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| author | hardythe1 | 2015-04-07 16:05:18 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-04-07 16:05:18 +0530 |
| commit | 67068710030ddd6b6c809518c34af2e04e0bf7ca (patch) | |
| tree | fd517673c5fc8a2e9931fff9fe07f6c6c62839e0 /Chemistry/Chapter_17.ipynb | |
| parent | cb2e12cd79d48ebbf281b9b118fd51f532f960f5 (diff) | |
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add book
Diffstat (limited to 'Chemistry/Chapter_17.ipynb')
| -rwxr-xr-x | Chemistry/Chapter_17.ipynb | 92 |
1 files changed, 92 insertions, 0 deletions
diff --git a/Chemistry/Chapter_17.ipynb b/Chemistry/Chapter_17.ipynb new file mode 100755 index 00000000..8debeae7 --- /dev/null +++ b/Chemistry/Chapter_17.ipynb @@ -0,0 +1,92 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 17:Chemistry in the Atmosphere"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:17.1,Page no:774"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "E=498.7*10**3/(6.022*10**23) #energy in J/molecule\n",
+ "h=6.63*10**-34 #plancks constant, J s\n",
+ "\n",
+ "#Calculation\n",
+ "v=E/h #frequency of the photon, s**-1\n",
+ "lamda=3*10**8/v #wavelength in m, since v*lambda=speed of light in vacuum\n",
+ "\n",
+ "#Variable declaration\n",
+ "print\"The maximum wavelength of the photon which can dissociate an O2 molecule is :\",round(lamda*10**9),\"nm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum wavelength of the photon which can dissociate an O2 molecule is : 240.0 nm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example no:17.3,Page no:792"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Rninitial=1 #initial mass of Rn, g\n",
+ "\n",
+ "#Calculation\n",
+ "Rnfinal=Rninitial*0.5**10 #final mass of Rn, g\n",
+ "\n",
+ "#Result\n",
+ "print\"The amount of Rn left after 10 half lives is :%.1e\"%Rnfinal,\"g\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The amount of Rn left after 10 half lives is :9.8e-04 g\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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