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diff --git a/Modern_Physics/chapter4.ipynb b/Modern_Physics/chapter4.ipynb deleted file mode 100755 index 9b118dda..00000000 --- a/Modern_Physics/chapter4.ipynb +++ /dev/null @@ -1,320 +0,0 @@ -{
- "metadata": {
- "name": "",
- "signature": "sha256:f3a8bba44ff010c0f1004529fad8de811f19168b5752929eb1e0272cfc5fc53c"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "heading",
- "level": 1,
- "metadata": {},
- "source": [
- "4: Crystallography"
- ]
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.2, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#import modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "d=2180; #density of NaCl(kg/m^3)\n",
- "M=23+35.5; #Molecular weight of NaCl(gm)\n",
- "Na=6.02*10**26; #Avgraodo no(per kg mole)\n",
- "n=4; #for f.c.c\n",
- "\n",
- "#calculation\n",
- "a=(n*M/(Na*d))**(1/3); #lattice constant(m)\n",
- "d=a/2; #distance(m)\n",
- "d=d*10**10; #distance(angstrom)\n",
- "d=math.ceil(d*10**3)/10**3; #rounding off to 3 decimals\n",
- "\n",
- "#Result\n",
- "print \"distance between two adajcent atoms is\",d,\"angstrom\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "distance between two adajcent atoms is 2.815 angstrom\n"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.3, Page number 70"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#import modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "d=2.163; #density(gm/cm^3)\n",
- "M=58.45; #molecular weight(gm)\n",
- "Na=6.02*10**23; #Avgraodo no.(molecules/gm mole)\n",
- "\n",
- "#calcualtion\n",
- "n=Na/M; #no. of molecules(per gram)\n",
- "n=n*d; #no. of molecules(per cm^3) \n",
- "n=2*n; #no. of atom(per cm^3)\n",
- "n=n**(1/3); #no. of atoms in a row 1cm long\n",
- "d=1/n; #spacing between atoms(cm)\n",
- "\n",
- "#Result\n",
- "print \"spacing between atoms is\",round(d/1e-8,2),\"angstrom\"\n",
- "print \"answer in the book varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "spacing between atoms is 2.82 angstrom\n",
- "answer in the book varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 17
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.4, Page number 74"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#import modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Variable declaration\n",
- "r=1.278; #radius(A.U)\n",
- "n=4; #structure is f.c.c\n",
- "M=63.54; #atomic weight(gm)\n",
- "Na=6.02*10**23; #Avgraodo no.(per gm mole)\n",
- "\n",
- "#calculation \n",
- "a=4*r/(math.sqrt(2)); #lattice constant(A.U)\n",
- "V=a**3; #volume(cm^3)\n",
- "rho=n*M/(Na*V); #density(gm/cm^3)\n",
- "rho=rho*(10**8)**3 #density(gm/m^3)\n",
- "\n",
- "#Result\n",
- "print \"Density is\",round(rho,2),\"g/m^3\"\n",
- "print \"answer in the textbook varies due to rounding off errors\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Density is 8.94 g/m^3\n",
- "answer in the textbook varies due to rounding off errors\n"
- ]
- }
- ],
- "prompt_number": 29
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.10, Page number 85"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#import modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#Vaiable declaration\n",
- "r=1.746; #atomic radius(AU)\n",
- "\n",
- "#calulation\n",
- "a=4*r/math.sqrt(2); #lattice constant(AU)\n",
- "#for (200)\n",
- "h=2;k=0;l=0;\n",
- "d=a/math.sqrt(h**2+k**2+l**2); #interplanar spacing(AU)\n",
- "#for (220)\n",
- "h=2;k=2;l=0; \n",
- "d1=a/math.sqrt(h**2+k**2+l**2); #interplanar spacing(AU)\n",
- "#for (111)\n",
- "h=1;k=1;l=1;\n",
- "d2=a/math.sqrt(h**2+k**2+l**2); #interplanar spacing(AU)\n",
- "d=math.ceil(d*10**4)/10**4; #rounding off to 4 decimals\n",
- "d1=math.ceil(d1*10**3)/10**3; #rounding off to 3 decimals\n",
- "d2=math.ceil(d2*10**3)/10**3; #rounding off to 3 decimals\n",
- "\n",
- "#Result\n",
- "print \"spacing for (200) is\",d,\"A.U\"\n",
- "print \"answer in the book varies in 3rd decimal due to rounding off errors\"\n",
- "print \"spacing for (220) is\",d1,\"A.U\"\n",
- "print \"spacing for (111) is\",d2,\"A.U\"\n"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "spacing for (200) is 2.4693 A.U\n",
- "answer in the book varies in 3rd decimal due to rounding off errors\n",
- "spacing for (220) is 1.746 A.U\n",
- "spacing for (111) is 2.852 A.U\n"
- ]
- }
- ],
- "prompt_number": 34
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.11, Page number 86"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#import modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#calculation\n",
- "#for (i)\n",
- "l=1;m=0;n=0;\n",
- "p=0;q=1;r=0;\n",
- "d=math.acos(((l*p)+(m*q)+(n*r))/(math.sqrt(l**2+m**2+n**2)*math.sqrt(p**2+q**2+r**2))); #angle between 2 normals(radian)\n",
- "theta=math.degrees(d); #angle between 2 normals(degrees) \n",
- "\n",
- "#for (ii)\n",
- "l=1;m=2;n=1;\n",
- "p=1;q=1;r=1;\n",
- "d1=math.acos(((l*p)+(m*q)+(n*r))/(math.sqrt(l**2+m**2+n**2)*math.sqrt(p**2+q**2+r**2))); #angle between 2 normals(radian)\n",
- "theta1=math.degrees(d1); #angle between 2 normals(degrees) \n",
- "deg=int(theta1); #angle(degrees)\n",
- "t=60*(theta1-deg);\n",
- "m=int(t); #angle(minutes)\n",
- " \n",
- "#Result\n",
- "print \"angle between the normal to pair of miller incdices (100) and (010) is\",theta,\"degrees\"\n",
- "print \"angle between the normal to pair of miller incdices (121) and (111) is\",deg,\"degrees\",m,\"minutes\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "angle between the normal to pair of miller incdices (100) and (010) is 90.0 degrees\n",
- "angle between the normal to pair of miller incdices (121) and (111) is 19 degrees 28 minutes\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "heading",
- "level": 2,
- "metadata": {},
- "source": [
- "Example number 4.13, Page number 87"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#import modules\n",
- "import math\n",
- "from __future__ import division\n",
- "\n",
- "#variable declaration\n",
- "a=3.61*10**-7; #lattice constant(mm)\n",
- "\n",
- "#Calcualtion\n",
- "#for plane (100)\n",
- "SA=a*a; #surface area(mm^2)\n",
- "tamc=2; #total atoms included\n",
- "ans=tamc/SA; #number of atoms per mm^2\n",
- "#for (ii) plane (110)\n",
- "A=a*(math.sqrt(2)*a); #area of the plane(mm^2)\n",
- "tamc=2; #total atoms included according to sketch\n",
- "ans1=tamc/A; #number of atoms per mm^2\n",
- "#for (iii) plane (111)\n",
- "A=0.866*a*a; #area of the plane(mm^2)\n",
- "tamc=2; #total atoms included according to sketch\n",
- "ans2=tamc/A; #number of atoms per mm^2\n",
- "\n",
- "#Result\n",
- "print \"atoms per mm^2 for (100) is\",round(ans/1e+13,3),\"*10**13 atoms/mm^2\"\n",
- "print \"atoms per mm^2 for (110) is\",round(ans1/1e+13,3),\"*10**13 atoms/mm^2\"\n",
- "print \"atoms per mm^2 for (111) is\",round(ans2/1e+13,3),\"*10**13 atoms/mm^2\"\n",
- "print \"answer for plane (111) given in the book is wrong\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "atoms per mm^2 for (100) is 1.535 *10**13 atoms/mm^2\n",
- "atoms per mm^2 for (110) is 1.085 *10**13 atoms/mm^2\n",
- "atoms per mm^2 for (111) is 1.772 *10**13 atoms/mm^2\n",
- "answer for plane (111) given in the book is wrong\n"
- ]
- }
- ],
- "prompt_number": 8
- }
- ],
- "metadata": {}
- }
- ]
-}
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