path: root/Nuclear_Physics/Chapter1.ipynb

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  {
   "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": {}
  }
 ]
}