Chapter 3 Ionization and Deionization Processes in gases

Exa 3.7.1 pg.no:103

In [1]:
#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 [2]:
#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 [3]:
#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 [4]:
#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 [5]:
#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 [6]:
#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 [7]:
#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 [8]:
#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 [18]:
#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 [10]:
#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 [11]:
#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 [12]:
#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 [13]:
#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 [14]:
#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 [15]:
#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 [16]:
#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 [17]:
#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