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| author | kinitrupti | 2017-05-12 18:53:46 +0530 |
|---|---|---|
| committer | kinitrupti | 2017-05-12 18:53:46 +0530 |
| commit | 6279fa19ac6e2a4087df2e6fe985430ecc2c2d5d (patch) | |
| tree | 22789c9dbe468dae6697dcd12d8e97de4bcf94a2 /Nuclear_Physics_by_D._C._Tayal/Chapter6.ipynb | |
| parent | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (diff) | |
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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": {}
+ }
+ ]
+}
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