{ "metadata": { "name": "", "signature": "sha256:a935d76914a9e4ae6d3e19bff37ec8b5eb26eace520b4d5bf46176dc57b307c5" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 6: Quantum Mechanics II" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.4, Page 205" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "import scipy\n", "import math\n", "from scipy.integrate import quad\n", "\n", "#Variable declaration\n", "alpha = 1; # For simplicity assume alpha to be unity\n", "\n", "#Calculations&Results\n", "p = lambda x: math.sqrt(alpha)*math.exp(-2*alpha*x) # Probability that the particle lies between 0 and 1/alpha\n", "integ, err = scipy.integrate.quad(p,0,1/alpha)\n", "print \"The probability that the particle lies between 0 and 1/alpha = %5.3f\"%integ\n", "p = lambda x: math.sqrt(alpha)*math.exp(-2*alpha*x)\n", "integ, err = scipy.integrate.quad(p,1/alpha,2/alpha) # Probability that the particle lies between 1/alpha and 2/alpha\n", "print \"The probability that the particle lies between 1/alpha and 2/alpha = %5.3f\"%integ\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The probability that the particle lies between 0 and 1/alpha = 0.432\n", "The probability that the particle lies between 1/alpha and 2/alpha = 0.059\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.9, Page 215" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "h = 6.62e-034; # Planck's constant, Js\n", "h_bar = h/(2*math.pi); # Reduced Planck's constant, Js\n", "c = 3.00e+008; # Speed of light, m/s\n", "e = 1.602e-019; # Energy equivalent of 1 eV, J\n", "m = 938.3e+006; # Energy equivalent of proton mass, eV\n", "L = 1e-005; # Diameter of the nucleus, nm\n", "\n", "#Calculations\n", "E1 = math.pi**2*(h_bar*c/(e*1e-009))**2/(2*L**2*m*1e+006); # Energy of the ground state of proton, MeV\n", "E2 = 4*E1; # Energy of first excited state of proton, MeV\n", "delta_E = E2 -E1; # Transition energy of the proton inside the nucleus, MeV\n", "\n", "#Result\n", "print \"The transition energy of the proton inside the nucleus = %1d MeV\"%delta_E" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The transition energy of the proton inside the nucleus = 6 MeV\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.14, Page 229" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "h = 6.62e-034; # Planck's constant, Js\n", "h_bar = h/(2*math.pi); # Reduced Planck's constant, Js\n", "c = 3.00e+008; # Speed of light, m/s\n", "e = 1.602e-019; # Energy equivalent of 1 eV, J\n", "m = 0.511e+006; # Energy equivalent of electron rest mass, eV\n", "V0 = 10; # Height of potential barrier, eV\n", "E = 5; # Energy of the incident electrons, eV\n", "L = 0.8e-009; # Width of the potential barrier, m\n", "\n", "#Calculations\n", "k = math.sqrt(2*m*(V0 - E))*e/(h_bar*c); # Schrodinger's constant, per m\n", "T = (1 + V0**2*math.sinh(k*L)**2/(4*E*(V0 - E)))**(-1); # Transmission electron probability\n", "\n", "#Result\n", "print \"The fraction of electrons tunnelled through the barrier = %3.1e\"%T" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The fraction of electrons tunnelled through the barrier = 4.4e-08\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.15, Page 229" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "h = 6.62e-034; # Planck's constant, Js\n", "h_bar = h/(2*math.pi); # Reduced Planck's constant, Js\n", "c = 3.00e+008; # Speed of light, m/s\n", "e = 1.602e-019; # Energy equivalent of 1 eV, J\n", "m = 0.511e+006; # Energy equivalent of electron rest mass, eV\n", "V0 = 10; # Height of potential barrier, eV\n", "\n", "#Calculations\n", "Sum_M = 0;\n", "i = 1;\n", "for E in range(1,5): # Range of energies of the incident electrons, eV\n", " M = 16*E/V0*(1-E/V0); # All the factors multiplying the exponential term\n", " Sum_M = Sum_M + M; # Accumulator\n", " i = i + 1;\n", "\n", "E = 5; # Given energy of the incident electrons, eV\n", "M = int(Sum_M/i); # Average value of M\n", "L = 0.8e-009; # Width of the potential barrier, m\n", "k = math.sqrt(2*m*(V0 - E))*e/(h_bar*c); # Schrodinger's constant, per m\n", "T = M*math.exp(-2*k*L); # Transmission electron probability\n", "\n", "#Result\n", "print \"The fraction of electrons tunnelled through the barrier = %3.1e\"%T" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The fraction of electrons tunnelled through the barrier = 2.2e-08\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.16, Page 230" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "h = 6.62e-034; # Planck's constant, Js\n", "h_bar = h/(2*math.pi); # Reduced Planck's constant, Js\n", "c = 3.00e+008; # Speed of light, m/s\n", "e = 1.602e-019; # Energy equivalent of 1 eV, J\n", "m = 0.511e+006; # Energy equivalent of electron rest mass, eV\n", "V0 = 10; # Height of potential barrier, eV\n", "E = 5; # Given energy of the incident electrons, eV\n", "\n", "#Calculations\n", "l = h_bar*c/(2*math.sqrt(2*m*(V0 - E))*1e-009*e); # Penetration distance into the barrier when the probability of the particle penetration drops to 1/e, nm\n", "\n", "#Result\n", "print \"The penetration distance for a %d eV electron approaching a step barrier of %d eV = %5.3f nm\"%(E, V0, l)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The penetration distance for a 5 eV electron approaching a step barrier of 10 eV = 0.044 nm\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.17, Page 234" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#Variable declaration\n", "h = 6.62e-034; # Planck's constant, Js\n", "h_bar = h/(2*math.pi); # Reduced Planck's constant, Js\n", "c = 3.00e+008; # Speed of light, m/s\n", "e = 1.602e-019; # Charge on an electron, C\n", "k = 9e+009; # Coulomb constant, N-Sq.m./C^2\n", "m = 3727; # Energy equivalent of alpha particle rest mass, MeV\n", "E = 5; # Given energy of the incident electrons, eV\n", "Z1 = 2; # Atomic number of an alpha particle\n", "Z2 = 92; # Atomic number of the U-238 nucleus\n", "r_N = 7e-015; # Nuclear radius, m\n", "K = 4.2; # Kinetic energy of alpha particle, MeV\n", "\n", "#Calculations\n", "V_C = Z1*Z2*e**2*k/(r_N*e*1e+006); # Coulomb Potential, MeV\n", "r_prime = V_C*r_N/(K*1e-015); # Distance through which the alpha particle must tunnel, fm\n", "kapa = math.sqrt(2*m*(V_C-E))*e/(h_bar*c*1e-006); # Schronginger's Constant, per m\n", "L = r_prime - r_N/1e-015; # Barrier width, fm\n", "T = 16*(K/V_C)*(1-K/V_C)*math.exp(-2*kapa*L*1e-015); # Tunnelling probability of alpha particle\n", "V_C_new = V_C/2; # Potential equal to half the Coulomb potential, MeV\n", "L = L/2; # Width equal to half the barrier width, fm\n", "kapa = math.sqrt(2*m*(V_C_new-E))*e/(h_bar*c*1e-006); # Schronginger's Constant, per m\n", "T_a = 16*(K/V_C_new)*(1-K/V_C_new)*math.exp(-2*kapa*L*1e-015); # Approximated tunnelling probability of alpha particle\n", "v = math.sqrt(2*K/m)*c; # Speed of the alpha particle, m/s\n", "t = r_N/v; # Time taken by alpha particle to cross the U-238 nucleus, s\n", "\n", "#Results\n", "print \"The barrier height = %2d MeV\"%(math.ceil(V_C))\n", "print \"The distance that alpha particle must tunnel = %2d fm\"%r_prime\n", "print \"The tunnelling potential assuming square-top potential = %1.0e\"%T\n", "print \"The approximated tunnelling potential = %3.1e\"%T_a\n", "print \"The speed of the alpha particle = %3.1e m/s\"%v\n", "print \"The time taken by alpha particle to cross the U-238 nucleus = %1.0e s\"%t\n", "\n", "# Some answers are given wrong in the textbook for this problem" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The barrier height = 38 MeV\n", "The distance that alpha particle must tunnel = 63 fm\n", "The tunnelling potential assuming square-top potential = 6e-123\n", "The approximated tunnelling potential = 3.8e-40\n", "The speed of the alpha particle = 1.4e+07 m/s\n", "The time taken by alpha particle to cross the U-238 nucleus = 5e-22 s\n" ] } ], "prompt_number": 7 } ], "metadata": {} } ] }