# Chapter 5 Atoms and Ion Moments in Materials¶

## Example 5_2 pgno:160¶

In [1]:
#EXAMPLE 5.2
#page 119
from math import log,exp
# Initialisation of Variables
R1=5*10**8;#The rate of moement of interstitial atoms in jumps/s 500 degree celsius
R2=8*10**10;#The rate of moement of interstitial atoms in jumps/s 800 degree celsius
T1=500;#Temperature at first jump in Degree celsius
T2=800;#Temperature at second jump in Degree  celsius
R=1.987;#Gas constant in cal/mol-K
#CALCULATIONS
Q=log(R2/R1)/(exp(1/(R*(T1+273)))-exp(1/(R*(T2+273))));#Activation Energy for Interstitial    Atoms in cal/mol
print "Activation Energy for Interstitial Atoms in cal/mol:",round(Q)

Activation Energy for Interstitial Atoms in cal/mol: 27865.0


## Example 5_3 pgno:166¶

In [2]:
# Initialisation of Variables
X=0.1;#Thickness of SIlicon Wafer in cm
n=8.;#No. of atoms in silicon per cell
ni=1.;#No of phosphorous atoms present for every 10**7 Si atoms
ns=400;#No of phosphorous atoms present for every 10**7 Si atoms
ci1=(ni/10**7)*100;#Initial compositions in atomic percent
cs1=(ns/10**7)*100;#Surface compositions in atomic percent
a0=1.6*10**-22;#The lattice parameter of silicon
v=(10**7/n)*a0;#volume of the unit cell in cm**3
ci2=ni/v;#The compositions in atoms/cm**3
cs2=ns/v;#The compositions in atoms/cm**3

concentration gradient in percent/cm: 0.0001


## Example 5_4 pgno:167¶

In [14]:
# Initialisation of Variables
cin=8.573*10**22;
dx=0.05;
d=9*10**-12;
j=d*cin/dx;
A=2*2;
tn=A*j;
print "total number no.of Ni atoms per second is ",tn
nm=tn/(8.573*10**22);
print "nickel atoms removed from the Ni/MgO interface is ",nm
thickness=nm/A;
print "the thickness is",thickness
t=10**-4/thickness;
print "for one micro meter of nickel to be removed,the treatment requires",round(t/3600)#SEC to be converted in hrs

total number no.of Ni atoms per second is  6.17256e+13
nickel atoms removed from the Ni/MgO interface is  7.2e-10
the thickness is 1.8e-10
for one micro meter of nickel to be removed,the treatment requires 154.0


## Example 5_5 pgno:171¶

In [3]:
from math import pi,log,exp
# Initialisation of Variables
N=1;#N0. of atoms on one side of iron bar
H=1;#No. of atoms onother side of iron bar
d=3;#Diameter of an impermeable cylinder in cm
l=10;#Length of an impermeable cylinder in cm
A1=50*10**18*N;# No. of gaseous Atoms per cm**3 on one side
A2=50*10**18*H;#No. of gaseous Atom per cm**3 on one  side
B1=1*10**18*N;#No. of gaseous atoms per cm**3 on another side
B2=1*10**18*H;#No. of gaseous atoms per cm**3 on another side
t=973;#The diffusion coefficient of nitrogen in BCC iron at 700 degree celsius in K
Q=18300;#The activation energy for diffusion of Ceramic
Do=0.0047;#The pre-exponential term of ceramic
R=1.987;#Gas constant in cal/mol.K
#CALCULATIONS
T=A1*(pi/4)*d**2*l;#The total number of nitrogen atoms in the container in N atoms
LN=0.01*T/3600;#The maximum number of atoms to be lost per second in N atoms per Second
JN=LN/((pi/4)*d**2);#The Flux of ceramic in Natoms per cm**2. sec.
Dn=Do*exp(-Q/(R*t));#The diffusion coefficient of Ceramic in cm**2/Sec
deltaX=Dn*(A1-B1)/JN;#minimum thickness of the membrane in cm
LH=0.90*T/3600;#Hydrogen atom loss per sec.
JH=LH/((pi/4)*d**2);#The Flux of ceramic in Hatoms per cm**2. sec.
Dh=Do*exp(-Q/(R*t));#The diffusion coeficient of Ceramic in cm**2/Sec
deltaX2=((1.86*10**-4)*(A2-B2))/JH;#Minimum thickness of  the membrane in cm
print "Minimum thickness of  the membrane of Natoms in cm", round(deltaX,3)
print "Minimum thickness of  the membrane of Hatoms in cm",round(deltaX2,3)

Minimum thickness of  the membrane of Natoms in cm 0.013
Minimum thickness of  the membrane of Hatoms in cm 0.073


## Example 5_6 pgno:174¶

In [4]:
from math import exp
# Initialisation of Variables
n=2;#no of atoms/ cell in BCC Tungsten
a0=3.165;#The lattice parameter of BCC tungsten in Angstromes
W=n/(a0*10**-8)**3;#The number of tungsten atoms per cm**3
Cth=0.01*W;#The number of thorium atoms per cm**3
Cg=-Cth/0.01;#The concentration gradient of Tungsten in atoms/cm**3.cm
Q=120000;#The activation energy for diffusion of Tungsten
Q2=90000;#The activation energy for diffusion of Tungsten
Q3=66400;#The activation energy for diffusion of Tungsten
Do=1.0;#The pre-exponential term of Tungsten
Do2=0.74;#The pre-exponential term of Tungsten
Do3=0.47;#The pre-exponential term of Tungsten
R=1.987;#Gas constant in cal/mol.K
t=2273;#The diffusion coefficient of nitrogen in BCC iron at 2000 degree celsius in K
#CALCULATIONS
D1=Do*exp(-Q/(R*t));#The diffusion coeficient  of Tungsten in cm**2/Sec
J1=-D1*Cg;#Volume Diffusion in Th atoms/cm**2.sec.
D2=Do2*exp(-Q2/(R*t));#The diffusion coeficient  of Tungsten in cm**2/Sec
J2=-D2*Cg;#Grain boundary Diffusion in Th atoms/cm**2.sec.
D3=0.47*exp(-66400/(1.987*2273));#The diffusion coeficient  of Tungsten in cm**2/Sec
J3=-D3*Cg;#Surfae Diffusion in Th atoms/cm**2.sec.

print "The number of tungsten atoms per cm**3:",W
print "The number of thorium atoms per cm**3:",Cth
print "The concentration gradient of Tungsten in atoms/cm**3.cm:",Cg
print "The diffusion coeficient  of Tungsten in cm**2/Sec:",D1
print "Volume Diffusion in Th atoms/cm**2.sec.:",J1
print "The diffusion coeficient  of Tungsten in cm**2/Sec:",D2
print "Grain boundry Diffusion in Th atoms/cm**2.sec.:",J2
print "The Surface diffusion coeficient of Tungsten in cm**2/Sec:",D3*10**7
print "Surface Diffusion in Th atoms/cm**2.sec.:",J3/10

The number of tungsten atoms per cm**3: 6.30824936432e+22
The number of thorium atoms per cm**3: 6.30824936432e+20
The concentration gradient of Tungsten in atoms/cm**3.cm: -6.30824936432e+22
The diffusion coeficient  of Tungsten in cm**2/Sec: 2.89066552915e-12
Volume Diffusion in Th atoms/cm**2.sec.: 1.82350389868e+11
The diffusion coeficient  of Tungsten in cm**2/Sec: 1.64051460984e-09
Grain boundry Diffusion in Th atoms/cm**2.sec.: 1.03487752447e+14
The Surface diffusion coeficient of Tungsten in cm**2/Sec: 1.93723957013
Surface Diffusion in Th atoms/cm**2.sec.: 1.22205902868e+15


## Example 5_7 pgno:178¶

In [17]:
import numpy
from math import exp
T=numpy.array([1173, 1273, 1373, 1473])#kelvins
t=numpy.array([0, 0, 0, 0])
#in K
t[0]=0.0861/exp(-16558/T[0])
t[1]=0.0861/exp(-16558/T[1])
t[2]=0.0861/exp(-16558/T[2])
t[3]=0.0861/exp(-16558/T[3])
print "the combution temperatures are",(0.5*t/3600)
#the difference in asnwer is due to round off error

the combution temperatures are [ 39.09194444  14.38111111   5.29041667   1.94625   ]


## Example 5_8 pgno:180¶

In [5]:
from math import exp
# Initialisation of Variables
H=10;#Required time to successfully carburize a batch of 500 steel gears
t1=1173;#Temperature at carburizing a batch of   500 steel gears in K
t2=1273;#Temperature at carburizing a batch of   500 steel gears in K
Q=32900;#The activation energy for diffusion  of BCC steel
R=1.987;#Gas constant in cal/mol.K
c1=1000;#cost per hour to operate the carburizing furnace at 900degree centigrades
c2=1500;#Cost per hour to operate the carburizing furnace at 1000 degree centigrade
H2=(exp(-Q /(R*t1))*H*3600)/exp(-Q /(R*t2));# Time requried to  successfully carburize a batch of 500 steel gears at 1000 degree centigrade
Cp1=c1*H/500;#The cost per Part of steel rods  at 900 degree centigrade
Cv=(c2*3.299)/500;#The cost per Part of steel rods  at 1000 degree centigrade
print "Time requried to  successfully carburize a batch of 500 steel gears at 1000 degree centigrade:",H2/3600
print "The cost of carburizing per Part of steel rods  at 900 degree centigrade",Cp1
print "The cost of carburizing per Part of steel rods  at 1000 degree centigrade",round(Cv,2)

Time requried to  successfully carburize a batch of 500 steel gears at 1000 degree centigrade: 3.2993917076
The cost of carburizing per Part of steel rods  at 900 degree centigrade 20
The cost of carburizing per Part of steel rods  at 1000 degree centigrade 9.9