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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/Chapter9.ipynb | |
| parent | d36fc3b8f88cc3108ffff6151e376b619b9abb01 (diff) | |
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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 |