{
"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": {}
}
]
}