Chapter 3 Ionization and Deionization Processes in Gases

Example 3_1 pgno:130

In [1]:
#Chapter 3, Exmaple 1, page 103
#Movement of oxygen molecule

#using equation 3.3
R = 3814 # J/Kg.mol.K
T = 300 # K
M = 32 # mol^-1
V2 = 3*R*(T/M)
V = (V2)**0.5
print"Velocity of Oxygen (O2)=  m^2/s^2",V2
#Since Oxygen is a diatomic gas
print"Velocity of Oxygen (O)=  m/s",V
#Velocity of oxygen is about 300 m/s

#Answer given in the textbook is wrong

Velocity of Oxygen (O2)=  m^2/s^2 102978
Velocity of Oxygen (O)=  m/s 320.901854155

Example 3_2 pgno:104

In [2]:
#Chapter 3, Exmaple 2, page 104
#Kinetic energy of oxygen molecule
#from Eq.3.2
G = (2*10**-3/32)*(8314*298*1.01*10**5)*10**-10
print"\nG =  m^3\t",G # Answer is is wrong in the text 
#From equation 3.1
mv2 = 3/2*1.01*10**5 # 1/2*m0*v^2
KE = mv2*G#total transalational K.E
print"K.E =  J\t",KE
#Answer may varry due to round off error

G =  m^3	0.001563967325
K.E =  J	157.960699825

Example 3_3 pgno:104

In [4]:
#Chapter 3, Exmaple 3, page 104
#Maximum pressure in the chamber
from math import pi
#Making use of equation 3.10
N1 = (4*pi*1.7*1.7*0.10*10**-10*10**-10)
N = 1/N1
#Using equation 3.2
R = 8314 # J/Kg*mol*K 
M = 28 # Mol**-1
N = 220*10**-8 # Kg
T = 300 # K
p =  N/M*R*T
print"\nN = ",N1
print"\nPressure = N/m**2",round(p,3)

#Answer vary due to round off error

N =  3.63168110755e-20

Pressure = N/m**2 0.196

Example 3_4 pgno:105

In [6]:
#Chapter 3, Exmaple 4, page 105
#Temperature & Average K.E of He atom

m0 = 1
v2 = 1.6*10**-19 # V**2
KE = m0*v2
#Using equation 3.3
T = 2*KE/(3*1.38*10**-23) 
print"\nK.E =  J",KE
print"\nTemperature = K",round(T)
  

K.E =  J 1.6e-19

Temperature = K 7729.0

Example 3_5 pgno:105

In [8]:
#Chapter 3, Exmaple 5, page 105
#Volume of Helium 

# Using equation 3.2
G = (1*8314*273)/(2.016*1.01*10**5)
print"\nVolume of He =  m^3",round(G,2)

Volume of He =  m^3 11.15

Example 3_6 pgno:105

In [11]:
#Chapter 3, Exmaple 6, page 105
#Determine mean free path 
from math import e
#(a) Mean free path
na = e**-1
#(b) 5 times mean free path
nb = e**-5
print"\n Mean free path = *n0 ",na
print"\n 5 times mean free path = *n0 ",round(nb,4)

 
 Mean free path = *n0  0.367879441171

 5 times mean free path = *n0  0.0067

Example 3_7 pgno:105

In [13]:
#Chapter 3, Exmaple 7, page 105
#Mean square velocity of Helium 

#based on equation 3.2 and 3.3 we derive the gas density
N = 178*10**-3 # kg/m**3
# calculating mean square velocity
v2 = (3*1.01*10**5)/N
print"\nV**2 =  m**2/s**2",v2
v = (v2)**0.5
print"\nMean square velocity = m/s",round(v)

V**2 =  m**2/s**2 1702247.19101

Mean square velocity = m/s 1305.0

Example 3_8 pgno:106

In [16]:
#Chapter 3, Exmaple 8, page 106
#Energy of free electron

#Using equation 3.3
mv2 = (3/2*1.38*10**-21*293) # 1/2*m*v^2
E = mv2*10**38/1.6*10**-19+0.79
print"\n1/2*m*v^2 =  J",mv2
print"\nEnergy of free electron =  eV",round(E,2)

#Answers may vary due to round off error

1/2*m*v^2 =  J 4.0434e-19

Energy of free electron =  eV 3.317125

Example 3_9 pgno:106

In [17]:
#Chapter 3, Exmaple 9, page 106
#Average separation and volume occupied by one atom

NA = 6.0244*10**23
NoA = NA*0.075 # Number of atoms/cm^3
V = 1/NoA # Average volume occupied by one atom
S = 2.807641e-08#nthroot(V,3) #  Average separation between atoms
print"\nNumber od atoms per cm^3 = ",NoA
print"\nAverage vloume occupied by one atom = cm^3",V
print"\nAverage separation between atoms = cm",S


#Answers may vary due to round off error

Number od atoms per cm^3 =  4.5183e+22

Average vloume occupied by one atom = cm^3 2.21322178696e-23

Average separation between atoms = cm 2.807641e-08

Example 3_10 pgno:106

In [18]:
#Chapter 3, Exmaple 10, page 106
#KE and velocity of photoelectron

h = 4.15*10**-15
c = 3*10**8
l = 200*10**-10
BE = 13.6 # Binding energy
PE = h*c/l
KE = PE-BE # Kinetic energy of photoelectron
Ve = ((2*KE*1.6*10**-19)/9.11*10**-31)**0.5*10**31
print"\nPhoton energy eV =  ",PE
print"\nKinetic energy eV =  ",KE
print"\nVelocity m/s =  ",Ve

#Answer may vary due to round off error

Photon energy eV =   62.25

Kinetic energy eV =   48.65

Velocity m/s =   4133873.88319

Example 3_11 pgno:107

In [20]:
#Chapter 3, Exmaple 11, page 107
#Find the absorption coefficient
from math import log
# Using equation 3.20
x = 20.
I0 = 6.
Mu = -1/x*log(1/I0)
print"\nLiquid photon absorption coefficient cm^-1 =  ",round(Mu,4)

 

Liquid photon absorption coefficient cm^-1 =   0.0896

Example 3_12 pgno:107

In [21]:
#Chapter 3, Exmaple 12, page 107
#Binding energy
h = 4.15*10**-15
c = 3*10**8
Imax = 1000*10**-10
We = h*c/Imax
print"\nBinding Energy =  eV ",We

#Answer may vary due to round off errorS

Binding Energy =  eV  12.45

Example 3_14 pgno:108

In [23]:
#Chapter 3, Exmaple 14, page 108
#Diameter of argon atom
from math import pi
#As derived from example 13
N = (1.01*10**5/760)/(1.38*10**-23*273)
print"\nN =  atoms/m**3 ",N
#Use equation 3.10
ra = ((85*10**2)/(pi*3.527*10**22))**0.5 
print"\nra =  m ",ra

#Answer may vary due to round off error

N =  atoms/m**3  3.52749208584e+22

ra =  m  2.7696940413e-10

Example 3_15 pgno:109

In [24]:
#Chapter 3, Exmaple 15, page 109
#Mobility of electrons

Ie = 3
d = 0.8
A = 8*10**-4
Vne = 20*10**17 #V*ne
e = 1.6*10**-19
ke = (Ie*d)/(A*Vne*e)
print"\Mobility of electrons =  m**2/s*V ",ke

\Mobility of electrons =  m**2/s*V  9375.0

Example 3_17 pgno:110

In [26]:
#Chapter 3, Exmaple 17, page 110
#Determine the ion density
from math import e
#Based on equation 3.50 and 3.52
nplus = 10**11*e**(-1.6*10**-19*5*0.02/(1.38*10**-23*293))
nminus = 10**10*e**(-1.6*10**-19*5*-0.02/(1.38*10**-23*293)) #textbook uses 0.02 inseatead of -0.02. In the program I have used -0.02
print"\n+(0.02) =  ions/m^3 ",nplus
print"\n+(-0.02) =  ions/m^3 ",nminus

#answers may vary due to round off error

+(0.02) =  ions/m^3  1911913074.19

+(-0.02) =  ions/m^3  5.23036331254e+11

Example 3_18 pgno:110

In [27]:
#Chapter 3, Exmaple 18, page 110
#Determine the diameter

#Based on the equation 3.40
k = 1.38*10**-23
T = 293
z2z1 = 0.05
e = 1.6*10**-19
E = 250
r1 = 0.09*10**-6
r1r2 = (6*k*T*z2z1)/(e*E)
r2 = (r1+r1r2)**0.5
print"\n r1^2-r2^2 =  ",r1r2
print"\n r2 =  m ",r2

#answers may vary due to round off error

 r1^2-r2^2 =   3.03255e-05

 r2 =  m  0.00551502493195

Example 3_19 pgno:111

In [30]:
#Chapter 3, Exmaple 19, page 111
#Determine mean free path and ionization

#(a)Mean free path
#Based on equation 3.14 and 3.15
lamb = 1/(9003*0.5);
#(b)Ionization potential
Vi = 256584./9003.
print"\n lambda =  m ",lamb
print"\n Vi =  V ",round(Vi,1)

#answers may vary due to round off error

 lambda =  m  0.000222148172831

 Vi =  V  28.5