Ch-3 : Electron Ballistics¶

Example 3.1 : Page 48¶

In [6]:
from math import sqrt
q=1.6*10**-19 #charge of electron
V=5000 #potential difference
m=9.1*10**-31 #mass of electron
v=sqrt(2*q*V/m) #speed of electron
print "Speed of the electron, v =sqrt(2*q*V/m) = %0.2e m/s"% v
ke=(q*V)/(1.6*10**-9) #kinetic energyin eV
x1=ke*10**10
print "The kinetic energy = q x V = %0.f eV"%x1

Speed of the electron, v =sqrt(2*q*V/m) = 4.19e+07 m/s
The kinetic energy = q x V = 5000 eV


Example 3.2 Page 48¶

In [1]:
from math import sqrt
me=1000*9.1*10**-31
print "Mass of the charged particle = 1000 times the mass of an electron = %0.2e kg"%me
print "The charge of the partical = 1.6*10**-19 C"
q=1.6*10**-19 #charge of the particle
V=1000 #potential difference
v=sqrt(2*q*V/me)
print "Therefore, The velocity, v = sqrt(2*q*V/me) = %0.2e m/s"%v
ke=(q*V)/(1.6*10**-19)  # in eV
print "Kinetic energy = q x V = %0.2f eV"%ke

Mass of the charged particle = 1000 times the mass of an electron = 9.10e-28 kg
The charge of the partical = 1.6*10**-19 C
Therefore, The velocity, v = sqrt(2*q*V/me) = 5.93e+05 m/s
Kinetic energy = q x V = 1000.00 eV


Example 3.3 : Page 50¶

In [2]:
from sympy import symbols, solve
from math import sqrt
d=6*10**-3
q=1.6*10**-19
m=9.1*10**-31
vax=3*10**6
E=350/d
print "Therefore, E = V / d = %0.2e "%E
ax=q*E/m
print "    ax = qE / m = %0.2e m/s**2"%ax
print "We know that,"
print "       x = vox*t + 0.5*a*t**2"
print "      vx = vox + ax*t"
print "(i) Consider x = 3*10**-3 m"
print "3*10**-3 = 3*10**6*t + 5.13*10**15*t**2"
print "Solving this equation,"
t=symbols('t')
p1=(5.13*10**15)*t**2+(3*10**6)*t-3*10**-3
t1=solve(p1,t)
ans1=t1[1]
print "t = %0.2e seconds "%ans1
vx=(3*10**6)+((1.026*10**16)*(5.264*10**-10))
print "vx = %0.2e m/s "%vx
print "(ii) Consider x = 6*10**-6 m"
print "t**2+(5.85*10**-10)*t-(1.17*10**-18) = 0"
print "Solving this equation,"
t=symbols('t')
p2=t**2+(5.85*10**-10)*t-1.17*10**-18
t2=solve(p2, t)
ans2=t2[1]
print "t = %0.2e seconds "%ans2
vx1=(3*10**6)+((8.28*10**-10)*(1.026*10**16))
print "vx = %0.2e m/s"%vx1

Therefore, E = V / d = 5.83e+04
ax = qE / m = 1.03e+16 m/s**2
We know that,
x = vox*t + 0.5*a*t**2
vx = vox + ax*t
(i) Consider x = 3*10**-3 m
3*10**-3 = 3*10**6*t + 5.13*10**15*t**2
Solving this equation,
t = 5.26e-10 seconds
vx = 8.40e+06 m/s
(ii) Consider x = 6*10**-6 m
t**2+(5.85*10**-10)*t-(1.17*10**-18) = 0
Solving this equation,
t = 8.28e-10 seconds
vx = 1.15e+07 m/s


Example 3.4 : Page 51¶

In [3]:
from math import sqrt
V=200
m=9.1*10**-31
v=sqrt(2*q*V/m)
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"
print "v = sqrt(2*q*V/m) = %0.2e m/s"%v
iv=0 #initial velocity
fv=8.38*10**6 #final velocity
va=(iv+fv)/2 #average velocity of electron in transit
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,"
print "vaverage = (Initial velocity + Final velocity) / 2 = %0.2e m/s"%va
sp=3*10**-3 #separation between the plates
time=sp/va
print "Therefore, time taken for travel is,"
print "Time = Separation between the plates / Average velocity = %0.2e seconds"%time
ke=q*V
print "(iii)Kinetic energy of the electron on reaching the plate B is"
print "Kinetic energy = q V = %0.2e Joules"%ke

(i)The electron starts from rest at plate A, therefore, the initial velocity is zero. The velocity of electron on reaching plate B is
v = sqrt(2*q*V/m) = 8.39e+06 m/s
(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,
vaverage = (Initial velocity + Final velocity) / 2 = 4.19e+06 m/s
Therefore, time taken for travel is,
Time = Separation between the plates / Average velocity = 7.16e-10 seconds
(iii)Kinetic energy of the electron on reaching the plate B is
Kinetic energy = q V = 3.20e-17 Joules


Example 3.5 : Page 51¶

In [22]:
from math import sqrt
vinitial=1*10**6
q=1.6*10**-19
V=300
m=9.1*10**-31
vfinal=10.33*10**6
sp=8*10**-3 #separation between plates
v=sqrt(vinitial**2+(2*q*V/m))
print "The speed acquired by electron due to the applied voltage is"
print "v = sqrt(vinitial**2+(2*q*V/m)) = %0.2e m/s"%v
va=(vinitial+vfinal)/2
print "The average velocity,"
print "vaverage = (vinitial + vfinal) / 2 = %0.2e  m/s"%va
time=sp/va
print "Therefore, time for travel = seperation between plates / vaverage = %0.2e seconds"%time

The speed acquired by electron due to the applied voltage is
v = sqrt(vinitial**2+(2*q*V/m)) = 1.03e+07 m/s
The average velocity,
vaverage = (vinitial + vfinal) / 2 = 5.66e+06  m/s
Therefore, time for travel = seperation between plates / vaverage = 1.41e-09 seconds


Example 3.6 : Page 52¶

In [24]:
from math import sqrt
d=(5*10**11*1.76*10**11)*(((1*10**-9)**3)/6)
x1=d*10**6
print "The electric field intensity,"
print "E = -5t / d*10*-9 = -5t / 10**-9*1*10**-2 = 5*10**11*t (for 0 < t < t1)"
print "  = 0     (for t1 < t < infinity)"
print "(i) The position of the electron after 1ns,"
print "     d(um) = (5*10**11)*(1.76*10**11)*((1*10**-9)**3/6) = %0.2f um"%x1
x2=0.8-(d*10**2)
print "(ii) The rest of the distance to be covered by the electron = 0.8cm - 14.7 um = %0.2f"%x2
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"
vx=(5*10**11*1.76*10**11)*(((1*10**-9)**2)/2)
print "vx = (5*10**11)*(q/m)*(t**2/2) = %0.2e m/s"%vx
x3=(x2/vx)*10**-2
print "Therefore, the time t2 = d / vx = %0.2e seconds"%x3
x4=(1*10**-9)+x3
print "The total time of transit of electron from cathode to anode = %0.2e seconds"%x4

The electric field intensity,
E = -5t / d*10*-9 = -5t / 10**-9*1*10**-2 = 5*10**11*t (for 0 < t < t1)
= 0     (for t1 < t < infinity)
(i) The position of the electron after 1ns,
d(um) = (5*10**11)*(1.76*10**11)*((1*10**-9)**3/6) = 14.67 um
(ii) The rest of the distance to be covered by the electron = 0.8cm - 14.7 um = 0.80
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
vx = (5*10**11)*(q/m)*(t**2/2) = 4.40e+04 m/s
Therefore, the time t2 = d / vx = 1.81e-07 seconds
The total time of transit of electron from cathode to anode = 1.82e-07 seconds


Example 3.7 : Page 56¶

In [27]:
from math import pi,sqrt
q=1.6*10**-19
Va=40
m=9.1*10**-31
B=0.91
ve=sqrt(2*q*Va/m)
print "The velocity of the electron is = sqrt(2qVa/m) = %0.2e m/s"%ve
tt=(2*pi*m)/(B*q)
print "The time taken for one revolution is T = 2*pi*m / B*q = %0.2e seconds"%tt
p=tt*ve*(sqrt(3)/2) #cos(30)=sqrt(3)/2
print "The pitch = T*v*cos(theta) = %0.2e meters"%p
print "Thus, the electron has travelled = %0.2e meters"%p

The velocity of the electron is = sqrt(2qVa/m) = 3.75e+06 m/s
The time taken for one revolution is T = 2*pi*m / B*q = 3.93e-11 seconds
The pitch = T*v*cos(theta) = 1.28e-04 meters
Thus, the electron has travelled = 1.28e-04 meters


Example 3.8 : Page 56¶

In [30]:
from math import radians as rdn, sin,pi,sqrt
q=1.6*10**-19
m=9.1*10**-31
V=50
Q=3*q
M=2*m
v=sqrt(2*Q*V/M)
print "(i) The velocity of the charged particle before entering the field is,"
print "v = sqrt(2aV/m) * sqrt(2(3q)V/2m) = sqrt(6qV/2m) = %0.2e m/s"%v
B=0.02
r1=r*10**3
print "(ii) The radius of the helical path is"
print "r = Mvsine(theta) / QB = 2mvsine(theta) / 3qB = %0.2f mm"%r1
T=(2*pi*M)/(B*Q)
print "(iii) Time for one revolution,"
print "T = 2*pi*M / B*Q = 2*pi*(2m) / B(3q) = %0.2e seconds"%T

(i) The velocity of the charged particle before entering the field is,
v = sqrt(2aV/m) * sqrt(2(3q)V/2m) = sqrt(6qV/2m) = 5.14e+06 m/s
(ii) The radius of the helical path is
r = Mvsine(theta) / QB = 2mvsine(theta) / 3qB = 0.41 mm
(iii) Time for one revolution,
T = 2*pi*M / B*Q = 2*pi*(2m) / B(3q) = 1.19e-09 seconds


Example 3.9 : Page 58¶

In [32]:
from math import pi,sqrt
print "Given,     T = 35.5/B *10**-12 s, B = 0.01 Wb/m**3, Va = 900V"
print "Therefore, T = 3.55*10**-9 s"
T = 3.55*10**-9
Va=900
v=sqrt(2*(1.76*10**11)*900)
print "Velocity,  v(m/s) = sqrt(2qVa/m) = %0.2e m/s"%v
r=(17.799*10**6)/(0.01*1.76*10**11)
x1=r*10**3
print "Radius,  r(mm) = mv/qB = v/(q/m)B = %0.2f mm"%x1

Given,     T = 35.5/B *10**-12 s, B = 0.01 Wb/m**3, Va = 900V
Therefore, T = 3.55*10**-9 s
Velocity,  v(m/s) = sqrt(2qVa/m) = 1.78e+07 m/s
Radius,  r(mm) = mv/qB = v/(q/m)B = 10.11 mm


Example 3.10 : Page 60¶

In [33]:
from math import pi,sqrt
Va=600
l=3.5
d=0.8
L=20
Vd=20
format(9)
q=1.6*10**-19
m=9.1*10**-31
v=sqrt(2*q*Va/m)
print "(i) The velocity of the electron, v = %0.2e m/s"%v
a=(q/m)*(Vd/d)
a1=a*10**2
print "(ii) ma = qE"
print "Thus, acceleration, a(m/s)= qE / m = (q/m)(Vd/d) = %0.2e m/s"%a1
D=(l*L*Vd)/(2*Va*d)
print"(iii) The deflection on the screen, D(cm)= ILVd / 2Vad = %0.2f cm"% D
Ds=D/Vd
print  "(iv) Deflection sensitivity(cm/V)= D / Vd = %0.2f cm/V"%Ds

(i) The velocity of the electron, v = 1.45e+07 m/s
(ii) ma = qE
Thus, acceleration, a(m/s)= qE / m = (q/m)(Vd/d) = 4.40e+14 m/s
(iii) The deflection on the screen, D(cm)= ILVd / 2Vad = 1.46 cm
(iv) Deflection sensitivity(cm/V)= D / Vd = 0.07 cm/V


Example 3.11 : Page 61¶

In [34]:
from math import pi,sqrt
q=1.6*10**-19
m=9.1*10**-31
Va=800
l=2
d=0.5
L=20
D=1
v=sqrt(2*q*Va/m)
print "(i) The velocity of the beam, v = sqrt(2qVa / m) = %0.2e m/s"%v
Vd=(D*2*d*Va)/(l*L)
print "(ii) The deflection of the beam, D = lLVd / 2dVa"
print "Therefore, the voltage that must be applied to the plates, Vd = %0.2f V"%Vd

(i) The velocity of the beam, v = sqrt(2qVa / m) = 1.68e+07 m/s
(ii) The deflection of the beam, D = lLVd / 2dVa
Therefore, the voltage that must be applied to the plates, Vd = 20.00 V


Example 3.12 : Page 61¶

In [50]:
from __future__ import division
from math import degrees, atan
v=sqrt((2*(1.6*10**-19)*1000)/(9.1*10**-31))
print "(i) Velocity of beam, v = sqrt(2qVa/m) = %0.2e m/s"%v
D=((2*10**-2)*(20*10**-2)*25)/(2*1000*(0.5*10**-2))
print "(ii) Deflection sensitivity = D/Vd"
print "where  D = l*L*Vd / 2*Va*d = %0.2f cm"%D
ds=D/25
print "Therefore, the deflection sensitivity = %0.2e cm/V"%ds
theta=degrees(atan(1/1800))
print "(iii) To find the angle of deflection, theta :"
print "  tan(theta) = D/L-l"
print "Therefore,  theta = tan**-1(D/L-l) = %0.3f degrees"%theta

(i) Velocity of beam, v = sqrt(2qVa/m) = 1.88e+07 m/s
(ii) Deflection sensitivity = D/Vd
where  D = l*L*Vd / 2*Va*d = 0.01 cm
Therefore, the deflection sensitivity = 4.00e-04 cm/V
(iii) To find the angle of deflection, theta :
tan(theta) = D/L-l
Therefore,  theta = tan**-1(D/L-l) = 0.032 degrees


Example 3.13 : Page 62¶

In [52]:
from math import cos,pi
v0=3*10**5
E=910
theta=60
m=9.109*10**-31
q=1.6*10**-19
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''"
v0y=v0*cos(theta*pi/180)
print "v0y = v0 * cos(theta) = %0.2e m/s"%v0y
ay=(q*E)/m
print "ay = qE / m = %0.2e m/s**2"%ay
tdash=v0y/ay
x1=tdash*10**9
print "t'' = v0y / ay = %0.2f ns"%x1

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''
v0y = v0 * cos(theta) = 1.50e+05 m/s
ay = qE / m = 1.60e+14 m/s**2
t'' = v0y / ay = 0.94 ns


Example 3.14 : Page 62¶

In [43]:
from math import pi,sqrt
D=(((2*10**-2)*(1*10**-4)*(20*10**-2))/sqrt(800))*sqrt((1.6*10**-19)/(2*9.1*10**-31))
x1=D*10**2
print "The deflection of the spot,"
print "D = (IBL/sqrt(Va))*sqrt(q/2m) = %0.2f cm"%x1

The deflection of the spot,
D = (IBL/sqrt(Va))*sqrt(q/2m) = 0.42 cm


Example 3.15 : Page 62¶

In [4]:
from math import pi,sqrt
print "The magnetostatic deflection, D = (IBL/sqrt(Va))*sqrt(q/2m)"
print "The electrostatic deflection, D = lLVd / 2dVa"
print "For returning the beam back to the centre, the electrostatic deflection and the magnetostatic deflection must be equal, i.e.,"
print "(IBL/sqrt(Va))*sqrt(q/2m) = lLVd / 2dVa"
print "Therefore,"
Vd=(1*10**-2*2*10**-4)*sqrt((2*800*1.6*10**-19)/(9.1*10**-31))
print "Vd = dB*sqrt(2*Va*q/m) = %0.2f V"%Vd

The magnetostatic deflection, D = (IBL/sqrt(Va))*sqrt(q/2m)
The electrostatic deflection, D = lLVd / 2dVa
For returning the beam back to the centre, the electrostatic deflection and the magnetostatic deflection must be equal, i.e.,
(IBL/sqrt(Va))*sqrt(q/2m) = lLVd / 2dVa
Therefore,
Vd = dB*sqrt(2*Va*q/m) = 33.55 V