{
"metadata": {
"name": "",
"signature": "sha256:308367390142ca69701c27a60b649f58dec3483f71d5fe87366628b17198229c"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter10-Nuclear reactions"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex1-pg455"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.1 : : Page-455 (2011) \n",
"#calculate The Q-value for the fromation\n",
"import math\n",
"M = 47.668; ## Total mass of reaction, MeV\n",
"E = 44.359; ## Total energy, MeV\n",
"Q = M-E; ## Q-value, MeV\n",
"print'%s %.2f %s'%(\"\\nThe Q-value for the formation of P30 = \",Q,\" MeV\");\n",
"\n",
"## Result\n",
"## The Q-value for the formation of P30 = 3.309 MeV "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The Q-value for the formation of P30 = 3.31 MeV\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2-pg455"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.2 : : Page-455 (2011) \n",
"#calculate The Q-value for the reaction\n",
"import math\n",
"E_x = 7.70; ## Energy of the alpha particle, MeV\n",
"E_y = 4.44; ## Energy of the proton, MeV\n",
"m_x = 4.0; ## Mass number of alpha particle\n",
"m_y = 1.0; ## Mass number of protium ion\n",
"M_X = 14; ## Mass number of nitrogen nucleus\n",
"M_Y = 17; ## Mass number of oxygen nucleus\n",
"theta = 90*3.14/180.; ## Angle between incident beam direction and emitted proton, degree\n",
"A_x = 4.0026033; ## Atomic mass of alpha particle, u\n",
"A_X = 14.0030742; ## Atomic mass of nitrogen nucleus, u\n",
"A_y = 1.0078252; ## Atomic mass of proton, u\n",
"Q = ((E_y*(1+m_y/M_Y))-(E_x*(1-m_x/M_Y))-2./M_Y*math.sqrt((m_x*m_y*E_x*E_y))*math.cos(theta))/931.5; ## Q-value, u\n",
"A_Y = A_x+A_X-A_y-Q; ## Atomic mass of O-17, u\n",
"print'%s %.4f %s %.2f %s '%(\"\\nThe Q-value of the reaction = \",Q,\" u\" \" \\nThe atomic mass of the O-17 =\",A_Y,\" u\");\n",
"\n",
"## Result\n",
"## The Q-value of the reaction = -0.0012755 u \n",
"## The atomic mass of the O-17 = 16.9991278 u \n",
"## Atomic mass of the O-17 : 16.9991278 u "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The Q-value of the reaction = -0.0013 u \n",
"The atomic mass of the O-17 = 17.00 u \n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3-pg455"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.3 : : Page-455 (2011) \n",
"#calculate The kinetic energy of the emitted neutron \n",
"import math\n",
"m_p = 1.007276; ## Atomic mass of the proton, u\n",
"m_H = 3.016049; ## Atomic mass of the tritium, u \n",
"m_He = 3.016029; ## Atomic mass of the He ion, u \n",
"m_n = 1.008665; ## Atomic mass of the emitted neutron, u\n",
"Q = (m_p+m_H-m_He-m_n)*931.5; ## Q-value in MeV\n",
"E_p = 3.; ## Kinetic energy of the proton, MeV \n",
"theta = 30*3.14/180; ## angle, radian\n",
"u = math.sqrt(m_p*m_n*E_p)/(m_He+m_n)*math.cos(theta); ##\n",
"v = ((m_He*Q)+E_p*(m_He-m_p))/(m_He+m_n); ##\n",
"E_n = (u+math.sqrt(u**2+v))**2; ## Kinetic energy of the emitted neutron,MeV\n",
"print'%s %.2f %s'%(\"\\nThe kinetic energy of the emitted neutron = \",E_n,\" MeV\");\n",
"\n",
"## Result\n",
"## The kinetic energy of the emitted neutron = 1.445 MeV "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The kinetic energy of the emitted neutron = 1.45 MeV\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4-pg456"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.4 : : Page-456 (2011)\n",
"#findThe temperature in the fusion reaction \n",
"import math\n",
"r_min = 4e-015; ## Distance between two deutrons, metre\n",
"k = 1.3806504e-023; ## Boltzmann's constant, Joule per kelvin\n",
"alpha = 1/137.; ## Fine structure constant\n",
"h_red = 1.05457168e-034; ## Reduced planck's constant, Joule sec\n",
"C = 3e+08; ## Velocity of light, meter per second\n",
"T = alpha*h_red*C/(r_min*k);\n",
"print'%s %.2e %s'%(\"\\nThe temperature in the fusion reaction is =\",T,\" K\");\n",
"\n",
"## Result\n",
"## The temperature in the fusion reaction is = 4.2e+009 K "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The temperature in the fusion reaction is = 4.18e+09 K\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5-pg456"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa11.5 : : Page-456 (2011) \n",
"#find the excitation energy of the compound nucleus\n",
"import math\n",
"E_0 = 4.99; ## Energy of the proton, MeV \n",
"m_p = 1.; ## Mass number of the proton\n",
"m_F = 19.; ## Mass number of the flourine\n",
"E = E_0/(1+m_p/m_F); ## Energy of the relative motion, MeV\n",
"A_F = 18.998405; ## Atomic mass of the fluorine, amu\n",
"A_H = 1.007276; ## Atomic mass of the proton, amu\n",
"A_Ne = 19.992440; ## Atomic mass of the neon, amu\n",
"del_E = (A_F+A_H-A_Ne)*931.5; ## Binding energy of the absorbed proton, MeV\n",
"E_exc = E+del_E; ## Excitation energy of the compound nucleus, MeV\n",
"print'%s %.2f %s'%(\"\\nThe excitation energy of the compound nucleus = \",E_exc,\" MeV\");\n",
"\n",
"## Result\n",
"## The excitation energy of the compound nucleus = 17.074 MeV "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The excitation energy of the compound nucleus = 17.07 MeV\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex6-pg457"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.6 : : Page-457 (2011) \n",
"#find The excitation energy of the compound nucleus and The parity of the compound nucleus\n",
"import math\n",
"E_d = 0.6; ## Energy of the deutron, MeV \n",
"m_d = 2.; ## Mass number of the deutron\n",
"m_Li = 19.; ## Mass number of the Lithium\n",
"E = E_d/(1.+m_d/m_Li); ## Energy of the relative motion, MeV\n",
"A_Li = 6.017; ## Atomic mass of the Lithium, amu\n",
"A_d = 2.015; ## Atomic mass of the deutron, amu\n",
"A_Be = 8.008; ## Atomic mass of the Beryllium, amu\n",
"del_E = (A_Li+A_d-A_Be)*931.5; ## Binding energy of the absorbed proton, MeV\n",
"E_exc = E+del_E; ## Excitation energy of the compound nucleus, MeV\n",
"l_f = 2.; ## orbital angular momentum of two alpha particle\n",
"P = (-1)**l_f*(+1.)**2; ## Parity of the compound nucleus\n",
"print'%s %.2f %s %.2f %s '%(\"\\nThe excitation energy of the compound nucleus = \",E_exc,\" MeV\" and \"\\nThe parity of the compound nucleus = \",P,\"\");\n",
"\n",
"## Result\n",
"## The excitation energy of the compound nucleus = 22.899 MeV\n",
"## The parity of the compound nucleus = 1 \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The excitation energy of the compound nucleus = 22.90 \n",
"The parity of the compound nucleus = 1.00 \n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7-pg457"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.7 : : Page-457 (2011)\n",
"#find The cross section for neutron induced fission\n",
"import math\n",
"D = 1e-016; ## Disintegration constant, per sec\n",
"phi = 10**11; ## Neutron flux, neutrons per square cm per sec\n",
"sigma = 5*D/(phi*10**-27); ## Cross section, milli barns\n",
"print'%s %.2f %s'%(\"\\nThe cross section for neutron induced fission = \",sigma,\" milli barns\");\n",
"\n",
"## Result\n",
"## The cross section for neutron induced fission = 5 milli barns "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The cross section for neutron induced fission = 5.00 milli barns\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8-pg457"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.8 : : Page-457 (2011)\n",
"#find The number of nuclei of Co60 produced and The initial activity\n",
"import math\n",
"N_0 = 6.02252e+026; ## Avogadro's constant \n",
"rho = 8.9*10**3; ## Nuclear density of Co-59, Kg per cubic metre\n",
"M = 59.; ## Mass number\n",
"sigma = 30e-028; ## Cross section, per square metre\n",
"phi = 10**16; ## Neutron flux, neutrons per square metre per sec\n",
"d = 0.04e-02; ## Thickness of Co-59 sheet, metre\n",
"t = 3*60*60; ## Total reaction time, sec\n",
"t_half = 5.2*365*86400; ## Half life of Co-60, sec \n",
"D = 0.693/t_half; ## Disintegration constant, per sec\n",
"N_nuclei = round(N_0*rho/M*sigma*phi*d*t); ## Number of nuclei of Co-60 produced\n",
"Init_activity = D*N_nuclei; ## Initial activity, decays per sec\n",
"print'%s %.2e %s %.2f %s '%(\"\\nThe number of nuclei of Co60 produced =\",N_nuclei,\" \" and \"\\nThe initial activity per Sq. metre = \",Init_activity,\" decays per sec\");\n",
"\n",
"## Result\n",
"## The number of nuclei of Co60 produced = 1.18e+019 \n",
"## The initial activity per Sq. metre = 5e+010 decays per sec "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The number of nuclei of Co60 produced = 1.18e+19 \n",
"The initial activity per Sq. metre = 49755893720.42 decays per sec \n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex9-pg458"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.9 : : Page-458 (2011)\n",
"#find The number of events detected\n",
"import math\n",
"d = 0.1; ## Thickness of Fe-54 sheet, Kg per squre metre\n",
"M = 54.; ## Mass number of Fe \n",
"m = 1.66e-027; ## Mass of the proton, Kg\n",
"n = d/(M*m); ## Number of nuclei in unit area of the target, nuclei per square metre\n",
"ds = 10**-5; ## Area, metre square\n",
"r = 0.1; ## Distance between detector and target foil, metre\n",
"d_omega =ds/r**2; ## Solid angle, steradian\n",
"d_sigma = 1.3e-03*10**-3*10**-28; ## Differential cross section, square metre per nuclei\n",
"P = d_sigma*n; ## Probablity, event per proton\n",
"I = 10**-7; ## Current, ampere\n",
"e = 1.6e-19; ## Charge of the proton, C\n",
"N = I/e; ## Number of protons per second in the incident beam, proton per sec\n",
"dN = P*N; ## Number of events detected per second, events per sec\n",
"print'%s %.2f %s'%(\"\\nThe number of events detected =\",dN,\" events per sec\");\n",
"\n",
"## Result\n",
"## The number of events detected = 90 events per sec "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The number of events detected = 90.64 events per sec\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex10-pg458"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.10 : : Page-458 (2011)\n",
"#find The fractional attenuation of neutron beam on passing through nickel sheet\n",
"import math\n",
"N_0 = 6.02252e+26; ## Avogadro's constant\n",
"sigma = 3.5e-28; ## Cross section, square metre\n",
"rho = 8.9e+03; ## Nuclear density, Kg per cubic metre\n",
"M = 58.; ## Mass number \n",
"summation = rho/M*N_0*sigma; ## Macroscopic cross section, per metre\n",
"x = 0.01e-02; ## Thickness of nickel sheet, metre\n",
"I0_ratio_I = math.exp(summation*x/2.3026); ## Fractional attenuation of neutron beam on passing through nickel sheet\n",
"print'%s %.2f %s ' %(\"\\nThe fractional attenuation of neutron beam on passing through nickel sheet =\", I0_ratio_I,\"\");\n",
"print(\"Wrong answer given in the textbook\")\n",
"## Result\n",
"## The fractional attenuation of neutron beam on passing through nickel sheet = 1.0014 \n",
"## Wrong answer given in the textbook\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The fractional attenuation of neutron beam on passing through nickel sheet = 1.00 \n",
"Wrong answer given in the textbook\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex11-pg458"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.11 : : Page-458 (2011)\n",
"#find he scattering contribution to the resonance \n",
"import math\n",
"D = math.sqrt(1.45e-021/(4.*math.pi)); ## Wavelength, metre\n",
"W_ratio = 2.3e-07; ## Width ratio\n",
"sigma = W_ratio*(4*math.pi)*D**2*10**28; ## Scattering contribution, barn\n",
"print'%s %.2f %s'%(\"\\nThe scattering contribution to the resonance = \",sigma,\" barns\");\n",
"\n",
"## Result\n",
"## The scattering contribution to the resonance = 3.33 barns \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The scattering contribution to the resonance = 3.33 barns\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex12-pg458"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.12 : : Page-458 (2011)\n",
"#find The relative probabilities\n",
"import math\n",
"sigma = 2.8e-024; ## Cross section, metre square\n",
"D = 2.4e-11; ## de Broglie wavelength, metre\n",
"R_prob = math.pi*sigma/D**2; ## Relative probabilities of (n,n) and (n,y) in indium\n",
"print'%s %.2f %s'%(\"\\nThe relative probabilities of (n,n) and (n,y) in indium = \", R_prob,\"\");\n",
"\n",
"## Result\n",
"## The relative probabilities of (n,n) and (n,y) in indium = 0.015 \n",
" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The relative probabilities of (n,n) and (n,y) in indium = 0.02 \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex13-pg459"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.13 : : Page-459 (2011)\n",
"#find The cross section of neutron capture\n",
"import math\n",
"h = 6.625e-34; ## Planck's constant, joule sec \n",
"m_n = 1.67e-27; ## Mass of neutron, Kg\n",
"E = 4.906; ## Energy, joule \n",
"w_y = 0.124; ## radiation width, eV\n",
"w_n = 0.007*E**(1/2.); ## Probability of elastic emission of neutron, eV\n",
"I = 3.; ## Total angular momentum\n",
"I_c = 2.; ## Total angular momentum in the compound state\n",
"sigma = ((h**2)*(2*I_c+1)*w_y*w_n)*10**28/(2.*math.pi*m_n*E*1.602e-019*(2*I+1)*(w_y+w_n)**2); ## Cross section, barns\n",
"print'%s %.3e %s'%(\"\\nThe cross section of neutron capture = \",sigma,\" barns\");\n",
"\n",
"## Result\n",
"## The cross section of neutron capture = 3.755e+004 barns \n",
" \n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The cross section of neutron capture = 3.755e+04 barns\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex14-pg459"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.14 : : Page-459 (2011)\n",
"#fid the value of theta max\n",
"import math\n",
"import numpy\n",
"R = 5.; ## Radius, femto metre\n",
"k_d = 0.98; ## The value of k for deutron \n",
"k_p = 0.82; ## The value of k for triton\n",
"theta = numpy.zeros((1,5)); ## Angles at which differetial cross section is maximum, degree\n",
"## Use of for loop for angles calculation(in degree)\n",
"for l in range (1,4):\n",
" theta = round((math.acos((k_d**2+k_p**2)/(2.*k_d*k_p)-l**2/(2.*k_d*k_p*R**2)))*180./3.14,0);\n",
" #theta = round((acos((k_d^2+k_p^2)/(2*k_d*k_p)-l^2/(2*k_d*k_p*R^2)))*180/3.14);\n",
" print'%s %.2f %s'%(\"\\nFor l = \", l,\"\");\n",
" print'%s %.2f %s'%(\"the value of theta_max = \", math.ceil(theta),\"degree\");\n",
" \n",
"\n",
"## Result\n",
"## For l = 0,the value of theta_max = 0 degree\n",
"## For l = 1,the value of theta_max = 8 degree\n",
"## For l = 2,the value of theta_max = 24 degree\n",
"## For l = 3,the value of theta_max = 38 degree\n",
"## For l = 4,the value of theta_max = 52 degree "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"For l = 1.00 \n",
"the value of theta_max = 8.00 degree\n",
"\n",
"For l = 2.00 \n",
"the value of theta_max = 24.00 degree\n",
"\n",
"For l = 3.00 \n",
"the value of theta_max = 38.00 degree\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex15-pg459"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa10.15 : : Page-459 (2011)\n",
"#calculate total angular momentum transfer\n",
"import math\n",
"k_d = 2.02e+30; ## The value of k for deutron\n",
"k_t = 2.02e+30; ## The value of k for triton\n",
"theta = 23*3.14/180; ## Angle, radiams\n",
"q = math.sqrt (k_d+k_t-2*k_t*math.cos(theta))*10**-15; ## the value of q in femto metre\n",
"R_0 = 1.2; ## Distance of closest approach, femto metre\n",
"A = 90.; ## Mass number of Zr-90\n",
"z = 4.30; ## Deutron size, femto metre\n",
"R = R_0*A**(1/3.)+1/2.*z; ## Radius of the nucleus, femto metre\n",
"l = round(q*R); ## Orbital angular momentum\n",
"I = l+1./2. ## Total angular momentum\n",
"print'%s %.2f %s'%(\"\\nThe total angular momentum transfer = \", I,\"\");\n",
"\n",
"## Result\n",
"## The total angular momentum transfer = 4.5 "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The total angular momentum transfer = 4.50 \n"
]
}
],
"prompt_number": 1
}
],
"metadata": {}
}
]
}