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-{
- "metadata": {
- "name": "",
- "signature": "sha256:df9996e09d849b24524dd415b626cbe4279b4acdbe25d68bb407e2d42467c7a7"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "11: Lattice dynamics"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.1, Page number 238"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #planck's constant(Js)\n",
- "k=1.38*10**-23; #boltzmann constant(J/K)\n",
- "thetaD=350; #temperature for Cu(K)\n",
- "theetaD=550; #temperature for Si(K)\n",
- "\n",
- "#Calculation\n",
- "newDCu=k*thetaD/h; #highest possible frequency for Cu(per sec)\n",
- "newDSi=k*theetaD/h; #highest possible frequency for Si(per sec)\n",
- "\n",
- "#Result\n",
- "print \"highest possible frequency for Cu is\",round(newDCu/10**11,3),\"*10**11 per sec\"\n",
- "print \"highest possible frequency for Si is\",round(newDSi/10**11,2),\"*10**11 per sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "highest possible frequency for Cu is 72.895 *10**11 per sec\n",
- "highest possible frequency for Si is 114.55 *10**11 per sec\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 11.2, Page number 238"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#importing modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "h=6.626*10**-34; #planck's constant(Js)\n",
- "k=1.38*10**-23; #boltzmann constant(J/K)\n",
- "N=6.02*10**26; #avagadro number(k/mole)\n",
- "T=10; #temperature(K)\n",
- "thetaD=105; #debye temperature(K)\n",
- "\n",
- "#Calculation\n",
- "C=(12/5)*(math.pi**4)*N*k*(T/thetaD)**3; #specific heat of lead(J/K kmol)\n",
- "newD=k*thetaD/h; #highest frequency(per sec)\n",
- "\n",
- "#Result\n",
- "print \"specific heat of lead is\",round(C,1),\"J/K kmol\"\n",
- "print \"answer varies due to rounding off errors\"\n",
- "print \"highest frequency is\",round(newD/10**11,2),\"*10**11 per sec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "specific heat of lead is 1677.7 J/K kmol\n",
- "answer varies due to rounding off errors\n",
- "highest frequency is 21.87 *10**11 per sec\n"
- ]
- }
- ],
- "prompt_number": 6
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file