Chapter No 1 : Semiconductor Materials and Their Properties

Example 1.3, Page No 37

In [2]:
# Finding Density of silicon crystal

#initialisation of variables
a=5.3   #in Å
a=a*10**-10     #in m
N_A=6.023*10**23  
At_Si=28    #atomic weight of Si
n = 4
m=At_Si/N_A  #in gm
m= m*10**-3  #in kg
V=a**3       #in m^3

#CALCULATIONS
Rho=((m*n)/V)/1000  #in kg/m^3

# Note: There is calculation error to find the value of density. So the answer in the book is wrong.#RESULTS
#RESULTS
print('Density of silicon crystal in kg/m^3 is = %.2f kg/m^3' %Rho)

Density of silicon crystal in kg/m^3 is = 1.25 kg/m^3

Example 1.4, Page No 37

In [3]:
import math
# Finding Density of the crystal 

#initialisation of variables
n = 4
r = 1.278  #in Å

#CALCULATIONS
a = (4*r)/(math.sqrt(2)) #inÅ
a = a * 10**-10   #in m
V = (a)**3  #in m^3
At_W = 63.5   #atomic weight
N_A = 6.023*10**23
m = At_W /N_A   #in gm
m = m*10**-3  #in kg
Rho = ((m*n)/V)/1000  #in kg/m^3

#RESULTS
print('Density of the crystal in kg/m^3 is =%.2f kg/m^3' %Rho)

Density of the crystal in kg/m^3 is =8.93 kg/m^3

Example 1.5 Page No 43

In [4]:
# Finding Wavaelength of X ray

import math
from numpy import *
#initialisation of variables
d=2.82   #in Å
d=d*10**-10    #in m
n=1
theta1=10  #n degree

#CALCULATIONS
lembda=2*d*math.sin(math.radians(theta1))   #in m
lembda=lembda*10**10       #in Å

#RESULTS
print('Wavaelength of X ray  is =%.2f Å degrees' %lembda)

Wavaelength of X ray  is =0.98 Å degrees

Example 1.6, Page No 43

In [5]:
# Finding The spacing of atomic layer

import math
#initialisation of variables
lembda = 1.6  #in Å
theta = 14.2  #in degree
n = 1

#CALCULATIONS
d = (n*lembda)/(2*math.sin(math.radians(theta)))   #in Å

#RESULTS
print('The spacing of atomic layer in crystal is =%.2f in Å degrees' %d)

The spacing of atomic layer in crystal is =3.26 in Å degrees

Example 1.7 Page No 43

In [6]:
#Finding The interplanner spacing

import math
#initialisation of variables
n = 1
theta1 = 30  #in degree
lembda = 1.78  #in Å

#CALCULATIONS
d = (n*lembda)/(2*math.sin(math.radians(theta1)))  #in Å

#RESULTS
print('The interplanner spacing is =%.2f in Å degrees' %d)

The interplanner spacing is =1.78 in Å degrees

Example 1.8 Page No 43

In [7]:
#Finding Interplaner spacing of the crystal

import math

#initialisation of variables
lembda = 0.58  #in Å
n = 1;
theta1 = 6.45   # in degree

#CALCULATIONS
d = (n*lembda)/(2*math.sin(math.radians(theta1))) #in Å 
print('Part (i) : At angle of 6.45°, Interplaner spacing of the crystal  is =%.2f Å' %d)
theta2 = 9.15  #in degree
d1 = (n*lembda)/(2*math.sin(math.radians(theta2))) #in Å 
print('Part (ii) : At angle of 9.15°, Interplaner spacing of the crystal is =%.2f Å' %d1)
theta3 = 13 #in degree
n2 = 1
d2 = (n2*lembda)/(2*math.sin(math.radians(theta3)))  #in Å
print('Part (iii) : At angle of 13°, Interplaner spacing of the crystal is =%.2f Å' %d2)
n=2
d2 = (n*lembda)/(2*math.sin(math.radians(theta3)))  #in Å 

#RESULTS
print('Part (iv) : : The interplaner spacing is : =%.4f Å' %d2)
print('The interplaner spacing for some other set of reflecting  is : =%.2f Å' %d1)

Part (i) : At angle of 6.45°, Interplaner spacing of the crystal  is =2.58 Å
Part (ii) : At angle of 9.15°, Interplaner spacing of the crystal is =1.82 Å
Part (iii) : At angle of 13°, Interplaner spacing of the crystal is =1.29 Å
Part (iv) : : The interplaner spacing is : =2.5783 Å
The interplaner spacing for some other set of reflecting  is : =1.82 Å

Example 1.9, Page No 44

In [8]:
#Finding The glancing angle for a cubic in degree

import math
from sympy.mpmath import *
import cmath

#initialisation of variables
a = 2.814 #in Å
l = 0
h = l
k = 0
d=a  #in Å
n = 2

#CALCULATIONS
lembda=0.710     #in Å
theta = math.degrees(math.asin(n*lembda/(2*d)))

#RESULTS
print('The glancing angle for a cubic in degree is : =%.2f  Å degrees' %theta)
print('Anwser is in points i.e = 14 36"40"..')

The glancing angle for a cubic in degree is : =14.61  Å degrees
Anwser is in points i.e = 14 36"40"..

Example 1.10 Page No 44

In [9]:
#Finding Wavelength of X ray

import math 

#initialisation of variables
a = 3.65  #in Å
a = 3.65*10**-10  #in m
h = 1
k = 0
l = 0

#CALCULATIONS
d= a/(math.sqrt(h**2+k**2+l**2))   #in m
n = 1
theta=60  #in degree
lembda = 2*d*math.sin(math.radians(theta))   #in m
lembda = lembda * 10**10   #in Å

#RESULTS
print('Wavelength of X ray is : =%.2f  Å' %lembda)

Wavelength of X ray is : =6.32  Å

Example 1.11 Page No 44

In [10]:
#Finding The glancing angle in degree

import math

#initialisation of variables
lembda = 1.54   #in Å
density = 9.024   #in gm/cc
n = 1
MI = 100
At_W = 63.54   #atomic weight
N_A = 6.023*10**23   
m = At_W/N_A   #in gm
a =(density*m)**(1.0/3)  #in cm
h = 1
k = 0
l = 0

#CALCULATIONS
d= a/(math.sqrt(h**2+k**2+l**2))
theta =math.degrees((lembda*10**-8)/(2*d))   #in degree


#RESULTS
print('The glancing angle is = %.2f degrees' %theta)

The glancing angle is = 4.48 degrees

Example 1.12 Page No 45

In [11]:
#Finding To get the 2nd order spectrum the position of the detector in degree

import math

#initialisation of variables
a = 3.615   #in Å
theta=22   #in degree
n=1
h=1
k=h
l=k

#CALCULATIONS
d = a/(math.sqrt( ((h)**2) + ((k)**2) + ((l)**2) ))   #in Å
lembda = 2*d*math.sin(math.radians(theta))   #in Å
print('The wavelength of X ray is = %.2f Å ' %lembda)
theta2 =math.degrees(math.asin(lembda/d))  #in degree
             #in degree

#RESULTS
print('To get the 2nd order spectrum the position of the detector  is = %.2f degrees ' %theta2)

The wavelength of X ray is = 1.56 Å 
To get the 2nd order spectrum the position of the detector  is = 48.52 degrees

Example 1.13, Page No 45

In [12]:
import math

#initialisation of variables
n = 1
lembda = 1.54      # in Å
theta = 21.7      # in degree

#CALCULATIONS
d = lembda/(2*math.sin(math.radians(theta)))     # in Å
h = 1
k = h
l = k
a = d*math.sqrt(h**2+k**2+l**2)      # in Å

#RESULTS
print('Lattice constant is = %.2f  in Å ' %a)

Lattice constant is = 3.61  in Å

Example 1.14, Page No 47

In [13]:
import math

#initialisation of variables
h = 2
k = 1
l = 1
a = 4.8      # in Å

#CALCULATIONS
d_211 = a/(math.sqrt(h**2+k**2+l**2))       #in Å

#RESULTS
print('The distance between planes is =%.2f Å ' %d_211)

The distance between planes is =1.96 Å

Example 1.15, Page No 47

In [14]:
import math

#initialisation of variables
r = 1.28       # in Å
n = 4
M = 63.5

#CALCULATIONS
a = (4*r)/(math.sqrt(2))        #in Å
a = a * 10**-8       # in cm
N_A = 6.023*10**23     
Rho = (n*M)/( N_A*((a)**3) )       # in gm/cc

#RESULTS
print('Density is =%.2f gm/cc  ' %Rho)

Density is =8.89 gm/cc

Example 1.16, Page No 47

In [15]:
#initialisation of variables
M = 55.85
a = 2.9        # in Å
Rho = 7.87    #in gm/cc

#CALCULATIONS
a = a * 10**-8    # in cm
N_A = 6.023*10**23
n = (Rho*N_A*((a)**3))/M            # atom per unit

#RESULTS
print('A lattice having =%.2f atom per unit cell is a BCC structure ' %n)

A lattice having =2.07 atom per unit cell is a BCC structure

Example 1.17, Page No 47

In [16]:
import math

#initialisation of variables
M = 60.0           # in gm/mole
Rho = 6.23      # in gm/cc
n = 4
N_A = 6.023*10**23 

#CALCULATIONS
a = ((n*M)/(N_A * Rho))**(1.0/3)         # in cm
r = (a*math.sqrt(2))/n               #radius of atom in cm
r = r * 10**8            # in Å

#RESULTS
print('Radius of atom is =%.2f Å  ' %r)

Radius of atom is =1.41 Å

Example 1.18, Page No 47

In [17]:
import math

#initialisation of variables
Rho = 5.96          # in gm/cc
M = 50.0
n = 2
N_A = 6.023*10**23 

#CALCULATIONS
a =((n*M)/(Rho*N_A))**(1.0/3)          # in cm
r = (a*math.sqrt(3))/4                 # in cm
P_f = (2*(4.0/3)*math.pi*((r)**3))/((a)**3)       # packing factor


#RESULTS
print('Packing factor is =%.2f  ' %P_f)

Packing factor is =0.68

Example 1.19, Page No 48

In [18]:
#initialisation of variables
M = 120.0
n = 2.0
N_A = 6.023*10**23

#CALCULATIONS
m1 = M/N_A              #mass of 1 atom in gm
m2 = n*m1               #mass of unit cell in gm

#RESULTS
print('Number of unit cell in 20 gms of element is : =%.2f X 10^22 unit cells ' %((20/m2)/(10**22)))

Number of unit cell in 20 gms of element is : =5.02 X 10^22 unit cells

Example 1.20, Page No 48

In [19]:
import math

#initialisation of variables
Rho = 2.48         # in gm/c.c
n = 4.0 
M = 58
N_A = 6.023*10**23

#CALCULATIONS
a = ( (n*M)/(Rho*N_A) )**(1.0/3)      # in cm
a = a * 10**8        # in Å
r = (a*math.sqrt(2))/n      # in Å
r = 2*r       # in Å

#RESULTS
print('The center to center distance between ions  is =%.2f  Å' %r)

The center to center distance between ions  is =3.80  Å