# Chapter 3 Ionization and Deionization Processes in gases¶

## Exa 3.7.1 pg.no:103¶

In :
#speed of air molecules
from math import sqrt

# given data
R=8314.;             # gas constant in J/kg . mol .K
T=300.;              # temperature 27 deg C, 27+293=300K
M=32.;               # oxygen is diatomic
v = sqrt(3*R*(T/M));
print "speed of oxygen molecule %f m/s" %v
# Note: Value of R is given wrong in book
# So answer in the book is wrong

speed of oxygen molecule 483.561010 m/s


## Exa 3.7.2 pg.no:104¶

In :
#total translational KE
# given data
R=8314;              # gas constant in J/kg . mol .K
T=298;               #in kelvin
M=32;                # oxygen is diatomic
m=2*10**-3;          # in kg
p=1.01*10**5;        # 1 atm=1.01∗10ˆ5 N/m2
G = (m*R*T)/(M*p);   #volume of gas
x=(3/2)*p;           #no. of molecules per unit volume where x=N∗0.5∗m∗vˆ2 is given as (3/2)∗p)
print"volume of gas %e mˆ3 "%G
KE = x*G;            #total translational kinetic energy
print"total translational kinetic energy is %f J "%KE
# Note: Value of G is calculated in book is wrong

volume of gas 1.533151e-03 mˆ3
total translational kinetic energy is 154.848250 J


## Exa 3.7.3 pg.no:104¶

In :
#max pressure in chamber
from math import pi

# given data
R=8314;           # gas constant in J/kg . mol .K
T=300;            # temperature 27 deg C, 27+293=300K
me=0.10;          #mean free path in meters
rm=1.7*10**-10    #molecular radius in angstrom
M=28 #im moleˆ−1
m0=4.8*10**-26     #mass of nitrogen molecule
N = 1/(4*pi*((rm)**2)*me); # no. of molecules in gas
print"no . of molecules %e"%N
p = ((N*m0)/M)*R*T; # max pressure in chamber in N/ m2
print"max pressure in chamber %f N/m2"%p
# Note: Calculation in the book is wrong So answer in the book is wrong

no . of molecules 2.753546e+19
max pressure in chamber 0.117735 N/m2


## Exa 3.7.4 pg.no:105¶

In :
#temperature at which avg KE of He atoms in gas become 1 eV
# given data
v = 1.6*10**-19; # avg kinetic energy in j
k = 1.38*10**-23 #boltzmann constant in J/K
T = (2*v)/(3*k);
print "temperature %e K"%T

temperature 7.729469e+03 K


## Exa 3.7.5 pg.no:105¶

In :
#volume of 1 kg of He
# given data
m = 1;          #in kg
M=2.016;        #molecular weight of helium
k =8314         # gas constant in J/kg
p = 1.01*10**5;
T = 273;        # in kelvin
G = m*k*T/(M*p);#volume of 1kg of helium in mˆ3
print"volume of 1kg of helium is %f mˆ3"%G

volume of 1kg of helium is 11.147071 mˆ3


## Exa 3.7.6 pg.no:105¶

In :
#density of ions at dist equal to mfp and five times mfp
from math import exp

# given data
z1=-1;         #ion at a distance equal to mean free path , −x=mfp
z2=-5;         #ion at a distance equal to five times the mean f r e e path , −x=5mfp
#n0 is the density of ions at the origin
n1 = exp(z1);  #density of ions at distance equal to the mean free path
n2 = exp(z2);  #density of ions at distance equal to five times the mean free path
print"density of ions at distance equal to the mean free path %0.2f n0"%n1
print"density of ions at distance equal to five times the mean free path %0.4f n0"%n2

density of ions at distance equal to the mean free path 0.37 n0
density of ions at distance equal to five times the mean free path 0.0067 n0


## Exa 3.7.7 pg.no:105¶

In :
#mean square velocity of He atoms
from math import sqrt

# given data
N = 178*10**-3     #gas density in kg/mˆ3
p = 1.01*10**5     # pressure
v = sqrt((3*p)/N); #mean square velosity of helium atoms
print"mean square velosity of helium atoms %d m/s"%round(v)

mean square velosity of helium atoms 1305 m/s


## Exa 3.7.8 pg.no:106¶

In :
#energy of free electrons
# given data
k =1.38e-21;     #boltzmanns constant
T = 293;         # temperature in K
e = 1.6*10** -19;
E =(1.5*k*T)/e;
print"energy of free electron %.2f eV"%E

energy of free electron 3.79 eV


## Exa 3.7.9 pg.no:106¶

In :
#avg separation of atoms and avg vol occupied by one atom
from math import pow

# given data
d = 0.075;            #density of solid atomic hydrogen in g/cmˆ3
N_A = 6.0224e23;      #1g of H consists of NA atoms
N = N_A*d;             # number of atoms/cmˆ3
print "no . of atoms %e /cmˆ3 "%N
x = 1/N;              #avg volume occupied by one atom in cmˆ3
y = pow(x,(1./3));        #avg seperation between atoms in cm
print "avg vokume occupied by one atom %e cmˆ3"%x
print "avg seperation between atoms %e cm"%y

#answers may vary due to round off errors

no . of atoms 4.516800e+22 /cmˆ3
avg vokume occupied by one atom 2.213957e-23 cmˆ3
avg seperation between atoms 2.807952e-08 cm


## Exa 3.7.10 pg.no:106¶

In :
#KE in eV and velocity of phototelectron
from math import sqrt

# given data
l=200*10**-10;# wavelength in angstrom
h=4.15*10**-15;#planks constant
c=3*10**8;#speed of light
me=9.11*10**-31;
BE=13.6;#binding energy in eV
PE=(h*c)/l;# in eV
print"photon enegy %f eV"%PE
KE = PE-BE;#in eV
print"kinetic energy of photoelectron %f ev"%KE
ve=sqrt((2*KE*1.6*10**-19)/me);
print" velosity of photoelectron %e m/s"%ve

photon enegy 62.250000 eV
kinetic energy of photoelectron 48.650000 ev
velosity of photoelectron 4.133874e+06 m/s


## Exa 3.7.11 pg.no:107¶

In :
#liquid photon absorption coefficient
from math import log

# given data
I = 1.;
I0 = 6.;
x=20;      #in cm
u = -(1./x)*log(I/I0);
print"absorption coefficient %.4f cmˆ−1"%u

absorption coefficient 0.0896 cmˆ−1


## Exa 3.7.12 pg.no:107¶

In :
#binding energy of the gas
# given data
c=3*10**8;
h=4.15*10**-15;
lmax =1000*10** -10;
We=(c*h)/lmax;
print"binding energy of gas %f eV"%We

binding energy of gas 12.450000 eV


## Exa 3.7.14 pg.no:108¶

In :
#diameter of the argon atom
from math import sqrt,pi

# given data
p=1.01*10**5/760;# 1 torr in N/m2
k=1.38*10**-23;
T=273; # in Kelvin
n=85*10**2;#no of collisions per meter
N=p/(k*T);
print "no of gas molecules %e atoms/mˆ3"%N
r_a=sqrt(n/(pi*N*1));
print "diameter of argon atom %e m"%r_a

no of gas molecules 3.527492e+22 atoms/mˆ3
diameter of argon atom 2.769501e-10 m


## Exa 3.7.15 pg.no:109¶

In :
#mobility of electrons
# given data
Ie=3;# current flow in amperes
A=8*10**-4;#area of the electrodes in mˆ2
V=20;#voltage across the electrodes
d=0.8;#spacing between the electrodes in meters
n_e=1*10**17;#electron density in mˆ−3
e=1.6*10**-19;
ke=(Ie*d)/(A*V*n_e*e);
print"mobility of electrons %f mˆ2/sV"%ke

mobility of electrons 9375.000000 mˆ2/sV


## Exa 3.7.17 pg.no:110¶

In :
#density point 02 m away in both directions at 25 deg C amperes
from math import exp

# given data
E = 5; #electric field in V/m
n_o = 10**11; #ion density in ions/m3
T = 293; # in kelvin
z = 0.02; #distance in meters
e = 1.6*10**-19; #in couloumb
k = 1.38*10**-23; # in m2 kg s−2 K−1
n1 = n_o*exp((-e*E*z)/(k*T));# ion density away
n2 = n_o*exp((e*E*z)/(k*T));# ion density away −0.02m
print"ion density 0.02m away %.2e ions/mˆ3 \n"%n1
print"ion density −0.02m away %.2e ions/mˆ3 \n"%n2

ion density 0.02m away 1.91e+09 ions/mˆ3

ion density −0.02m away 5.23e+12 ions/mˆ3



## Exa 3.7.18 pg.no:110¶

In :
#diameter of cloud after drifting a distance of point 05
from math import sqrt

# given data
E = 250; #electric field in V/m
r1 = 0.3*10**-3#intial diameter of cloud in meters
k = 1.38*10**-23;#in m2 kg s−2 K−1
T = 293; #in kelvin
e = 1.6*10**-19;# in couloumb
z = 0.05;#drift distance in meters
r = (6*k*T*z)/(e*E);#diameter before drift
print"diameter before drift %.2e m \n"%r
r2 = sqrt (r1**2 + r );#diamter after drifting a distance
print"diameter after drift %.3e m \n"%r2
# round off value calculated for r and r2

diameter before drift 3.03e-05 m

diameter after drift 5.515e-03 m



## Exa 3.7.19 pg.no:111¶

In :
#a mean free path of electrons in nitrogen and b ionization potential of nitrogen
# given data
a = 9003.;#constant in m−1kPa−1
B = 256584.;#in V/m.kPa
p = 0.5;#in kPa
M = 1/(a*p);#mean free path in meters
print"mean free path of electron in nitrogen %.2e m"%M
Vi = B/a; #ionization potential of nitrogen
print"ionization potential of nitrogen %.1f V"%Vi

mean free path of electron in nitrogen 2.22e-04 m
ionization potential of nitrogen 28.5 V