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{
"metadata": {
"name": "",
"signature": "sha256:bb0d2851c786dd28ae3ad8cbd1e6a7fc5db7cc1384cc2031f2bb6380ed7515e2"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 14: Nuclear Physics Apllications"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.1, page no. 505"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#Variable declaration\n",
"\n",
"MLi = 7.016003 #atomic mass of Lithium\n",
"MH = 1.007825 #atomic mass of Hydrogen\n",
"MHe = 4.002603 #atomic mass of Helium\n",
"c2 = 931.50 #Square of speed of light (MeV/u)\n",
"\n",
"#Calculation\n",
"\n",
"Q = (MLi + MH - 2*MHe) * c2\n",
"\n",
"#Results\n",
"\n",
"print \"(a) The Q value is\",round(Q,1),\"MeV.\"\n",
"\n",
"\n",
"\n",
"#Variable declaration\n",
"\n",
"Kincident = 0.6 #kinetic energy of incident particle (MeV)\n",
"\n",
"#Calculation\n",
"\n",
"Kproducts = Q + Kincident\n",
"\n",
"#Results\n",
"\n",
"print \"(b) The kinetic energy of the products is\",round(Kproducts,1),\"MeV.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a) The Q value is 17.3 MeV.\n",
"(b) The kinetic energy of the products is 17.9 MeV.\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.2, page no. 509"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#Variable declaration\n",
"\n",
"A = 27 #Atomic number of Aluminum\n",
"d = 2.7 #density of aluminum\n",
"Av = 6.02 * 10 ** 23 #Avogadro number nuclei/mol\n",
"sigma = 2.0 * 10 **-31 #capture cross section (m^2)\n",
"x = 0.3 * 10 ** -3 #thickness of the foil(m)\n",
"R0 = 5.0 * 10 ** 12 #rate of incident particles(neutrons/cm^2.s)\n",
"\n",
"#Calculation\n",
"\n",
"n = Av * d / A * 10**6\n",
"R = R0 * sigma * x * n\n",
"\n",
"#Result\n",
"\n",
"print \"The number of neutrons captured by 1.0cm^2/s is \",round(R/10**7,1),\"X 10^7 neutrons/cm^2.s\"\n",
"\n",
"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The number of neutrons captured by 1.0cm^2/s is 1.8 X 10^7 neutrons/cm^2.s\n"
]
}
],
"prompt_number": 9
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.3, page no. 512"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration\n",
"\n",
"A = 235 #atomic number of uranium\n",
"m = 10**3 #mass of 235U\n",
"Av = 6.02*10**23#avogadro number (nuclei/mol)\n",
"Q = 208 #disintegration energy per event (MeV)\n",
"\n",
"#Calculation\n",
"\n",
"N = Av * m / A\n",
"E = N * Q\n",
"\n",
"#Result\n",
"\n",
"print \"The disintegration energy is\",round(E/10**26,2),\"X 10^26 MeV.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The disintegration energy is 5.33 X 10^26 MeV.\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.4, page no. 513"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#Variable declaration\n",
"\n",
"A = 235 #atomic mass of Uranium\n",
"Av = 6.02 * 10**23 # Avagadro's number\n",
"Q = 208 # disintegration energy per event (MeV)\n",
"\n",
"#Calculation\n",
"\n",
"N = Av / A *10**3 #No of nuclei in 1kg of 235U\n",
"E = N*Q #disintegration energy (MeV)\n",
"\n",
"#result\n",
"\n",
"print \"The disintegration energy is\",round(E/10**26,2),\" X 10^26 MeV.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The disintegration energy is 5.33 X 10^26 MeV.\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.5, page no. 513"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"#Variable declaration\n",
"\n",
"rBa = 6.2 * 10 **-15 #nuclear radius of Ba(m)\n",
"rKr = 5.4 * 10**-15 #nuclear radius of Kr(m)\n",
"Z1 = 56 #atomic number of Ba\n",
"Z2 = 36 #atomic number of Kr\n",
"k = 1.440 * 10 **-9 #Coulomb constant (eV.nm)\n",
"\n",
"#Calculation\n",
"\n",
"r = rBa + rKr\n",
"U = k * Z1 * Z2 /( round(r/10**-15) * 10**-15)\n",
"\n",
"#Result\n",
"\n",
"print \"The potential energy of the two nuclei is\",round(U/10**6),\"MeV.\" "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The potential energy of the two nuclei is 242.0 MeV.\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.6, page no. 519"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"\n",
"#variable declaration\n",
"\n",
"e = 1.6 * 10 ** -19 #electron charge (C)\n",
"r = 1.0 * 10**-14 #separation (m)\n",
"k = 8.99 * 10**9 #Coulomb constant(N.m^2/C^2)\n",
"\n",
"#Calculation\n",
"\n",
"U = k * e**2/ r\n",
"\n",
"#result\n",
"\n",
"print \"(a)The height of the potential barrier is\",round(U/e/10**6,2),\"MeV.\"\n",
"\n",
"\n",
"#Variable Declaration\n",
"\n",
"kB = 1.38 * 10**-23 #Boltzmann's constant (J/K)\n",
"\n",
"#Calculation\n",
"\n",
"T = (U/2.0)*(2.0/3.0)/(kB+.07*10**-23)\n",
"\n",
"#Result\n",
"\n",
"print\"(b)The effective temperature is\",round(T/10**8,1),\"X 10^8 K.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"(a)The height of the potential barrier is 0.14 MeV.\n",
"(b)The effective temperature is 5.3 X 10^8 K.\n"
]
}
],
"prompt_number": 22
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 14.7, page no. 530"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"\n",
"import math\n",
"\n",
"#variable declaration\n",
"\n",
"mu = 55 #mu for x-rays \n",
"\n",
"#Calculation\n",
"\n",
"x = math.log(2)/mu\n",
"\n",
"#Result\n",
"\n",
"print \"The half value thickness for lead is\",round(x/10**-2,2),\"X 10^-2 cm.\""
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The half value thickness for lead is 1.26 X 10^-2 cm.\n"
]
}
],
"prompt_number": 24
}
],
"metadata": {}
}
]
}
|
