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{
"metadata": {
"name": "Chapter1"
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 1:Introduction"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.1, Page 12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initiation of variable\n",
"Mn=1.008665;Mp=1.007276 #Given mass of an electron and a proton in terms of u\n",
"\n",
"#calculation\n",
"Md= Mn-Mp; #mass difference \n",
"Md2=Md*931.50; #converting u into Mev/c^2 by multiplying by 931.5 MeV/c^2\n",
"\n",
"#result\n",
"print \"Mass difference in terms of U is\",round(Md,4); \n",
"print\"which equals in Mev/c^2. :\",round(Md2,3);"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Mass difference in terms of U is 0.0014\n",
"which equals in Mev/c^2. : 1.294\n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.2, Page 12"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initiation of variable\n",
"Mp=1.007276 ; Me=5.4858*10**-4; #mass of proton and electron in terms of U\n",
"\n",
"#calculation\n",
"Mt=Mp+Me; #Total mass= sum of above masses \n",
"\n",
"#result\n",
"print\"The combined mass of an electron and a proton was found out to be in U.\",round(Mt,3);\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The combined mass of an electron and a proton was found out to be in U. 1.008\n"
]
}
],
"prompt_number": 5
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 1.3, Page 13"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"#initiation of variable\n",
"h=6.621*10**-34 ; c=2.9979*10**8; # h is in J/s and c is in m/s\n",
"hc=h*c*((10**9)/(1.6022*10**-19)); #1e=1.602*10^-19 J and 1 m=10^9 nm\n",
"\n",
"#result\n",
"print \"The value of hc in eV.nm is\",round(hc,4); \n",
"print 'Hence zero at the end is significant.';\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of hc in eV.nm is 1238.8651\n",
"Hence zero at the end is significant.\n"
]
}
],
"prompt_number": 8
}
],
"metadata": {}
}
]
}
|
