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{
"metadata": {
"name": "",
"signature": "sha256:6dfe8c3f0e395e550057fecf65810f55f95ea60853478f130ac485dd9541147f"
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"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": {}
}
]
}
|
