{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 6: X-rays" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.1, Page 369" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "i = 2e-003; # Current through X-ray tube, A\n", "e = 1.6e-019; # Charge on an electron, C\n", "V = 12.4e+003; # Potential difference applied across X-ray tube, V \n", "m0 = 9.1e-031; # Rest mass of the electron, Kg \n", "\n", "#Calculations&Results\n", "n = i/e; # Number of electrons striking the target per second\n", "print \"The number of electrons striking the target per sec = %4.2e electrons\"%n\n", "v = sqrt(2*e*V/m0); # Velocity of the electrons, m/s\n", "print \"The speed with which electrons strike the target = %4.2e m/s\"%v\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The number of electrons striking the target per sec = 1.25e+16 electrons\n", "The speed with which electrons strike the target = 6.60e+07 m/s\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.2, Page 370" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "e = 1.6e-019; # Charge on an electron, C\n", "V = 13.6e+003; # Potential difference applied across X-ray tube, V \n", "m0 = 9.1e-031; # Rest mass of the electron, Kg \n", "\n", "#Calculations\n", "v = sqrt(2*e*V/m0); # Velocity of the electron, m/s \n", "\n", "#Result\n", "print \"The maximum speed with which the electrons strike the target = %4.2e m/s\"%v\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The maximum speed with which the electrons strike the target = 6.92e+07 m/s\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.3, Page 370" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "d = 2.82e-010; # Spacing of the rock-salt, m \n", "n = 2; # Order of diffraction\n", "\n", "#Calculations\n", "theta = pi/2; # Angle of diffraction, radian\n", "# Braggs equation for X-rays of wavelength lambda is n*lambda = 2*d*sin(theta), solving for lambda\n", "lamda = 2*d*sin(theta)/n; # Wavelength of X-ray using Bragg's law, m\n", "\n", "#Result\n", "print \"The longest wavelength that can be analysed by a rock-salt crystal = %4.2f angstrom\"%(lamda/1e-010)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The longest wavelength that can be analysed by a rock-salt crystal = 2.82 angstrom\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.4, Page 371" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "lamda = 3e-011; # Wavelength of the X-ray, m\n", "d = 5e-011; # Lattice spacing, m \n", "\n", "#Calculations&Results\n", "# Bragg's equation for X-rays of wavelength lambda is n*lambda = 2*d*sin(theta), solving for thetas\n", "for n in range(2,4):\n", " theta = degrees(asin((n*lamda)/(2*d))); \n", " print \"For n = %d, theta = %.1f degrees\"%(n, theta)\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "For n = 2, theta = 36.9 degrees\n", "For n = 3, theta = 64.2 degrees\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.5, Page 371" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "lamda = 3.6e-011; # Wavelength of X-rays, m\n", "n = 1; # Order of diffraction\n", "theta = 4.8; # Angle of diffraction, degrees\n", "\n", "#Calculations\n", "# Braggs equation for X-rays is n*lambda = 2*d*sin(theta), solving for d\n", "d = n*lamda/(2*sin(theta*pi/180)); # Interplanar spacing, m\n", "\n", "#Result\n", "print \"The interplanar separation of atomic planes in the crystal = %4.2f angstrom\"%(d/1e-010)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The interplanar separation of atomic planes in the crystal = 2.15 angstrom\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.6, Page 371" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "lambda1 = 0.71; # Wavelength of k alpha line in molybdenum, angstrom\n", "Z1 = 42; # Atomic number of Mo\n", "Z2 = 29; # Atomic number of Cu\n", "\n", "#Calculations\n", "# Wavelength of characteristic X-ray for K-alpha spectral line is given by \n", "# 1/lambda = 3/4*R*(Z-1)^2 then\n", "lambda2 = lambda1*(Z1-1)**2/(Z2-1)**2; # The wavelength of K alpha radiation in copper, m\n", "\n", "#Result\n", "print \"The wavelength of K-alpha radiation in copper = %4.2f angstrom\"%lambda2\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The wavelength of K-alpha radiation in copper = 1.52 angstrom\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.7, Page 372" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "phi = pi/2; # Scattering angle, degrees\n", "m0 = 9.1e-031; # Rest mass of an electron, kg\n", "h = 6.62e-034; # Planck's constant, J-s\n", "c = 3e+008; # Speed of light in vacuum, m/s \n", "E = 8.16e-014; # Energy of gamma radiation, J\n", "\n", "#Calculations\n", "lamda = h*c/(E*1e-010); # Wavelength of incident photon, angstrom \n", "lambda_prime = lamda+h*(1-cos(phi*pi/180))/(m0*c*1e-010); # Wavelength of scattered photon, angstrom\n", "\n", "#Result\n", "print \"The wavelength of radiation at 90 degrees = %6.4f angstrom\"%(lambda_prime+lamda)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The wavelength of radiation at 90 degrees = 0.0487 angstrom\n" ] } ], "prompt_number": 56 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.8, Page 372" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "phi = 90; # Scattering angle, radian\n", "m0 = 9.1e-031; # Rest mass of the electron, kg\n", "h = 6.62e-034; # Planck's constant, J-s\n", "c = 3e+008; # Speed of light in vacuum, m/s \n", "lamda = 1.00 ; # Wavelength of incident photon,in angstrom\n", "\n", "#Calculations\n", "del_lambda = (h*(1-round(cos(degrees(phi))))/(m0*c))/10**-10; # Compton shift, angstrom\n", "\n", "#Result\n", "print \"The Compton shift = %.4f angstrom\"%del_lambda\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The Compton shift = 0.0242 angstrom\n" ] } ], "prompt_number": 54 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.9, Page 373" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import *\n", "\n", "#Variable declaration\n", "phi = pi/2; # Scattering angle, radian\n", "m0 = 9.1e-031; # Rest mass of the electron, kg\n", "h = 6.62e-034; # Planck's constant, J-s\n", "c = 3e+008; # Speed of light in vacuum, m/s \n", "\n", "#Calculations\n", "# As Compton shift = del_lambda = lambda, so\n", "lamda = h*(1-cos(phi))/(m0*c*1e-010); # Wavelength of incident photon, angstrom\n", "\n", "#Result\n", "print \"The wavelength of incident radiation = %6.4f angstrom\"%lamda\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The wavelength of incident radiation = 0.0242 angstrom\n" ] } ], "prompt_number": 55 } ], "metadata": {} } ] }