diff options
Diffstat (limited to 'Modern_Physics/Chapter7.ipynb')
| -rwxr-xr-x | Modern_Physics/Chapter7.ipynb | 251 |
1 files changed, 0 insertions, 251 deletions
diff --git a/Modern_Physics/Chapter7.ipynb b/Modern_Physics/Chapter7.ipynb deleted file mode 100755 index b8966938..00000000 --- a/Modern_Physics/Chapter7.ipynb +++ /dev/null @@ -1,251 +0,0 @@ -{ - "metadata": { - "name": "Chapter7" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter 7:The Hydrogen Atom in Wave Mechanics" - ] - }, - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Example 7.2 Page 213" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "from math import exp\n", - "import math\n", - "from scipy import integrate\n", - "# calculating radial probability P= (4/ao^3)*integral(r^2 * e^(-2r/ao)) between the limits 0 and ao for r\n", - "\n", - "#calculation\n", - "def integrand(x):\n", - " return ((x**2)*exp(-x))/2.0\n", - "Pr=integrate.quad(integrand,0,2,args=());#simplifying where as x=2*r/a0; hence the limits change between 0 to 2\n", - "\n", - "#result\n", - "print \"Hence the probability of finding the electron nearer to nucleus is\",round(Pr[0],3);\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Hence the probability of finding the electron nearer to nucleus is 0.323\n" - ] - } - ], - "prompt_number": 2 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.3 Page 213" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "from math import exp\n", - "import math\n", - "from scipy import integrate\n", - "# employing the formula for probability distribution similarly as done in Exa-7.2 \n", - "#calculation\n", - "def integrand(x):\n", - " return (1.0/8)*((4.0*x**2)-(4.0*x**3)+(x**4))*exp(-x)\n", - "Pr1= integrate.quad(integrand,0,1,args=()) #x=r/ao; similrly limits between 0 and 1.\n", - "\n", - "#result\n", - "print\"The probability for l=0 electron is\",round(Pr1[0],5)\n", - "\n", - "#part2\n", - "def integrand(x):\n", - " return (1.0/24)*(x**4)*(exp(-x))\n", - "Pr2=integrate.quad(integrand,0,1); #x=r/ao; similarly limits between 0 and 1.\n", - "\n", - "#result\n", - "print\"The probability for l=1 electron is\",round(Pr2[0],5)\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The probability for l=0 electron is 0.03432\n", - "The probability for l=1 electron is 0.00366\n" - ] - } - ], - "prompt_number": 4 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.4 Page 215" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "from math import exp, sqrt\n", - "import math\n", - "from scipy import integrate\n", - "l=1.0; #given value of l\n", - "\n", - "#calculation\n", - "am1=sqrt(l*(l+1)); #angular momentum==sqrt(l(l+1)) h\n", - "l=2.0 #given l\n", - "am2=sqrt(l*(l+1));\n", - "\n", - "#result\n", - "print\"The angular momenta are found out to be\", round(am1,3),\" h and\",round(am2,3),\" h respectively for l=1 and l=2.\";\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The angular momenta are found out to be 1.414 h and 2.449 h respectively for l=1 and l=2.\n" - ] - } - ], - "prompt_number": 6 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.5 Page 216" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "from math import sqrt\n", - "print \"The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.\";\n", - "print \"Length of the vector as found out previously is %.2f*h.\",round(sqrt(6),4);#angular momentum==sqrt(l(l+1)) h" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The possible values for m are [+2,-2] and hence any of the 5 components [-2h,2h] are possible for the L vector.\n", - "Length of the vector as found out previously is %.2f*h. 2.4495\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.6 Page 223" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "uz=9.27*10**-24; t=1.4*10**3; x=3.5*10**-2; #various constants and given values\n", - "m=1.8*10**-25;v=750; # mass and velocity of the particle\n", - "\n", - "#calculation\n", - "d=(uz*t*(x**2))/(m*(v**2)); #net separtion \n", - "\n", - "#result\n", - "print\"The distance of separation in mm is\",round(d*10**3,3);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The distance of separation in mm is 0.157\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 7.7 Page 227" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "#initiation of variable\n", - "n1=1.0;n2=2.0;hc=1240.0; #hc=1240 eV.nm\n", - "E=(-13.6)*((1/n2**2)-(1/n1**2)); #Energy calculation\n", - "\n", - "#calculation\n", - "w=hc/E; #wavelength\n", - "u=9.27*10**-24; B=2; #constants\n", - "delE= u*B/(1.6*10**-19); #change in energy\n", - "delw=((w**2/hc))*delE; #change in wavelength\n", - "\n", - "#result\n", - "print\"The change in wavelength in nm. is\",round(delw,4);" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The change in wavelength in nm. is 0.0014\n" - ] - } - ], - "prompt_number": 12 - } - ], - "metadata": {} - } - ] -}
\ No newline at end of file |