Ch-5,Material Characterization

example-5.1 page no-136

In [1]:
from math import pi, sin, asin
#given
#wavelength of X-rays beams of light
lamda=0.824*10**-10  #m
#glancing angle of the incident light
theta1=(8+35/60)*(pi)/180  #radians
n1=1
#to find theta3 i.e at
n3=3
#as we know that
#2*d*sin(theta)=n*lambda
#so for n1 and n3 we get in the same way and solving together we get 
theta3=asin(3*sin(theta1))
#so 
d=lamda/2/sin(theta1)
print "the galncing angle for the third order diffraction is and interplanar spacing of the crystal is 2.761 A"

the galncing angle for the third order diffraction is and interplanar spacing of the crystal is 2.761 A

example-5.2 page no-141

In [2]:
from math import sqrt, pi
#given
#bragg's angle of reflection
theta1=17.03*(pi)/180  #radians
#wavelength of light
lamda=0.71  #A
#according to bragg's equation 
#n*lambda=2*d*sin(theta)
#for n=1
d=lamda/2/sin(theta1)  #A
#given that h**2+k**2+l**2=8
#let (h**2+k**2+l**2)**1/2=H
#we get
H=sqrt(8)
a=d*H  #A
print "since h**2+k**2+l**2=8 ,hence the reflecting planes will be (220). family of planes (220) include (220), (202), (022) ,etc."

since h**2+k**2+l**2=8 ,hence the reflecting planes will be (220). family of planes (220) include (220), (202), (022) ,etc.

example-5.3 page n0-141

In [3]:
#given
#lattice constant
a=1.54  #A
#wavelength of beam of light
lamda=1.54 #A
#according to bragg's equation
#n*lambda=2*d*sin(theta)
#following angles are given
theta1=20.3*(pi)/180
theta2=29.2*(pi)/180
theta3=36.7*(pi)/180
theta4=43.6*(pi)/180
#interplaner spadcing is 
d1=lamda/(2*sin(theta1))  #A
d2=lamda/(2*sin(theta2))  #A
d3=lamda/(2*sin(theta3))  #A
d4=lamda/(2*sin(theta4))  #A
#magnitude of bragg's 
#we have h**2+k**2+l**2=a**2/d**2
#let a**2/d**2= D for notation only
#so
D1=2
D2=4
D3=6
D4=8
#so from bragg's magnitude we can get (hkl)
#(hkl1)=(110)
#(hkl3)=(200)
#(hkl3)=(211)
#(hkl4)=(220)
print "the reflection will take from {110},{200},{211} and (220)"

the reflection will take from {110},{200},{211} and (220)

example-5.5 page no-146

In [4]:
#given
#wavelength of X-ray
lamda=1.54  #A
#diameter of powder camera
D=114.6  #mm
#radius of powder camera
R=D/2  #mm
#value of l
l1=86
l2=100
l3=148
l4=180
l5=188
l6=232
l7=272
#we know that
#theta=l/4
#so
theta1=l1/4*(pi)/180   #radians
theta2=l2/4*(pi)/180   #radians
theta3=l3/4*(pi)/180   #radians
theta4=l4/4*(pi)/180   #radians
theta5=l5/4*(pi)/180   #radians
theta6=l6/4*(pi)/180   #radians
theta7=l7/4*(pi)/180   #radians
#now values of sin (theta) and sin(theta2)
S1=sin(theta1)
SS1=(sin(theta1))**2
S2=sin(theta1)
SS2=(sin(theta1))**2
S3=sin(theta1)
SS3=(sin(theta1))**2
S4=sin(theta1)
SS4=(sin(theta1))**2
S5=sin(theta1)
SS5=(sin(theta1))**2
S6=sin(theta1)
SS6=(sin(theta1))**2
S7=sin(theta1)
SS7=(sin(theta1))**2
#so the ratio can be expressed as
#3:4:8:11:12:16:19
print "from the extinction rule, we notice that this is an FCC Structure"
#the lattice parameter for highest bragg's angle is
#a=lambda*sqrt(h**2+k**2+l**2)/(2*sin(theta))
#here h**2+k**2+l**2=19
#and let h**2+k**2+l**2 =M for notation
M=19
a=lamda*sqrt(M)/(2*sin(theta6))  #A
print "lattice parameter of material is %0.2f A"%(a)

from the extinction rule, we notice that this is an FCC Structure
lattice parameter of material is 3.96 A

example-5.6 page no-158

In [5]:
#given
#ASTM number
n=12
#as we know that the number of grains N observed on photomicrograph is given by
N=2**(n-1)
#as we know that grain size diameter is given by
d=1/sqrt((N/645)*10**4)  #mm  because 1 square inch=645 mm**2
print "the grain diameter for an ASTM number 12 is %f mm"%(d)

the grain diameter for an ASTM number 12 is 0.005774 mm

example-5.7 page no-158

In [6]:
from math import log
#given
#no of grains within the view of a micrograph
n1=41
#no of grains cut by circumference
n2=42
#diameter of circular area
d=1  #inch
#area
A=(pi)/4*d**2  #inch**2
#the area density of grains

N=(n1+n2/2)/A  #grains/inch**2
#grain size
n=log(N)/log(2)+1  
print "the area density of grains is %0.2f grains/inch**2 and grain size is 8"%(N)

the area density of grains is 78.94 grains/inch**2 and grain size is 8

example-5.8 page no-159

In [7]:
from __future__ import division
#given
#ASTM no of grains
ASTM=5
#area density of grains
N=2**(ASTM-1)  #grains/inch**2 at magnification of 100*
#as we know that lineal and areal magnification are related as
#*100 lineal=*10000 areal
#therefore
Nnew=N/0.01/0.01  #grains/inch**2 at 1*
#average area of one grain
A=1/Nnew*(2.54)**2  #cm**2
#now 160000 grains/inch**2 of surface is sqrt(160000)=400 grains/inch of length and this is equal to=(400)**3==6.4*10**7 grains/m**3 of volume
#surface area of each cubic surface
S=(1/400)**2  #inch**2
#there are 6 surfaces in accubic grain
#thus total surface area of each grain
T=1/2*6*S*(400)**3/2.54  #cm**2 boundary per cubic cm of steel
print "the boumdary area per cubic centimeter of steel is %0.2e cm**2 boundary per cubic centimeter of steel"%T 

the boumdary area per cubic centimeter of steel is 4.72e+02 cm**2 boundary per cubic centimeter of steel