{ "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": {} } ] }