8: Quantum Physics

Example number 8.1, Page number 204

In [2]:
#importing modules
import math
from __future__ import division

#Variable declaration
W=0.1*10**-9;   #wavelength of photon(m)
h=6.62*10**-34;    #Planck's constant(m^2 Kg/sec)
c=3*10**8;      #velocity of light(m/s)
e=1.6*10**-19;   #charge of electron(c)

#Calculation
E=h*c/(W*e);    #energy of photon(eV)
P=h/W;       #momentum of the photon(Kgms^-1)

#Result
print "The energy of photon is",E,"eV"
print "The momentum of the photon is",P,"Kg m s^-1"

The energy of photon is 12412.5 eV
The momentum of the photon is 6.62e-24 Kg m s^-1

Example number 8.2, Page number 205

In [5]:
#importing modules
import math
from __future__ import division

#Variable declaration
w=5893*10**-10;      #wavelength of emitted light(m)
e=100;     #total energy emitted per sec
h=6.625*10**-34;    #Planck's constant(m^2 Kg/sec)
c=3*10**8;     #velocity of light(m/s)

#Calculation
E=h*c/w;    #energy of one photon(J)
N=e/E;      #The total numberof photons emitted(sec)

#Result
print "The total number of photons emitted per second is",round(N/10**20,3),"*10**20 per sec"
print "answer varies due to rounding off errors"

The total number of photons emitted per second is 2.965 *10**20 per sec
answer varies due to rounding off errors

Example number 8.3, Page number 205

In [2]:
#importing modules
import math
from __future__ import division

#Variable declaration
w=4000*10**-10;    #wavelength in black body(m)
t=1500;     #temperature of black body(K)
h=6.625*10**-34;   #Planck's constant(m^2 Kg/sec)
c=3*10**8;    #velocity of light(m/s)
Kb=1.38*10**-23;   #Boltzmann's constant(m^2 Kg s^-2 k^-1)

#Calculation
Edw=(8*math.pi*h*c/w**5)*(1/(math.exp(h*c/(w*Kb*t))-1));    #The energy density per unit wavelength in a black body cavity(J/m^4)

#Result
print "The energy density per unit wavelength in a black body cavity is",round(Edw,6),"J/m^4"
print "answer varies due to rounding off errors"

The energy density per unit wavelength in a black body cavity is 0.018349 J/m^4
answer varies due to rounding off errors

Example number 8.4, Page number 211

In [7]:
#importing modules
import math
from __future__ import division

#Variable declaration
h=6.625*10**-34;     #Planck's constant(m^2 Kg/sec)
c=3*10**8;     #velocity of light(m/s)
m=9.11*10**-31;    #mass of electron(Kg)

#Calculation
w=h/(c*m)*10**10;     #The compton wavelength for an electron(Armstrong)

#Result
print "The compton wavelength for an electron is",round(w,4),"Angstrom"

The compton wavelength for an electron is 0.0242 Angstrom

Example number 8.5, Page number 212

In [10]:
#importing modules
import math
from __future__ import division

#Variable declaration
theta=90;    #x ray photon scattered at a angle(degrees)
h=6.625*10**-34;   #Planck's constant(J-sec)
c=3*10**8;    #velocity of light(m/s)
m=9.11*10**-31;   #mass of electron(Kg)

#Calculation
theta=theta*math.pi/180;     #angle(radian)
deltalamda=((h/(c*m))*(1-math.cos(x)))/10**-10;    #The change in wavelength for Xray photon(Angstrom)

#Result
print "The change in wavelength for X ray photon is",round(deltalamda,4),"Angstrom"

The change in wavelength for X ray photon is 0.0242 Angstrom

Example number 8.6, Page number 212

In [12]:
#importing modules
import math
from __future__ import division

#Variable declaration
theta=180;     #x ray carbon scattered at a angle(degrees)
h=6.625*10**-34;   #Planck's constant(m^2 Kg/sec)
c=3*10**8;    #velocity of light(m/s)
m=9.11*10**-31;   #mass of electron(kg)
v=1.8*10**18;     #frequency of incident rays(s^-1)

#Calculation
theta=theta*math.pi/180;    #angle(radian)
w=c/v;    #wavelength(m)
tw=(h/(c*m))*(1-math.cos(theta));    #The change wavelength for Xray carbon(m)
lamda_dash=(w+tw)/10**-10;       #The wavelength of X-rays carbon(Angstrom)

#Result
print "The wavelength of X-rays carbon is",round(lamda_dash,2),"Angstrom"

The wavelength of X-rays carbon is 1.72 Angstrom

Example number 8.7, Page number 212

In [17]:
#importing modules
import math
from __future__ import division

#Variable declaration
w=3*10**-10;    #wavelength of incident photons(m)
theta=60;     #angle of view(degrees)
h=6.625*10**-34;    #Planck's constant(m^2 Kg/sec)
c=3*10**8;    #velocity of light(m/sec)
m=9.11*10**-31;   #mass of electron(Kg)

#Calculation
theta=theta*math.pi/180;    #angle(radian)
lamda_dash=(w+((h/(c*m))*(1-math.cos(theta))))/10**-10;    #The wavelength of scattered photons(Angstrom)

#Result
print "The wavelength of scattered photons is",round(lamda_dash,3),"Angstrom"

The wavelength of scattered photons is 3.012 Angstrom

Example number 8.8, Page number 213

In [3]:
#importing modules
import math
from __future__ import division

#Variable declaration
x=4;     #Total energy increase to 4 times of its initial rest energy
c=3*10**8;    #velocity of light(m/sec)

#Calculation
v=math.sqrt(c**2*(1-(1/x**2)));    #The Velocity of moving electron(m/sec)

#Result
print "The Velocity of moving electron is",round(v/10**8,4),"*10**8 m/sec"

The Velocity of moving electron is 2.9047 *10**8 m/sec

Example number 8.9, Page number 224

In [5]:
#importing modules
import math
from __future__ import division

#Variable declaration
a=0.1*10**-9;    #width of high potential box(m)
h=6.625*10**-34;    #Planck's constant(m^2 Kg/sec)
m=9.11*10**-31;     #mass of electron(Kg)
e=1.6*10**-19;      #charge of electron(c)
n=1;    #take n equal to one

#Calculation
E=(n**2*h**2)/(8*m*a**2*e);     #The least energy of the particle can be obtained(eV)

#Result
print "The least energy of the particle can be obtained is",round(E,3),"eV"
print "answer varies due to rounding off errors"

The least energy of the particle can be obtained is 37.639 eV
answer varies due to rounding off errors

Example number 8.10, Page number 224

In [8]:
#importing modules
import math
from __future__ import division

#Variable declaration
a=10**-14;    #length of impenerable box(m)
m=1.67*10**-27;    #mass of neutron(Kg)
n=1;    #for lowest energy
h=6.625*10**-34;    #Planck's constant(m^2 Kg/sec)

#Calculation
E=(n**2*h**2)/(8*m*a**2);     #The least energy of the neutron can be obtained(J)

#Result
print "The least energy of the neutron can be obtained is",round(E/(1.6*10**-19*10**6),3),"MeV"

The least energy of the neutron can be obtained is 2.053 MeV

Example number 8.11, Page number 225

In [10]:
#importing modules
import math
from __future__ import division

#Variable declaration
a=4*10**-10;    #width of electron box(m)
h=6.625*10**-34;   #Planck's constant(m^2 Kg/sec)
m=9.11*10**-31;    #mass of electron(kg)
e=1.6*10**-19;     #charge of electron(c)
n=1;     #first permitted level

#Calculation
E1=((n**2*h**2)/(8*m*a**2*e));    #The first permitted energy level by taking n=1(eV)
E2=4*E1;    #The second permitted energy level by taking n=2(eV)
E3=9*E1;    #The third permitted energy level by taking n=3(eV)

#Result
print "The first permitted energy level by taking n=1 is",round(E1,3),"eV"
print "The second permitted energy level by taking n=2 is",round(E2,2),"eV"
print "The third permitted energy level by taking n=3 is",round(E3,3),"eV"
print "answer varies due to rounding off errors"

The first permitted energy level by taking n=1 is 2.352 eV
The second permitted energy level by taking n=2 is 9.41 eV
The third permitted energy level by taking n=3 is 21.172 eV
answer varies due to rounding off errors

Example number 8.12, Page number 226

In [12]:
#importing modules
import math
from __future__ import division

#Variable declaration
a=1.5*10**-10;     #each side of cubical box(m)
n1=1;    #for lowest energy
n2=1;    #for lowest energy
n3=1;    #for lowest energy
h=6.625*10**-34;    #Planck's constant(m^2 Kg/sec)
m=9.11*10**-31;    #mass of electron(Kg)
e=1.6*10**-19;     #charge of electron(c)

#Calculation
n=(n1**2+n2**2+n3**2);     #total value of n
E=((n*h**2)/(8*m*a**2*e));    #The lowest energy of electron ina cubical box(eV)

#Result
print "The lowest energy of electron in a cubical box is",round(E,3),"eV"

The lowest energy of electron in a cubical box is 50.186 eV

Example number 8.13, Page number 226

In [16]:
#importing modules
import math
from __future__ import division

#Variable declaration
a=4*10**-9;    #width of potential well(m)
n=1;    #For minimum energy n value
h=6.625*10**-34;    #Planck's constant(m^2 Kg/sec)
m=9.11*10**-31;     #mass of electron(Kg)
e=1.6*10**-19;      #charge of electron(c)

#Calculation
E=((n**2*h**2)/(8*m*a**2*e));     #The lowest energy of electron in deep potential well(eV)

#Result
print "The lowest energy of electron in deep potential well is",round(E,5),"eV"

The lowest energy of electron in deep potential well is 0.02352 eV

Example number 8.14, Page number 227

In [18]:
#importing modules
import math
from __future__ import division

#Variable declaration
a=0.1*10**-9;    #length of one dimensional box(m)
n=1;        #first permitted level
h=6.625*10**-34;    #Planck's constant(m^2 Kg/sec)
m=9.11*10**-31;     #mass of electron(kg)
e=1.6*10**-19;      #charge of electron(c)

#Calculation
E1=((n**2*h**2)/(8*m*a**2*e));     #The ground state of electron in an one dimensional box(eV)
E6=36*E1;    #The fifth exited state of electron(eV)
E=E6-E1;     #The energy required the electron from its ground state to the fifth exited state(eV)

#Result
print "The energy required the electron from its ground state to the fifth exited state is",int(E),"eV"

The energy required the electron from its ground state to the fifth exited state is 1317 eV

Example number 8.15, Page number 227

In [20]:
#importing modules
import math
from __future__ import division

#Variable declaration
a=0.1*10**-9;    #length of one dimensional box(m)
n=1;     #first permitted level
h=6.625*10**-34;    #Planck's constant(m^2 Kg/sec)
m=9.11*10**-31;     #mass of electron(Kg)
e=1.6*10**-19;      #charge of electron(c)
ne=3;      #the number of electrons

#Calculation
E=((n**2*h**2)/(8*m*a**2*e))*ne;    #The lowest energy of the system consisting of three electron ia a one dimensional box(eV)

#Result
print "The lowest energy of the system consisting of three electron ia a one dimensional box is",round(E,4),"eV"
print "answer varies due to rounding off errors"

The lowest energy of the system consisting of three electron ia a one dimensional box is 112.9184 eV
answer varies due to rounding off errors