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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 13 - Atomic structure"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example I2 - Pg 302"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the probability of finding the electron and the chance that the electron would be found\n",
"#Initialization of variables\n",
"import math\n",
"dv=1 #pm^3\n",
"a0=52.9 #pm\n",
"#calculations\n",
"Probability=dv/(math.pi*a0*a0*a0)\n",
"#results\n",
"print '%s %.1e' %(\"probability of finding electron = \",Probability)\n",
"print '%s %d %s' %(\"\\n Chance that electron would be found is one in\",1./Probability,\"times\")\n",
"#The answer is a bit different due to rounding off error in textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"probability of finding electron = 2.2e-06\n",
"\n",
" Chance that electron would be found is one in 465068 times\n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example I3 - Pg 304"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#calculate the probability of finding the electron\n",
"#Initialization of variables\n",
"import math\n",
"dr=1 #pm\n",
"r=52.9 #pm\n",
"#calculations\n",
"Probability=4*math.pow(math.e,(-2)) *dr/r\n",
"#results\n",
"print '%s %.1f' %(\"About 1 inspection in \",1./Probability)\n",
"#The answer is a bit different due to rounding off error in textbook"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"About 1 inspection in 97.7\n"
]
}
],
"prompt_number": 2
}
],
"metadata": {}
}
]
}
|