{ "metadata": { "name": "", "signature": "sha256:6dfe8c3f0e395e550057fecf65810f55f95ea60853478f130ac485dd9541147f" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 12: The Solid State" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.1, page no. 418" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "#Variable Declaration\n", "\n", "kb = 1.38 * 10 **-23 #Boltzmann constant (J/K)\n", "T = 300 # room temperature (K)\n", "me = 9.11 * 10 ** -31 # mass of electron (kg)\n", "d = 8.96 * 10**6 # density of copper (g/m^3)\n", "N = 6.023 * 10 **23 #avagodro's number (atoms/mole)\n", "mw = 63.5 #molar weight (g)\n", "A = 4 * 10 ** -6 #area (m^2)\n", "I = 10 # current (A)\n", "e = 1.6 * 10 ** -19 # charge of electron(C)\n", "\n", "#Calculation\n", "\n", "Vrms = (3*kb*T/me)**0.5\n", "n = N * d / mw\n", "Vd = I /(n*e*A)\n", "\n", "#Results\n", "\n", "print \"(a) The ratio of drift speed to rms speed is \",round(Vd/Vrms/10**-9,2),\"X 10^-9.\"\n", "\n", "\n", "#Variable Declaration\n", "\n", "L = 2.6 * 10 ** -10 #interatomic distance(A') \n", "\n", "#Calculation\n", "\n", "t = L / Vrms \n", "\n", "#Results\n", "\n", "print \"(b) The average time between collisions\",round(t/10**-15,1),\"X 10^-15 m/s.\"\n", "\n", "\n", "#Variable Declaration\n", "\n", "T = 300 #Room temperature (K)\n", "\n", "#Calculation\n", "\n", "sigma = n*e**2 * L /(3*kb*T*me)**0.5\n", "\n", "#Results\n", "\n", "print \"(c) The conductivity is\",round(sigma/10**6,1),\"X 10^6 (ohm.m)^-1.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a) The ratio of drift speed to rms speed is 1.57 X 10^-9.\n", "(b) The average time between collisions 2.2 X 10^-15 m/s.\n", "(c) The conductivity is 5.3 X 10^6 (ohm.m)^-1.\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.2, page no. 429" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "#Variable Declaration\n", "\n", "V = 7 # voltage(V)\n", "L = 5 * 10 ** -8 # mean free path (m)\n", " \n", "#Calculation\n", "\n", "E = V/L\n", "\n", "#Results\n", "\n", "print \"The electric field required is\",round(E/10**8,1),\"X 10^8 V/m.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The electric field required is 1.4 X 10^8 V/m.\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 12.3, page no. 436" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "import math\n", "\n", "#Variable Declaration\n", "\n", "V = 1 #voltage(eV)\n", "kBT = 0.025 # kB * T (eV)\n", "\n", "#Calculation\n", "\n", "ratio = (math.exp(V/kBT)-1)/(math.exp(-V/kBT)-1)\n", "\n", "#Results\n", "\n", "print \"The ratio of forward to reverse current is\",round(ratio/10**17,1),\"X 10^17.\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The ratio of forward to reverse current is -2.4 X 10^17.\n" ] } ], "prompt_number": 6 } ], "metadata": {} } ] }