{
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter12-Neutrons"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Exx1-pg573"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa12.1 : : Page-573 (2011)\n",
"#calculate activity for Cu-63 and disintegrations\n",
"import math \n",
"N_0 = 6.23e+23; ## Avogadro's number, per mole\n",
"m = 0.1; ## Mass of copper foil, Kg\n",
"phi = 10**12; ## Neutron flux density, per square centimetre sec\n",
"a_63 = 0.691; ## Abundance of Cu-63\n",
"a_65 = 0.309; ## Abundance of Cu-65\n",
"W_m = 63.57; ## Molecular weight, gram\n",
"sigma_63 = 4.5e-24; ## Activation cross section for Cu-63, square centi metre\n",
"sigma_65 = 2.3e-24; ## Activation cross section for Cu-65, square centi metre\n",
"A_63 = phi*sigma_63*m*a_63/W_m*N_0; ## Activity for Cu-63, disintegrations per sec\n",
"A_65 = phi*sigma_65*m*a_65/W_m*N_0; ## Activity for Cu-65, disintegrations per sec\n",
"print'%s %.2e %s %.2e %s'%(\"\\nThe activity for Cu-63 is = \",A_63,\" disintegrations per sec\" and \"\\nThe activity for Cu-65 is = \",A_65,\" disintegrations per sec\");\n",
"\n",
"## Result\n",
"## The activity for Cu-63 is = 3.047e+009 disintegrations per sec \n",
"## The activity for Cu-65 is = 6.97e+008 disintegrations per sec "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The activity for Cu-63 is = 3.05e+09 \n",
"The activity for Cu-65 is = 6.97e+08 disintegrations per sec\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex2-pg573"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa12.2 : : Page-573 (2011)\n",
"import math \n",
"#calculate enegy loss \n",
"A_Be = 9.; ## Mass number of beryllium\n",
"A_U = 238.; ## Mass number of uranium\n",
"E_los_Be = (1-((A_Be-1)**2/(A_Be+1)**2))*100.; ## Energy loss for beryllium\n",
"E_los_U = round((1-((A_U-1)**2/(A_U+1)**2))*100.); ## Energy loss for uranium\n",
"print'%s %.2f %s %.2f %s '%(\"\\nThe energy loss for beryllium is = \",E_los_Be,\" percent\"and \" \\nThe energy loss for uranium is = \",E_los_U,\" percent\");\n",
"\n",
"## Check for greater energy loss !!!!\n",
"if E_los_Be >= E_los_U :\n",
" print(\"\\nThe energy loss is greater for beryllium\");\n",
"else:\n",
" print(\"\\nThe energy loss is greater for uranium\");\n",
"\n",
"\n",
"## Result\n",
"## The energy loss for beryllium is = 36 percent \n",
"## The energy loss for uranium is = 2 percent\n",
"## The energy loss is greater for beryllium \n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The energy loss for beryllium is = 36.00 \n",
"The energy loss for uranium is = 2.00 percent \n",
"\n",
"The energy loss is greater for beryllium\n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex3-pg574"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa12.3 : : Page-574 (2011)\n",
"#calculate energy loss of neutron\n",
"import math \n",
"A = 12.; ## Mass number of Carbon\n",
"alpha = (A-1)**2/(A+1)**2; ## Scattering coefficient\n",
"E_loss = 1/2.*(1-alpha)*100.; ## Energy loss of neutron\n",
"print'%s %.2f %s'%(\"\\nThe energy loss of neutron = \",E_loss,\" percent\")\n",
"\n",
"## Result\n",
"## The energy loss of neutron = 14.201 percent \n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The energy loss of neutron = 14.20 percent\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex4-pg574"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa12.4 : : Page-574 (2011)\n",
"#calculate number of collisions of neutrons \n",
"import math \n",
"zeta = 0.209; ## Moderated assembly\n",
"E_change = 100./1.; ## Change in energy of the neutron\n",
"E_thermal = 0.025; ## Thermal energy of the neutron, electron volts\n",
"E_n = 2*10**6; ## Energy of the neutron, electron volts\n",
"n = 1/zeta*math.log(E_change); ## Number of collisions of neutrons to loss 99 percent of their energies \n",
"n_thermal = 1/zeta*math.log(E_n/E_thermal); ## Number of collisions of neutrons to reach thermal energies\n",
"print'%s %.2f %s %.2f %s'%(\"\\nThe number of collisions of neutrons to loss 99 percent of their energies = \",n,\" \\nThe number of collisions of neutrons to reach thermal energies = \",n_thermal,\"\")\n",
"\n",
"## Result\n",
"## The number of collisions of neutrons to loss 99 percent of their energies = 22 \n",
"## The number of collisions of neutrons to reach thermal energies = 87 \n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The number of collisions of neutrons to loss 99 percent of their energies = 22.03 \n",
"The number of collisions of neutrons to reach thermal energies = 87.07 \n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5-pg574"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa12.5 : : Page-574 (2011)\n",
"#calculate average distance travelled by the neutron\n",
"import math\n",
"import scipy\n",
"from scipy import integrate\n",
"L = 1.; ## For simplicity assume thermal diffusion length to be unity, unit\n",
"def fun(x):\n",
" y=x*math.exp(-x/L)\n",
" return y\n",
"x_b = scipy.integrate.quad(fun, 0, 100); ## Average distance travelled by the neutron, unit\n",
"x_b1=x_b[0]\n",
"def fun2(x): \n",
" y1=x**2*math.exp(-x/L)\n",
" return y1\n",
"X=scipy.integrate.quad(fun2, 0, 100)\n",
"x_rms = math.sqrt(X[0]); ## Root mean square of the distance trvelled by the neutron, unit\n",
"print'%s %.2f %s'%(\"\\nThe average distance travelled by the neutron = \", x_b1,\"*L\");\n",
"print'%s %.2f %s %.2f %s '%(\"\\nThe root mean square distance travelled by the neutron = \",x_rms,\"\"and \"\",x_rms,\"x_bar\")\n",
"\n",
"## Result\n",
"## The average distance travelled by the neutron = 1*L\n",
"## The root mean square distance travelled by the neutron = 1.414L = 1.414x_bar \n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The average distance travelled by the neutron = 1.00 *L\n",
"\n",
"The root mean square distance travelled by the neutron = 1.41 1.41 x_bar \n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex6-pg574"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa12.6 : : Page-574 (2011)\n",
"#calculate neutron flux through water \n",
"import math\n",
"Q = 5e+08; ## Rate at which neutrons produce, neutrons per sec\n",
"r = 20.; ## Distance from the source, centi metre\n",
"## For water\n",
"lambda_wtr = 0.45; ## Transport mean free path, centi metre\n",
"L_wtr = 2.73; ## Thermal diffusion length, centi metre\n",
"phi_wtr = 3*Q/(4.*math.pi*lambda_wtr*r)*math.exp(-r/L_wtr); ## Neutron flux for water, neutrons per square centimetre per sec\n",
"## For heavy water\n",
"lambda_h_wtr = 2.40; ## Transport mean free path, centi metre\n",
"L_h_wtr = 171.; ## Thermal diffusion length, centi metre\n",
"phi_h_wtr = 3*Q/(4.*math.pi*lambda_h_wtr*r)*math.exp(-r/L_h_wtr); ## Neutron flux for heavy water, neutrons per square centimetre per sec\n",
"print'%s %.2e %s %.2e %s '%(\"\\nThe neutron flux through water = \",phi_wtr,\" neutrons per square cm per sec\"and \"\\nThe neutron flux through heavy water = \",phi_h_wtr,\" neutrons per square cm per sec\")\n",
"\n",
"## Result\n",
"## The neutron flux through water = 8.730e+003 neutrons per square cm per sec \n",
"## The neutron flux through heavy water = 2.212e+006 neutrons per square cm per sec \n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The neutron flux through water = 8.73e+03 \n",
"The neutron flux through heavy water = 2.21e+06 neutrons per square cm per sec \n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex7-pg575"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa12.7 : : Page-575 (2011)\n",
"#calculate neutron flux and diffusion length\n",
"import math\n",
"k = 1.38e-23; ## Boltzmann constant, joules per kelvin\n",
"T = 323.; ## Temperature, kelvin\n",
"E = (k*T)/1.6e-19; ## Thermal energy, joules\n",
"sigma_0 = 13.2e-28; ## Cross section, square metre\n",
"E_0 = 0.025; ## Energy of the neutron, electron volts\n",
"sigma_a = sigma_0*math.sqrt(E_0/E); ## Absorption cross section, square metre\n",
"t_half = 2.25; ## Half life, hours\n",
"D= 0.69/t_half; ## Decay constant, per hour\n",
"N_0 = 6.023e+026; ## Avogadro's number, per \n",
"m_Mn = 55.; ## Mass number of mangnese\n",
"w = 0.1e-03; ## Weight of mangnese foil, Kg\n",
"A = 200.; ## Activity, disintegrations per sec\n",
"N = N_0*w/m_Mn; ## Number of mangnese nuclei in the foil\n",
"x1 = 1.5; ## Base, metre\n",
"x2 = 2.0; ## Height, metre\n",
"phi = A/(N*sigma_a*0.416); ## Neutron flux, neutrons per square metre per sec\n",
"phi1 = 1.; ## For simplicity assume initial neutron flux to be unity, neutrons/Sq.m-sec\n",
"phi2 = 1/2.*phi1; ## Given neutron flux, neutrons/Sq.m-sec\n",
"L1 = 1/math.log(phi1/phi2)/(x2-x1); ## Thermal diffusion length for given neutron flux, m\n",
"L = math.sqrt(1./((1./L1)**2+(math.pi/x1)**2+(math.pi/x2)**2)); ## Diffusion length, metre\n",
"print'%s %.2e %s %.2f %s '%(\"\\nThe neutron flux = \",phi,\" neutrons per square metre per sec\"and \" \\nThe diffusion length = \",L,\" metre\");\n",
"\n",
"## Result\n",
"## The neutron flux = 3.51e+008 neutrons per square metre per sec \n",
"## The diffusion length = 0.38 metre\n",
"## Note: the difussion length is solved wrongly in the testbook\n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The neutron flux = 3.51e+08 \n",
"The diffusion length = 0.38 metre \n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex8-pg575"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa12.8 : : Page-575(2011)\n",
"#find diffusion length for thermal neutron\n",
"import math\n",
"N_0 = 6.023e+026; ## Avogadro's number, per mole\n",
"rho = 1.62e+03; ## Density, kg per cubic metre\n",
"sigma_a = 3.2e-31; ## Absorption cross section, square metre\n",
"sigma_s = 4.8e-28; ## Scattered cross section, square metre\n",
"A = 12.; ## Mass number\n",
"lambda_a = A/(N_0*rho*sigma_a); ## Absorption mean free path, metre\n",
"lambda_tr = A/(N_0*rho*sigma_s*(1.-2./(3.*A))); ## Transport mean free path, metre\n",
"L = math.sqrt(lambda_a*lambda_tr/3.); ## Diffusion length for thermal neutron\n",
"print'%s %.2f %s'%(\"\\nThe diffusion length for thermal neutron = \",L,\" metre \")\n",
"\n",
"## Result\n",
"## The diffusion length for thermal neutron = 0.590 metre \n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The diffusion length for thermal neutron = 0.59 metre \n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex9-pg575"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa12.9 : : Page-575 (2011)\n",
"#calculate graphite and neutron age and slowing down length and same as berylliums\n",
"import math\n",
"E_0 = 2e+06; ## Average energy of the neutron, electron volts\n",
"E = 0.025; ## Thermal energy of the neutron, electron volts\n",
"## For graphite\n",
"A = 12. ## Mass number\n",
"sigma_g = 33.5; ## The value of sigma for graphite\n",
"tau_0 = 1./(6.*sigma_g**2)*(A+2./3.)/(1.-2./(3.*A))*math.log(E_0/E); ## Age of neutron for graphite, Sq.m\n",
"L_f = math.sqrt(tau_0); ## Slowing down length of neutron through graphite, m\n",
"print'%s %.2f %s'%(\"\\nFor Graphite, A = \", A,\"\");\n",
"print'%s %.2f %s'%(\"\\nNeutron age = \",tau_0*1e+004,\" Sq.cm\");\n",
"print'%s %.2f %s'%(\"\\nSlowing down length =\",L_f,\" m\");\n",
"## For beryllium\n",
"A = 9. ## Mass number\n",
"sigma_b = 57.; ## The value of sigma for beryllium\n",
"tau_0 = 1/(6.*sigma_b**2)*(A+2./3.)/(1.-2./(3.*A))*math.log(E_0/E); ## Age of neutron for beryllium, Sq.m\n",
"L_f = math.sqrt(tau_0); ## Slowing down length of neutron through graphite, m\n",
"print'%s %.2f %s'%(\"\\n\\nFor Beryllium, A = \", A,\"\");\n",
"print'%s %.2f %s'%(\"\\nNeutron age = \",tau_0*1e+004,\" Sq.cm\");\n",
"print'%s %.2e %s'%(\"\\nSlowing down length = \",L_f,\" m\");\n",
"\n",
"## Result\n",
"## For Graphite, A = 12\n",
"## Neutron age = 362 Sq.cm\n",
"## Slowing down length = 0.190 m\n",
"\n",
"## For Beryllium, A = 9\n",
"## Neutron age = 97 Sq.cm\n",
"## Slowing down length = 9.9e-002 m "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"For Graphite, A = 12.00 \n",
"\n",
"Neutron age = 362.46 Sq.cm\n",
"\n",
"Slowing down length = 0.19 m\n",
"\n",
"\n",
"For Beryllium, A = 9.00 \n",
"\n",
"Neutron age = 97.46 Sq.cm\n",
"\n",
"Slowing down length = 9.87e-02 m\n"
]
}
],
"prompt_number": 13
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex10-pg576"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"## Exa12.10 : : Page-576 (2011)\n",
"#find enegy of the neutrons\n",
"import math\n",
"theta = 3.5*math.pi/180.; ## Reflection angle, radian\n",
"d = 2.3e-10; ## Lattice spacing, metre\n",
"n = 1.; ## For first order\n",
"h = 6.6256e-34; ## Planck's constant, joule sec\n",
"m = 1.6748e-27; ## Mass of the neutron, Kg\n",
"E = n**2*h**2/(8.*m*d**2*math.sin(theta)**2*1.6023e-19); ## Energy of the neutrons, electron volts\n",
"print'%s %.2f %s'%(\"\\nThe energy of the neutrons = \",E,\" eV\");\n",
"\n",
"## Result\n",
"## The energy of the neutrons = 1.04 eV \n",
" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"\n",
"The energy of the neutrons = 1.04 eV\n"
]
}
],
"prompt_number": 14
}
],
"metadata": {}
}
]
}