{ "metadata": { "name": "", "signature": "sha256:64e0a844a3ba33dead720cfb202f40f7dab27e1a9fab7ab4975bb30624a36195" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 31 Radioactivity" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 31.1 Page no 855" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "T=138.0 #days\n", "E=0.693\n", "N=12.5\n", "N0=100\n", "\n", "#Calculation\n", "import math\n", "L=E/T\n", "N1=N/N0\n", "t=(2.303*math.log10(8))/L\n", "\n", "#Result\n", "print\"Time is\",round(t,2),\"days\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Time is 414.16 days\n" ] } ], "prompt_number": 71 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 31.2 Page no 855" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "T=1.42*10**17 #s\n", "A=6.02*10**23 #mol**-1\n", "E=0.693\n", "L=238.0\n", "\n", "#Calculation\n", "N=A/L\n", "L1=E/T\n", "Z=L1*N\n", "\n", "#Result\n", "print\"Disintegrations per second occur in 1g of U**238 is\",round(Z*10**-4,3),\"*10**4 s**-1\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Disintegrations per second occur in 1g of U**238 is 1.234 *10**4 s**-1\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 31.3 Page no 855" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "K=-40 #Cl\n", "M=0.075 #kg mole**-1\n", "m=1.2*10**-6 #kg\n", "A=6.0*10**23\n", "D=170 #s**-1\n", "E=0.693\n", "\n", "#Calculation\n", "N=(A/M)*m\n", "L=D/N\n", "T=(E/L)\n", "O=T/31536000.0\n", "\n", "#Result\n", "print\"Half life of K40 atom is\",round(O*10**-9,3),\"*10**9 years\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Half life of K40 atom is 1.241 *10**9 years\n" ] } ], "prompt_number": 81 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 31.4 Page no 856" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "T=6.0 #hours\n", "E=0.693\n", "A=6.025*10**23\n", "W=99.0\n", "S=10**-12\n", "t=1 #hours\n", "\n", "#Calculation\n", "import math\n", "L=E/T\n", "N0=(A/W)*S\n", "R0=L*N0\n", "N=N0*math.exp(-L)\n", "R=L*N\n", "R1=L*N\n", "\n", "#Result\n", "print\"Activity at beginning is\", round(R0*10**-8,2),\"*10**8 /h\"\n", "print\"Activity at the end is\",round(R1*10**-8,3),\"*10**8 /h\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Activity at beginning is 7.03 *10**8 /h\n", "Activity at the end is 6.262 *10**8 /h\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 31.5 Page no 856" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "t=30.0 #years\n", "a=16\n", "\n", "#Calculation\n", "import math\n", "l=(2.3026*math.log10(a))/t\n", "T=0.693/l\n", "\n", "#Result\n", "print\"Half life period is\", round(T,1),\"years\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Half life period is 7.5 years\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 31.6 Page no 856" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "m1=238.05081\n", "m2=234.04363\n", "m3=4.00260\n", "A=931.5 #Mev\n", "\n", "#Calculation\n", "Ea=(m1-m2-m3)*A\n", "\n", "#Result\n", "print\"Kinetic energy is\",round(Ea,2),\"Mev\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Kinetic energy is 4.27 Mev\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 31.7 Page no 856" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "m1=22.994466\n", "m2=22.989770\n", "A=931.5\n", "\n", "#Calculation\n", "Eb=(m1-m2)*A\n", "\n", "#Result\n", "print\"Maximum kinetic energy is\",round(Eb,3),\"Mev\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Maximum kinetic energy is 4.374 Mev\n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 31.8 Page no 856" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "E1=0\n", "E2=0.412*1.6*10**-13\n", "E3=1.088*1.6*10**-13\n", "h=6.62*10**-34\n", "m1=197.968233\n", "m2=197.966760\n", "a=931.5\n", "A=1.088\n", "A1=0.412\n", "\n", "#Calculation\n", "v1=(E3-E1)/h\n", "v2=(E2-E1)/h\n", "v3=(E3-E2)/h\n", "Eb1=((m1-m2)*a)-A\n", "Eb2=((m1-m2)*a)-A1\n", "\n", "#Result\n", "print\"Radiation frequencies are\",round(v1*10**-20,2)*10**20,\",\",round(v2*10**-20,3)*10**20,\"and\",round(v3*10**-20,2)*10**20\n", "print\"Maximum kinetic energies are\",round(Eb1,3),\"Mev and \",round(Eb2,3),\"Mev\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Radiation frequencies are 2.63e+20 , 9.96e+19 and 1.63e+20\n", "Maximum kinetic energies are 0.284 Mev and 0.96 Mev\n" ] } ], "prompt_number": 42 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 31.9 Page no 857" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "a=15\n", "b=9.0\n", "T=5730\n", "\n", "#Calculation\n", "import math\n", "A=a/b\n", "l=2.303*math.log10(A)\n", "t=(l*T)/0.693\n", "\n", "#Result\n", "print\"Approximation age of the indus valley civilization is\",round(t,2),\"years\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Approximation age of the indus valley civilization is 4224.47 years\n" ] } ], "prompt_number": 46 } ], "metadata": {} } ] }