summaryrefslogtreecommitdiff
path: root/Modern_Physics_by_K_S_Krane/4-The_Wavelike_properties_of_particles.ipynb
diff options
context:
space:
mode:
Diffstat (limited to 'Modern_Physics_by_K_S_Krane/4-The_Wavelike_properties_of_particles.ipynb')
-rw-r--r--Modern_Physics_by_K_S_Krane/4-The_Wavelike_properties_of_particles.ipynb322
1 files changed, 322 insertions, 0 deletions
diff --git a/Modern_Physics_by_K_S_Krane/4-The_Wavelike_properties_of_particles.ipynb b/Modern_Physics_by_K_S_Krane/4-The_Wavelike_properties_of_particles.ipynb
new file mode 100644
index 0000000..3a27ca5
--- /dev/null
+++ b/Modern_Physics_by_K_S_Krane/4-The_Wavelike_properties_of_particles.ipynb
@@ -0,0 +1,322 @@
+{
+"cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 4: The Wavelike properties of particles"
+ ]
+ },
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.10: Group_velocity_of_a_wave_packet_in_terms_of_phase_velocity.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clc\n",
+"clear\n",
+"disp('Ex-4.10');\n",
+"printf(' Group velocity is found out from Eq. 4.18.\n Since k=2*pi/w ; Vphase= w/k \n w/k = sqrt(g/k) /n w=sqrt(g*k)');\n",
+"printf('\ndifferetiating on both sides\n');\n",
+"printf('dw=1/2 * sqrt(g) * k^-1/2 * dk\n dw= 1/2 * sqrt(g/k)\n Hence Vgroup= Vphase/2');"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.1: Solution_for_a_b_c_d_and_e.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear\n",
+"clc\n",
+"disp('Ex: 4.1 ');\n",
+"h=6.6*10^-34; // h(planck's constant)= 6.6*10^-34 \n",
+"m1= 10^3;v1=100;; // for automobile\n",
+"w1= h/(m1*v1); // ['w'-wavelength in metre'm'-mass in Kg 'v'-velocity in metres/sec.] of the particles \n",
+"printf('Wavelength of the automobile is %1.2e m\n',w1 );\n",
+"m2=10*(10^-3);v2= 500; // for bullet\n",
+"w2=h/(m2*v2);\n",
+"printf('Wavelength of the bullet is %1.2e m\n ',w2 );\n",
+"m3=(10^-9)*(10^-3); v3=1*10^-2;\n",
+"w3=h/(m3*v3);\n",
+"printf('Wavelength of the smoke particle is %1.2e m\n',w3 );\n",
+"m4=9.1*10^-31;k=1*1.6*10^-19; // k- kinetic energy of the electron & using 1ev = 1.6*10^-19 joule\n",
+"p=sqrt(2*m4*k); // p=momentum of electron ;from K=1/2*m*v^2\n",
+"w4=h/p;\n",
+"printf('Wavelength of the electron(1ev) is %1.2fnm\n',w4*10^9 );\n",
+"hc=1240;pc=100 // In the extreme relativistc realm, K=E=pc; Given pc=100MeV,hc=1240MeV \n",
+"w5= hc/pc;\n",
+"printf('Wavelength of the electron (100Mev) is %1.2f fm\n',w5);"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.2: Minimum_uncertainity_in_wavelength.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear \n",
+"clc \n",
+"disp('Ex-4.2');\n",
+"// w=wavelength; consider k=2*(pi/w); \n",
+"// differentiate k w.r.t w and replace del(k)/del(w) = 1 for equation.4.3\n",
+"// which gives del(w)= w^2 /(2*pi*del(x)), hence \n",
+"w=20; delx=200; // delx=200cm and w=20cm\n",
+"delw=(w^2)/(delx*2*%pi);\n",
+"printf('Hence uncertainity in length is %1.2f cm',delw);"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.3: Validity_of_the_claim.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear\n",
+"clc\n",
+"disp('Ex-4.3')\n",
+"delt=1; //consider time interval of 1 sec\n",
+"delw=1/delt; // since delw*delt =1 from equation 4.4\n",
+"delf=0.01 //calculated accuracy is 0.01Hz\n",
+"delwc =2*%pi*delf // delwc-claimed accuracy from w=2*pi*f\n",
+"printf('The minimum uncertainity calculated is 1rad/sec. The claimed accuracy is %.3f rad/sec\n',delwc);\n",
+"if delw==delwc then disp('Valid claim');\n",
+"end\n",
+"if delw~=delwc then disp('Invalid claim');\n",
+"end"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.4: Solution_for_a_and_b.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear\n",
+"clc\n",
+"disp('Ex-4.4(a)');\n",
+"m=9.11*10^-31;v=3.6*10^6; //'m','v' - mass an velocity of the electron in SI units\n",
+"h=1.05*10^-34; //planck's constant in SI\n",
+"p=m*v; //momentum\n",
+"delp=p*0.01;//due to 1% precision in p\n",
+"delx = h/delp//uncertainity in position\n",
+"printf('Uncertainity in position is %1.2f nm',delx*10^9);\n",
+"disp('Ex-4.4((b)')\n",
+"printf('Since the motion is strictly along X-direction, its velocity in Y direction is absolutely zero.\n So uncertainity in velocity along y is zero=> uncertainity in position along y is infinite. \nSo nothing can be said about its position/motion along Y')"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.5: Solution_for_a_and_b.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear\n",
+"clc\n",
+"disp('Ex-4.5(a)');\n",
+"m=0.145;v=42.5; //'m','v' - mass an velocity of the electron in SI units\n",
+"h=1.05*10^-34; //planck's constant in SI\n",
+"p=m*v; //momentum\n",
+"delp=p*0.01;//due to 1% precision in p\n",
+"delx = h/delp//uncertainity in position\n",
+"printf('Uncertainity in position is %1.2e',delx);\n",
+"disp('Ex-4.5(b)');\n",
+"printf('Motion along y is unpredictable as long as the veloity along y is exactly known(as zero).');"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.6: Uncertainity_in_x_component.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clc\n",
+"clear\n",
+"disp('Ex-4.6')\n",
+"printf('The uncertainity in the poisition of electron after it passes through the slit is reduced to width of the slit\n delx=a\n');\n",
+"printf('The uncertainity in momentum = h/a\n');\n",
+"printf('Position of landing(angle t) = sin t = tan t = delz/dely =(h/a)/2*pi*a= w/2*pi*a \nwhere w=wavelenghth\n');\n",
+"printf('Rewriting the above expression a*sint = w/(2*pi)\n which is similar to a*sint = w (neglect 2*pi)as found out by first minimum in diffraction by a slit of width a');\n",
+"disp('It proves a close connection between wave behaviour and uncertainity principle');"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.7: Range_of_kinetic_energy_of_an_electron.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear\n",
+"clc\n",
+"disp('Ex-4.7');\n",
+"mc2=2.15*10^-4; //mc2 is the mass of the electron, concidered in Mev for the simplicity in calculations\n",
+"hc=197 // The value of h*c in Mev.fm for simplicity\n",
+"delx= 10 // Given uncertainity in position=diameter of nucleus= 10 fm\n",
+"delp= hc/delx ; //Uncertainiy in momentum per unit 'c' i.e (Mev/c) delp= h/delx =(h*c)/(c*delx);hc=197 Mev.fm 1Mev=1.6*10^-13 Joules')\n",
+"p=delp; // Equating delp to p as a consequence of equation 4.10\n",
+"K1=[[p]^2]+[mc2]^2 // The following 3 steps are the steps invlolved in calculating K.E= sqrt((p*c)^2 + (mc^2)^2)- m*c^2\n",
+"K1=sqrt(K1)\n",
+"K1= K1-(mc2);\n",
+"printf('Kinetic energy was found out to be %d Mev', K1)"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.8: Solution_for_a_b_and_c.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear \n",
+"clc\n",
+"disp('Ex-4.8')\n",
+"h=6.58*10^-16; // plack's constant\n",
+"delt1=26*10^-9;E=140*10^6 //given values of lifetime and rest energy of charged pi meson\n",
+"delE=h/delt1; k=delE/E; // k is the measure of uncertainity\n",
+"printf('Uncertainity in energy of charged pi meson is %1.2e\n',k);\n",
+"delt2=8.3*10^-17;E=135*10^6; //given values of lifetime and rest energy of uncharged pi meson\n",
+"delE=h/delt2; k=delE/E;\n",
+"printf('Uncertainity in energy of uncharged pi meson is %1.2e\n',k);\n",
+"delt3=4.4*10^-24;E=765*10^6; //given values of lifetime and rest energy of rho meson\n",
+"delE=h/delt3; k=delE/E;\n",
+"printf('Uncertainity in energy of rho meson is %.1f\n',k);"
+ ]
+ }
+,
+{
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4.9: minimum_velocity_of_the_billiard_ball.sce"
+ ]
+ },
+ {
+"cell_type": "code",
+ "execution_count": null,
+ "metadata": {
+ "collapsed": true
+ },
+ "outputs": [],
+"source": [
+"clear\n",
+"clc\n",
+"disp('Ex-4.9')\n",
+"h=1.05*10^-34; //value of planck's constant in J.sec\n",
+"delx= 1; // uncertainity in positon= dimension of the ball\n",
+"delp=h/delx; // uncertainity in momentum \n",
+"m=0.1; //mass of the ball in kg\n",
+"delv=delp/m; // uncertainity in velocity\n",
+"printf('The value of minimum velocity was found out to be %1.2e m/sec',delv);"
+ ]
+ }
+],
+"metadata": {
+ "kernelspec": {
+ "display_name": "Scilab",
+ "language": "scilab",
+ "name": "scilab"
+ },
+ "language_info": {
+ "file_extension": ".sce",
+ "help_links": [
+ {
+ "text": "MetaKernel Magics",
+ "url": "https://github.com/calysto/metakernel/blob/master/metakernel/magics/README.md"
+ }
+ ],
+ "mimetype": "text/x-octave",
+ "name": "scilab",
+ "version": "0.7.1"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}
generated by cgit (git 2.34.1) at 2025-03-12 01:14:59 +0000