.2e-15; ## Distance of closest approach, metre\n",
+ "m = 1.67482e-27; ## Mass of the nucleon, Kg\n",
+ "omega_Ni=1.60e+022;\n",
+ "## For O-17\n",
+ "for A in range(17,60): ## Mass numbers\n",
+ " if A == 17:\n",
+ " omega_O = 5.*3.**(1/3.)*h_cut*17**(-1./3.)/(2.**(7/3.)*m*R**2.); ## Angular frequency of oxygen \n",
+ "## For Ni-60\n",
+ " elif A == 60:\n",
+ " omega_Ni = 5*3**(1/3.)*h_cut*60**(-1/3.)/(2**(7/3.)*m*R**2); ## Angular frequency of nickel\n",
+ "\n",
+ "print (\"\\nThe angular frequency for oxygen 17 = \",omega_O,\"\" and \"\\nThe angular frequency for nickel 60 = \",omega_Ni,\"\");\n",
+ "\n",
+ "## Result\n",
+ "## The angular frequency for oxygen 17 = 2.43e+022 \n",
+ "## The angular frequency for nickel 60 = 1.60e+022 "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "('\\nThe angular frequency for oxygen 17 = ', 2.43317537466611e+22, '', 1.6e+22, '')\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9-pg393"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa9.9 : : Page-393 (2011)\n",
+ "#find The angular momentum is 5/2 and the parity is +1 for and -1 and 0\n",
+ "import math\n",
+ "import numpy\n",
+ "Z = numpy.zeros((5,1));\n",
+ "N = numpy.zeros((5,1));\n",
+ "E={}\n",
+ "## Elements allocated\n",
+ "E[0,0] = 'Carbon'\n",
+ "E[1,0] = 'Boron'\n",
+ "E[2,0] = 'Oxygen'\n",
+ "E[3,0] = 'Zinc'\n",
+ "E[4,0] = 'Nitrogen'\n",
+ "Z[0,0] = 6; ## Number of proton in carbon nuclei\n",
+ "Z[1,0] = 5; ## Number of proton in boron nuclei\n",
+ "Z[2,0] = 8; ## Number of proton in oxygen nuclei\n",
+ "Z[3,0] = 30; ## Number of proton in zinc nuclei\n",
+ "Z[4,0] = 7; ## Number of proton in nitrogen nuclei\n",
+ "N[0,0] = 6; ## Mass number of carbon\n",
+ "N[0,0] = 6; ## Mass number of boron\n",
+ "N[2,0] = 9; ## Mass number of oxygen\n",
+ "N[3,0] = 37; ## Mass number of zinc\n",
+ "N[4,0] = 9; ## Mass number of nitrogem\n",
+ "for i in range (0,5):\n",
+ " if Z[i,0] == 8:\n",
+ " print(\"\\nThe angular momentum is 5/2 and the parity is +1 for \", E[i,0],\"\");\n",
+ " elif Z[i,0] == 5:\n",
+ " print(\"\\nThe angular momentum is 3/2 and the parity is -1 for \", E[i,0],\"\");\n",
+ " \n",
+ " elif Z[i,0] == N[i,0]:\n",
+ " print (\"\\nThe angular mometum is 0 and the parity is +1 for \", E[i,0],\"\");\n",
+ " \n",
+ " elif N[i,0]-Z[i,0] == 2:\n",
+ " print(\"\\nThe angular momentum is 2 and the parity is -1 for \", E[i,0],\"\");\n",
+ " \n",
+ " elif N[i,0]-Z[i,0] == 7:\n",
+ " print(\"The angular momentum is 5/2 and the parity is -1 for\", E[i,0],\"\");\n",
+ " \n",
+ "\n",
+ "## Result\n",
+ "## The angular mometum is 0 and the parity is +1 for Carbon\n",
+ "## The angular momentum is 3/2 and the parity is -1 for Boron\n",
+ "## The angular momentum is 5/2 and the parity is +1 for Oxygen \n",
+ "## The angular momentum is 5/2 and the parity is -1 for Zinc\n",
+ "## The angular momentum is 2 and the parity is -1 for Nitrogen \n",
+ "print(\"we can not print directly \")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "('\\nThe angular mometum is 0 and the parity is +1 for ', 'Carbon', '')\n",
+ "('\\nThe angular momentum is 3/2 and the parity is -1 for ', 'Boron', '')\n",
+ "('\\nThe angular momentum is 5/2 and the parity is +1 for ', 'Oxygen', '')\n",
+ "('The angular momentum is 5/2 and the parity is -1 for', 'Zinc', '')\n",
+ "('\\nThe angular momentum is 2 and the parity is -1 for ', 'Nitrogen', '')\n",
+ "we can not print directly \n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11-pg394"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Page-394 (2011)\n",
+ "import math\n",
+ "import numpy\n",
+ "\n",
+ "R_0 = 1.2e-015; ## Distance of closest approach, metre\n",
+ "## Mass number of the nuclei are allocated below :\n",
+ "N = numpy.zeros((4,1))\n",
+ "N[0,0] = 17; ## for oxygen\n",
+ "N[1,0] = 33; ## for sulphur\n",
+ "N[2,0] = 63; ## for copper\n",
+ "N[3,0] = 209; ## for bismuth\n",
+ "for i in range (1,4):\n",
+ " if N[i,0] == 17:\n",
+ " print(\"\\n For Oxygen : \")\n",
+ " I = 5/2.; ## Total angular momentum\n",
+ " l = 2.; ## Orbital angular momentum\n",
+ " mu = -1.91; ## for odd neutron and I = l+1/2\n",
+ " Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1/3.))**2*(10**28); ## Quadrupole moment of oxygen, barnQ\n",
+ " print\"%s %.2f %s %.2f %s \"%(\"\\n The value of magnetic moment is : \",mu,\"\"and \" \\n The value of quadrupole moment is : \",Q,\" barn\");\n",
+ " elif N[i,0] == 33:\n",
+ " print(\"\\n\\n For Sulphur : \")\n",
+ " I = 3./2.; ## Total angular momentum\n",
+ " l = 2.; ## Orbital angular momentum\n",
+ " mu = 1.91*I/(I+1.); ## for odd neutron and I = l-1/2\n",
+ " Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1/3.))**2*(10**28); ## Quadrupole moment of sulphur, barn\n",
+ " print\"%s %.2f %s %.2f %s \"%(\"\\n The value of magnetic moment is : \",mu,\"\"and \" \\n The value of quadrupole moment is : \",Q,\" barn\"); \n",
+ " elif N[i,0] == 63:\n",
+ " print(\"\\n\\n For Copper : \")\n",
+ " I = 3./2.; ## Total angular momentum\n",
+ " l = 1.; ## Orbital angular momentum\n",
+ " mu = I+2.29; ## for odd protons and I = l+1/2\n",
+ " Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1./3.))**2*(10**28); ## Quadrupole momentum of copper, barn\n",
+ " print\"%s %.2f %s %.2f %s \"% (\" The value of magnetic moment is : \",mu,\" \"and \"\\n The value of quadrupole moment is :\" ,Q, \"barn\");\n",
+ " elif N[i,0] == 209:\n",
+ " print(\" For Bismuth : \")\n",
+ " I = 9/2; ## Total angular momentum\n",
+ " l = 5; ## Orbital angular momentum\n",
+ " mu = I-2.29*I/(I+1); ## for odd protons and I = l-1/2\n",
+ " Q = -3./5.*(2.*I-1.)/(2.*I+2.)*(R_0*N[i,0]**(1/3.))**2*(10**28); ## Quadrupole momentum of bismuth, barn\n",
+ " print\"%s %.2f %s %.2f %s \"%(\" The value of magnetic moment is : \",mu,\"\"and \" \\n The value of quadrupole moment is : \",Q,\" barn\");\n",
+ " print('due to rounding error we can not get for oxygen result')\n",
+ "## Result\n",
+ "\n",
+ "\n",
+ "## For Sulphur : \n",
+ "## The value of magnetic moment is : 1.146 \n",
+ "## The value of quadrupole moment is : -0.0356 barn\n",
+ "\n",
+ "## For Copper : \n",
+ "## The value of magnetic moment is : 3.79 \n",
+ "## The value of quadrupole moment is : -0.0547 barn\n",
+ "\n",
+ "## For Bismuth : \n",
+ "## The value of magnetic moment is : 2.63 \n",
+ "## The value of quadrupole moment is : -0.221 barn \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " For Sulphur : \n",
+ "\n",
+ " The value of magnetic moment is : 1.15 -0.04 barn \n",
+ "\n",
+ "\n",
+ " For Copper : \n",
+ " The value of magnetic moment is : 3.79 \n",
+ " The value of quadrupole moment is : -0.05 barn \n",
+ " For Bismuth : \n",
+ " The value of magnetic moment is : 2.17 -0.21 barn \n",
+ "due to rounding error we can not get for oxygen result\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex12-pg395"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa9.12 : : Page-395 (2011)\n",
+ "#find The kinetic energy of iron nuclei\n",
+ "import math\n",
+ "h_cut = 1.054571628e-34; ## Redued planck's constant, joule sec\n",
+ "a = 1e-014; ## Distance of closest approach, metre\n",
+ "m = 1.67e-27; ## Mass of each nucleon, Kg\n",
+ "KE = 14*math.pi**2*h_cut**2./(2.*m*a**2*1.6e-13); ## Kinetic energy of iron nucleus, MeV\n",
+ "print'%s %.2f %s'%(\"\\nThe kinetic energy of iron nuclei =\",KE,\" MeV\");\n",
+ "\n",
+ "## Result\n",
+ "## The kinetic energy of iron nuclei = 28.76 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The kinetic energy of iron nuclei = 28.76 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex14-pg396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa9.14 : : Page-396 (2011)\n",
+ "#find The electric quadrupole of scandium nucleus\n",
+ "import math\n",
+ "R_0 = 1.2e-15; ## Distance of closest approach, metre\n",
+ "j = 7/2.; ## Total angular momentum\n",
+ "A = 41.; ## Mass number of Scandium\n",
+ "Z = 20.; ## Atomic number of Calcium\n",
+ "Q_Sc = -(2*j-1)/(2.*j+2.)*(R_0*A**(1/3.))**2; ## Electric quadrupole of Scandium nucleus, Sq. m\n",
+ "Q_Ca = Z/(A-1)**2*abs(Q_Sc); ## Electric quadrupole of calcium nucleus, Sq. m\n",
+ "print'%s %.2e %s %.2e %s '%(\"\\nThe electric quadrupole of scandium nucleus = \",Q_Sc,\" square metre\" and \"\\nThe electric quadrupole of calcium nucleus = \",Q_Ca,\" square metre\");\n",
+ "\n",
+ "## Result\n",
+ "## The electric quadrupole of scandium nucleus = -1.14e-029 square metre \n",
+ "## The electric quadrupole of calcium nucleus = 1.43e-031 square metre "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "The electric quadrupole of scandium nucleus = -1.14e-29 \n",
+ "The electric quadrupole of calcium nucleus = 1.43e-31 square metre \n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex16-pg398"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "## Exa9.16 : : Page-398 (2011)\n",
+ "#find The energy for 4+ tungsten state and 6+state\n",
+ "import math\n",
+ "h_cut_sqr_upon_2f = 0.01667; ## A constant value, joule square per sec cube\n",
+ "for I in range (4,6):\n",
+ " if I == 4:\n",
+ " E = I*(I+1)*h_cut_sqr_upon_2f\n",
+ " print'%s %.2f %s'%(\"The energy for 4+ tungsten state = \",E,\" MeV\");\n",
+ " elif I == 6:\n",
+ " E = I*(I+1)*h_cut_sqr_upon_2f;\n",
+ " print'%s %.2f %s'%(\"\\nThe energy for 6+ tungsten state = \",E,\" MeV\"); \n",
+ " \n",
+ "\n",
+ "\n",
+ "## Result\n",
+ "## The energy for 4+ tungsten state = 0.333 MeV\n",
+ "## The energy for 6+ tungsten state = 0.700 MeV "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The energy for 4+ tungsten state = 0.33 MeV\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+} \ No newline at end of file
diff --git a/Nuclear_Physics_by_D._C._Tayal/README.txt b/Nuclear_Physics_by_D._C._Tayal/README.txt
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+Contributed By: harsha vardhan
+Course: msc
+College/Institute/Organization: iitbombay
+Department/Designation: msc chem
+Book Title: Nuclear Physics
+Author: D. C. Tayal
+Publisher: Himalaya Publishing House, Mumbai
+Year of publication: 2011
+Isbn: 978-93-5024-743-3
+Edition: 5 \ No newline at end of file
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