diff options
Diffstat (limited to 'sample_notebooks/LalitKumar/Ch3.ipynb')
| -rw-r--r-- | sample_notebooks/LalitKumar/Ch3.ipynb | 683 |
1 files changed, 683 insertions, 0 deletions
diff --git a/sample_notebooks/LalitKumar/Ch3.ipynb b/sample_notebooks/LalitKumar/Ch3.ipynb new file mode 100644 index 00000000..9c58334b --- /dev/null +++ b/sample_notebooks/LalitKumar/Ch3.ipynb @@ -0,0 +1,683 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-3 Electron Ballistics" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1 : Page 225" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Speed of the electron, v =sqrt(2*q*V/m) = 4.19e+07 m/s\n", + "The kinetic energy = q x V = 5000 eV\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "q=1.6*10**-19 #charge of electron\n", + "V=5000 #potential difference\n", + "m=9.1*10**-31 #mass of electron\n", + "v=sqrt(2*q*V/m) #speed of electron\n", + "print \"Speed of the electron, v =sqrt(2*q*V/m) = %0.2e m/s\"% v\n", + "ke=(q*V)/(1.6*10**-9) #kinetic energyin eV\n", + "x1=ke*10**10\n", + "print \"The kinetic energy = q x V = %0.f eV\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2 Page 225" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Mass of the charged particle = 1000 times the mass of an electron = 9.10e-28 kg\n", + "The charge of the partical = 1.6*10**-19 C\n", + "Therefore, The velocity, v = sqrt(2*q*V/me) = 5.93e+05 m/s\n", + "Kinetic energy = q x V = 1000.00 eV\n" + ] + } + ], + "source": [ + "me=1000*9.1*10**-31\n", + "print \"Mass of the charged particle = 1000 times the mass of an electron = %0.2e kg\"%me\n", + "print \"The charge of the partical = 1.6*10**-19 C\"\n", + "q=1.6*10**-19 #charge of the particle\n", + "V=1000 #potential difference\n", + "format(8)\n", + "v=sqrt(2*q*V/me)\n", + "print \"Therefore, The velocity, v = sqrt(2*q*V/me) = %0.2e m/s\"%v\n", + "ke=(q*V)/(1.6*10**-19) # in eV\n", + "print \"Kinetic energy = q x V = %0.2f eV\"%ke" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.3 : Page 226" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Therefore, E = V / d = 5.83e+04 \n", + " ax = qE / m = 1.03e+16 m/s**2\n", + "We know that,\n", + " x = vox*t + 0.5*a*t**2\n", + " vx = vox + ax*t\n", + "(i) Consider x = 3*10**-3 m\n", + "3*10**-3 = 3*10**6*t + 5.13*10**15*t**2\n", + "Solving this equation,\n", + "t = 5.26e-10 seconds \n", + "vx = 8.40e+06 m/s \n", + "(ii) Consider x = 6*10**-6 m\n", + "t**2+(5.85*10**-10)*t-(1.17*10**-18) = 0\n", + "Solving this equation,\n", + "t = 8.28e-10 seconds \n", + "vx = 1.15e+07 m/s\n" + ] + } + ], + "source": [ + "d=6*10**-3\n", + "q=1.6*10**-19\n", + "m=9.1*10**-31\n", + "vax=3*10**6\n", + "E=350/d\n", + "print \"Therefore, E = V / d = %0.2e \"%E\n", + "ax=q*E/m\n", + "print \" ax = qE / m = %0.2e m/s**2\"%ax\n", + "print \"We know that,\"\n", + "print \" x = vox*t + 0.5*a*t**2\"\n", + "print \" vx = vox + ax*t\"\n", + "print \"(i) Consider x = 3*10**-3 m\"\n", + "print \"3*10**-3 = 3*10**6*t + 5.13*10**15*t**2\"\n", + "print \"Solving this equation,\"\n", + "from sympy import symbols, solve\n", + "t=symbols('t')\n", + "p1=(5.13*10**15)*t**2+(3*10**6)*t-3*10**-3\n", + "t1=solve(p1,t)\n", + "ans1=t1[1]\n", + "print \"t = %0.2e seconds \"%ans1\n", + "vx=(3*10**6)+((1.026*10**16)*(5.264*10**-10))\n", + "print \"vx = %0.2e m/s \"%vx\n", + "print \"(ii) Consider x = 6*10**-6 m\"\n", + "print \"t**2+(5.85*10**-10)*t-(1.17*10**-18) = 0\"\n", + "print \"Solving this equation,\"\n", + "t=symbols('t')\n", + "p2=t**2+(5.85*10**-10)*t-1.17*10**-18\n", + "t2=solve(p2, t)\n", + "ans2=t2[1]\n", + "print \"t = %0.2e seconds \"%ans2\n", + "vx1=(3*10**6)+((8.28*10**-10)*(1.026*10**16))\n", + "print \"vx = %0.2e m/s\"%vx1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4 : Page 227" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)The electron starts from rest at plate A, therefore, the initial velocity is zero. The velocity of electron on reaching plate B is\n", + "v = sqrt(2*q*V/m) = 8.39e+06 m/s\n", + "(ii)Time taken by the electron to travel from plate A to plate B can be calculated from the average velocity of the electron in transit.The average velocity is,\n", + "vaverage = (Initial velocity + Final velocity) / 2 = 4.19e+06 m/s\n", + "Therefore, time taken for travel is,\n", + "Time = Separation between the plates / Average velocity = 7.16e-10 seconds\n", + "(iii)Kinetic energy of the electron on reaching the plate B is\n", + "Kinetic energy = q V = 3.20e-17 Joules\n" + ] + } + ], + "source": [ + "V=200\n", + "m=9.1*10**-31\n", + "format(8)\n", + "v=sqrt(2*q*V/m)\n", + "print \"(i)The electron starts from rest at plate A, therefore, the initial velocity is zero. The velocity of electron on reaching plate B is\"\n", + "print \"v = sqrt(2*q*V/m) = %0.2e m/s\"%v\n", + "iv=0 #initial velocity\n", + "fv=8.38*10**6 #final velocity\n", + "va=(iv+fv)/2 #average velocity of electron in transit\n", + "print \"(ii)Time taken by the electron to travel from plate A to plate B can be calculated from the average velocity of the electron in transit.The average velocity is,\"\n", + "print \"vaverage = (Initial velocity + Final velocity) / 2 = %0.2e m/s\"%va\n", + "sp=3*10**-3 #separation between the plates\n", + "time=sp/va\n", + "print \"Therefore, time taken for travel is,\"\n", + "print \"Time = Separation between the plates / Average velocity = %0.2e seconds\"%time\n", + "ke=q*V\n", + "print \"(iii)Kinetic energy of the electron on reaching the plate B is\"\n", + "print \"Kinetic energy = q V = %0.2e Joules\"%ke" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5 : Page 228" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed acquired by electron due to the applied voltage is\n", + "v = sqrt(vinitial**2+(2*q*V/m)) = 1.03e+07 m/s\n", + "The average velocity,\n", + "vaverage = (vinitial + vfinal) / 2 = 5.66e+06 m/s\n", + "Therefore, time for travel = seperation between plates / vaverage = 1.41e-09 seconds\n" + ] + } + ], + "source": [ + "vinitial=1*10**6\n", + "q=1.6*10**-19\n", + "V=300\n", + "m=9.1*10**-31\n", + "vfinal=10.33*10**6\n", + "sp=8*10**-3 #separation between plates\n", + "v=sqrt(vinitial**2+(2*q*V/m))\n", + "print \"The speed acquired by electron due to the applied voltage is\"\n", + "print \"v = sqrt(vinitial**2+(2*q*V/m)) = %0.2e m/s\"%v\n", + "va=(vinitial+vfinal)/2\n", + "print \"The average velocity,\"\n", + "print \"vaverage = (vinitial + vfinal) / 2 = %0.2e m/s\"%va\n", + "time=sp/va\n", + "print \"Therefore, time for travel = seperation between plates / vaverage = %0.2e seconds\"%time" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6 : Page 229" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The electric field intensity,\n", + "E = -5t / d*10*-9 = -5t / 10**-9*1*10**-2 = 5*10**11*t (for 0 < t < t1)\n", + " = 0 (for t1 < t < infinity)\n", + "(i) The position of the electron after 1ns,\n", + " d(um) = (5*10**11)*(1.76*10**11)*((1*10**-9)**3/6) = 14.67 um\n", + "(ii) The rest of the distance to be covered by the electron = 0.8cm - 14.7 um = 0.80\n", + "Since, the potential difference drops to zero volt, after 1ns, the electron will travel the distance of 0.799 cm with a constant velocity of\n", + "vx = (5*10**11)*(q/m)*(t**2/2) = 4.40e+04 m/s\n", + "Therefore, the time t2 = d / vx = 1.81e-07 seconds\n", + "The total time of transit of electron from cathode to anode = 1.82e-07 seconds\n" + ] + } + ], + "source": [ + "d=(5*10**11*1.76*10**11)*(((1*10**-9)**3)/6)\n", + "x1=d*10**6\n", + "print \"The electric field intensity,\"\n", + "print \"E = -5t / d*10*-9 = -5t / 10**-9*1*10**-2 = 5*10**11*t (for 0 < t < t1)\"\n", + "print \" = 0 (for t1 < t < infinity)\"\n", + "print \"(i) The position of the electron after 1ns,\"\n", + "print \" d(um) = (5*10**11)*(1.76*10**11)*((1*10**-9)**3/6) = %0.2f um\"%x1\n", + "x2=0.8-(d*10**2)\n", + "print \"(ii) The rest of the distance to be covered by the electron = 0.8cm - 14.7 um = %0.2f\"%x2\n", + "print \"Since, the potential difference drops to zero volt, after 1ns, the electron will travel the distance of 0.799 cm with a constant velocity of\"\n", + "vx=(5*10**11*1.76*10**11)*(((1*10**-9)**2)/2)\n", + "print \"vx = (5*10**11)*(q/m)*(t**2/2) = %0.2e m/s\"%vx\n", + "x3=(x2/vx)*10**-2\n", + "print \"Therefore, the time t2 = d / vx = %0.2e seconds\"%x3\n", + "x4=(1*10**-9)+x3\n", + "print \"The total time of transit of electron from cathode to anode = %0.2e seconds\"%x4" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.7 : Page 230" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The velocity of the electron is = sqrt(2qVa/m) = 3.75e+06 m/s\n", + "The time taken for one revolution is T = 2*pi*m / B*q = 3.93e-11 seconds\n", + "The pitch = T*v*cos(theta) = 1.28e-04 meters\n", + "Thus, the electron has travelled = 1.28e-04 meters\n" + ] + } + ], + "source": [ + "from math import pi\n", + "q=1.6*10**-19\n", + "Va=40\n", + "m=9.1*10**-31\n", + "B=0.91\n", + "ve=sqrt(2*q*Va/m)\n", + "print \"The velocity of the electron is = sqrt(2qVa/m) = %0.2e m/s\"%ve\n", + "tt=(2*pi*m)/(B*q)\n", + "print \"The time taken for one revolution is T = 2*pi*m / B*q = %0.2e seconds\"%tt\n", + "p=tt*ve*(sqrt(3)/2) #cos(30)=sqrt(3)/2\n", + "print \"The pitch = T*v*cos(theta) = %0.2e meters\"%p\n", + "print \"Thus, the electron has travelled = %0.2e meters\"%p" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.8 : Page 231" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The velocity of the charged particle before entering the field is,\n", + "v = sqrt(2aV/m) * sqrt(2(3q)V/2m) = sqrt(6qV/2m) = 5.14e+06 m/s\n", + "(ii) The radius of the helical path is\n", + "r = Mvsine(theta) / QB = 2mvsine(theta) / 3qB = 0.41 mm\n", + "(iii) Time for one revolution,\n", + "T = 2*pi*M / B*Q = 2*pi*(2m) / B(3q) = 1.19e-09 seconds\n" + ] + } + ], + "source": [ + "from math import radians as rdn, sin\n", + "radians=rdn(25)\n", + "q=1.6*10**-19\n", + "m=9.1*10**-31\n", + "V=50\n", + "Q=3*q\n", + "M=2*m\n", + "v=sqrt(2*Q*V/M)\n", + "print \"(i) The velocity of the charged particle before entering the field is,\"\n", + "print \"v = sqrt(2aV/m) * sqrt(2(3q)V/2m) = sqrt(6qV/2m) = %0.2e m/s\"%v\n", + "B=0.02\n", + "r=(M*v*sin(radians))/(Q*B)\n", + "r1=r*10**3\n", + "print \"(ii) The radius of the helical path is\"\n", + "print \"r = Mvsine(theta) / QB = 2mvsine(theta) / 3qB = %0.2f mm\"%r1\n", + "T=(2*pi*M)/(B*Q)\n", + "print \"(iii) Time for one revolution,\"\n", + "print \"T = 2*pi*M / B*Q = 2*pi*(2m) / B(3q) = %0.2e seconds\"%T" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.9 : Page 232" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Given, T = 35.5/B *10**-12 s, B = 0.01 Wb/m**3, Va = 900V\n", + "Therefore, T = 3.55*10**-9 s\n", + "Velocity, v(m/s) = sqrt(2qVa/m) = 1.78e+07 m/s\n", + "Radius, r(mm) = mv/qB = v/(q/m)B = 10.11 mm\n" + ] + } + ], + "source": [ + "print \"Given, T = 35.5/B *10**-12 s, B = 0.01 Wb/m**3, Va = 900V\"\n", + "print \"Therefore, T = 3.55*10**-9 s\"\n", + "T = 3.55*10**-9\n", + "Va=900\n", + "v=sqrt(2*(1.76*10**11)*900)\n", + "print \"Velocity, v(m/s) = sqrt(2qVa/m) = %0.2e m/s\"%v\n", + "r=(17.799*10**6)/(0.01*1.76*10**11)\n", + "x1=r*10**3\n", + "print \"Radius, r(mm) = mv/qB = v/(q/m)B = %0.2f mm\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.10 : Page 232" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The velocity of the electron, v = 1.45e+07 m/s\n", + "(ii) ma = qE\n", + "Thus, acceleration, a(m/s)= qE / m = (q/m)(Vd/d) = 4.40e+14 m/s\n", + "(iii) The deflection on the screen, D(cm)= ILVd / 2Vad = 1.46 cm\n", + "(iv) Deflection sensitivity(cm/V)= D / Vd = 0.07 cm/V\n" + ] + } + ], + "source": [ + "Va=600\n", + "l=3.5\n", + "d=0.8\n", + "L=20\n", + "Vd=20\n", + "format(9)\n", + "q=1.6*10**-19\n", + "m=9.1*10**-31\n", + "v=sqrt(2*q*Va/m)\n", + "print \"(i) The velocity of the electron, v = %0.2e m/s\"%v\n", + "a=(q/m)*(Vd/d)\n", + "a1=a*10**2\n", + "print \"(ii) ma = qE\"\n", + "print \"Thus, acceleration, a(m/s)= qE / m = (q/m)(Vd/d) = %0.2e m/s\"%a1\n", + "D=(l*L*Vd)/(2*Va*d)\n", + "print\"(iii) The deflection on the screen, D(cm)= ILVd / 2Vad = %0.2f cm\"% D\n", + "Ds=D/Vd\n", + "print \"(iv) Deflection sensitivity(cm/V)= D / Vd = %0.2f cm/V\"%Ds" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.11 : Page 233" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The velocity of the beam, v = sqrt(2qVa / m) = 1.68e+07 m/s\n", + "(ii) The deflection of the beam, D = lLVd / 2dVa\n", + "Therefore, the voltage that must be applied to the plates, Vd = 20.00 V\n" + ] + } + ], + "source": [ + "q=1.6*10**-19\n", + "m=9.1*10**-31\n", + "Va=800\n", + "l=2\n", + "d=0.5\n", + "L=20\n", + "D=1\n", + "v=sqrt(2*q*Va/m)\n", + "print \"(i) The velocity of the beam, v = sqrt(2qVa / m) = %0.2e m/s\"%v\n", + "Vd=(D*2*d*Va)/(l*L)\n", + "print \"(ii) The deflection of the beam, D = lLVd / 2dVa\"\n", + "print \"Therefore, the voltage that must be applied to the plates, Vd = %0.2f V\"%Vd" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.12 : Page 234" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Velocity of beam, v = sqrt(2qVa/m) = 1.88e+07 m/s\n", + "(ii) Deflection sensitivity = D/Vd\n", + "where D = l*L*Vd / 2*Va*d = 0.01 cm\n", + "Therefore, the deflection sensitivity = 4.00e-04 cm/V\n", + "(iii) To find the angle of deflection, theta :\n", + " tan(theta) = D/L-l\n", + "Therefore, theta = tan**-1(D/L-l) = 0.032 degrees\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import degrees, atan\n", + "v=sqrt((2*(1.6*10**-19)*1000)/(9.1*10**-31))\n", + "print \"(i) Velocity of beam, v = sqrt(2qVa/m) = %0.2e m/s\"%v\n", + "D=((2*10**-2)*(20*10**-2)*25)/(2*1000*(0.5*10**-2))\n", + "print \"(ii) Deflection sensitivity = D/Vd\"\n", + "print \"where D = l*L*Vd / 2*Va*d = %0.2f cm\"%D\n", + "ds=D/25\n", + "print \"Therefore, the deflection sensitivity = %0.2e cm/V\"%ds\n", + "theta=degrees(atan(1/1800))\n", + "print \"(iii) To find the angle of deflection, theta :\"\n", + "print \" tan(theta) = D/L-l\"\n", + "print \"Therefore, theta = tan**-1(D/L-l) = %0.3f degrees\"%theta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.13 : Page 235" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The electron starts moving in the +y direction, but, since acceleration is along the -y direction, its velocity isreduced to zero at time t=t''\n", + "v0y = v0 * cos(theta) = 1.50e+05 m/s\n", + "ay = qE / m = 1.60e+14 m/s**2\n", + "t'' = v0y / ay = 0.94 ns\n" + ] + } + ], + "source": [ + "from math import cos\n", + "v0=3*10**5\n", + "E=910\n", + "theta=60\n", + "m=9.109*10**-31\n", + "q=1.6*10**-19\n", + "print \"The electron starts moving in the +y direction, but, since acceleration is along the -y direction, its velocity isreduced to zero at time t=t''\"\n", + "v0y=v0*cos(theta*pi/180)\n", + "print \"v0y = v0 * cos(theta) = %0.2e m/s\"%v0y\n", + "ay=(q*E)/m\n", + "print \"ay = qE / m = %0.2e m/s**2\"%ay\n", + "tdash=v0y/ay\n", + "x1=tdash*10**9\n", + "print \"t'' = v0y / ay = %0.2f ns\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.14 : Page 235" + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The deflection of the spot,\n", + "D = (IBL/sqrt(Va))*sqrt(q/2m) = 0.42 cm\n" + ] + } + ], + "source": [ + "D=(((2*10**-2)*(1*10**-4)*(20*10**-2))/sqrt(800))*sqrt((1.6*10**-19)/(2*9.1*10**-31))\n", + "x1=D*10**2\n", + "print \"The deflection of the spot,\"\n", + "print \"D = (IBL/sqrt(Va))*sqrt(q/2m) = %0.2f cm\"%x1" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.15 : Page 236" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnetostatic deflection, D = (IBL/sqrt(Va))*sqrt(q/2m)\n", + "The electrostatic deflection, D = lLVd / 2dVa\n", + "For returning the beam back to the centre, the electrostatic deflection and the magnetostatic deflection must be equal, i.e.,\n", + "(IBL/sqrt(Va))*sqrt(q/2m) = lLVd / 2dVa\n", + "Therefore,\n", + "Vd = dB*sqrt(2*Va*q/m) = 33.55 V\n" + ] + } + ], + "source": [ + "print \"The magnetostatic deflection, D = (IBL/sqrt(Va))*sqrt(q/2m)\"\n", + "print \"The electrostatic deflection, D = lLVd / 2dVa\"\n", + "print \"For returning the beam back to the centre, the electrostatic deflection and the magnetostatic deflection must be equal, i.e.,\"\n", + "print \"(IBL/sqrt(Va))*sqrt(q/2m) = lLVd / 2dVa\"\n", + "print \"Therefore,\"\n", + "Vd=(1*10**-2*2*10**-4)*sqrt((2*800*1.6*10**-19)/(9.1*10**-31))\n", + "print \"Vd = dB*sqrt(2*Va*q/m) = %0.2f V\"%Vd" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |