# Chapter 4: Quantum Physics¶

## Example 1, Page No: 4.52¶

In [5]:
import math;

# Variable Declaration
lamda   = 3*10**-10;         # wavelength of incident photons in m
theta   = 60;                # viewing angle in degrees
h       = 6.625*10**-34       # plancks constant
mo      = 9.11*10**-31        # mass in Kg
c       = 3*10**8;            # vel. of light

# Calculatioms
# from Compton theory ,Compton shift is given by
# lamda' - lamda = (h/(mo*c))*(1-cosθ)

theta_r = theta*math.pi/180;     # degree to radian conversion
lamda1  = lamda+( (h/(mo*c))*(1-math.cos(theta_r)))  # wavelength of scattered photons

# Result
print 'Wavelength of Scattered photons = %3.4f'%(lamda1*10**10),'Å';

Wavelength of Scattered photons = 3.0121 Å


## Example 2, Page No:4.52¶

In [10]:
import math;

# Variable declaration
theta   = 135;               #  angle in degrees
h       = 6.625*10**-34       # plancks constant
mo      = 9.1*10**-31         # mass in Kg
c       = 3*10**8;            # vel. of light in m/s

# Calculatioms
# from Compton theory ,Compton shift is given by
# lamda' - lamda = (h/(mo*c))*(1-cosθ)
theta_r = theta*math.pi/180;     # degree to radian conversion
c_lamda = ( (h/(mo*c))*(1-math.cos(theta_r))) # Change in wavelength in m

# Result
print 'Change in Wavelength  = %3.5f' %(c_lamda*10**10),' Å';

Change in Wavelength  = 0.04143  Å


## Example 3, Page No:4.53¶

In [12]:
import math;

# Variable Declaration
lamda   = 0.1*10**-9;         # wavelength of X-rays in m
theta   = 90;                # angle with incident beam in degrees
h       = 6.625*10**-34       # plancks constant
mo      = 9.11*10**-31        # mass in Kg
c       = 3*10**8;            # vel. of light

# Calculatioms
# from Compton theory ,Compton shift is given by
# lamda' - lamda = (h/(mo*c))*(1-cosθ)
theta_r = theta*math.pi/180;     # degree to radian conversion
lamda1  = lamda+( (h/(mo*c))*(1-math.cos(theta_r)))  #wavelength of scattered beam

# Result
print 'Wavelength of Scattered beam = %3.4f' %(lamda1*10**10),' Å';

Wavelength of Scattered beam = 1.0242  Å


## Example 4, Page No:4.53¶

In [15]:
import math;

# Variable Declaration
h       = 6.625*10**-34         # plancks constant
m       = 9.11*10**-31          #  mass of electron in Kg
e       = 1.6*10**-19           # charge of electron
V       = 150;                  # potential difference in volts

# Calculations

lamda   = h/(math.sqrt(2*m*e*V))  # de Broglie wavelength

#Result
print 'The de-Broglie wavelength = %d' %(lamda*10**10), 'Å';

The de-Broglie wavelength = 1 Å


## Example 5, Page No:4.54¶

In [17]:
import math;

# Variable Declaration
h       = 6.625*10**-34        # plancks constant
m       = 9.11*10**-31         # mass of electron in Kg
e       = 1.6*10**-19          # charge of electron
V       = 5000;                # potential in volts

# Calculations

lamda   = h/(math.sqrt(2*m*e*V))     #de Broglie wavelength

# Result
print 'The de-Broglie wavelength of electron = %3.5f' %(lamda*10**10),' Å';

The de-Broglie wavelength of electron = 0.17353  Å


## Example 6, Page No:4.55¶

In [21]:
import math;

# Variable Declaration
E       = 100                 # Energy of electron in eV
h       = 6.625*10**-34       # plancks constant
m       = 9.11*10**-31        # mass of electron in Kg
e       = 1.6*10**-19         # Charge of electron in Columbs

# Calculations

E1      = E*e                    # Energy conversion from eV to Joule
lamda   = h/(math.sqrt(2*m*E1))  # de Broglie wavelength

# Result
print 'The de-Broglie wavelength  = %3.3f' %(lamda*10**10),' Å';

The de-Broglie wavelength  = 1.227  Å


## Example 7, Page No:4.55¶

In [22]:
import math;

# Variable Declaration
m       = 1.675*10**-27;     # Mass of proton in kg
c       = 3*10**8;           # velocity of light in m/s
h       = 6.625*10**-34      # plancks constant

# Calculations

vp      = c/20;             # velocity of proton in m/s
lamda   = h/(m*vp)          # de-Broglie wavelength in m

# Result
print 'de-Broglie wavelength = %e'%(lamda),'lamda';

de-Broglie wavelength = 2.636816e-14 lamda


## Example 8, Page No:4.56¶

In [23]:
import math;

# Variable declaration
E       = 10000               # Energy of neutron in eV
h       = 6.625*10**-34       # plancks constant
m       = 1.675*10**-27       # mass of neutron in Kg
e       = 1.6*10**-19

# Calculations

E1      = E*e               # Energy conversion from eV to Joule
lamda   = h/(math.sqrt(2*m*E1))  # de Broglie wavelength

# Result
print 'The de-Broglie wavelength of neutron = %3.3e' %lamda,' m';

The de-Broglie wavelength of neutron = 2.862e-13  m


## Example 10, Page No:4.58¶

In [2]:
import math;

# Variable decalaration
l       = 0.1*10**-9;        # side of cubical box
h       = 6.625*10**-34      # plancks constant in Jsec
m       = 9.11*10**-31       # mass of electron in Kg
Kb      = 1.38*10**-23       # Boltzmann constant

# Calculations
# for cubical box the energy eigen value is Enx ny nz = (h^2/(8*m*l^2))*(nx^2 + ny^2 +nz^2)
# For the next energy level to the lowest energy level nx = 1 , ny = 1 and nz = 2
nx      = 1
ny      = 1
nz      = 2
E112    = (h**2/(8*m*l**2))*( nx**2 + ny**2 + nz**2);

# We know the average energy of molecules of aperfect gas = (3/2)*(Kb*T)
T       = (2*E112)/(3*Kb);      # Temperature in kelvin

# Result
print 'E112 = %3.4e' %E112,'Joules'',\n','Temperature of the molecules T = %3.4e' %T, 'K';

E112 = 3.6134e-17 Joules,
Temperature of the molecules T = 1.7456e+06 K


## Example 11, Page No:4.59¶

In [3]:
import math;

# Variable declaration
l       = 4*10**-9;           # width of infinitely deep potential
h       = 6.625*10**-34       # plancks constant in Jsec
m       = 9.11*10**-31        # mass of electron in Kg
n       = 1;                  # minimum energy
e       = 1.6*10**-19         # charge of electron in columbs

# Calculations
E       = (h**2 * n**2)/(8*m*l**2)     # Energy of electron in an infinitely deep potential well
E1      = E/e                          #energy conversion from joules to eV

# Result
print 'Minimum energy of an electron = %3.4f' %E1,' eV';

Minimum energy of an electron = 0.0235  eV


## Example 12, Page No:4.61¶

In [8]:
import math;

# Variable Declaration
l       = 0.1*10**-9;         # length of one dimensional box
h       = 6.625*10**-34       # plancks constant in Jsec
m       = 9.11*10**-31        # mass of electron in Kg
n       = 1;                  # for ground state
n5      = 6;                  # n value for fifth excited state
e       = 1.6*10**-19         # charge of electron in columbs

# Calculations
Eg      = (h**2 * n**2)/(8*m*l**2 * e )   # Energy in ground state in eV
Ee      = (h**2 * n5**2)/(8*m*l**2 * e)   # Energy in excited state  in eV
E       = Ee - Eg;                       # energy req to excite electrons from ground state to fifth excited state

# Result
print 'Energy required to excite an electron from ground state to fifth excited state = %3.2f' %E, 'eV';

Energy required to excite an electron from ground state to fifth excited state = 1317.38 eV


## Example 13, Page No:4.62¶

In [10]:
import math;

# Variable decalration
l       = 0.1*10**-9;          # length of one dimensional box
h       = 6.625*10**-34        # plancks constant in Jsec
m       = 9.11*10**-31         # mass of electron in Kg
n       = 1;                   # for ground state
e       = 1.6*10**-19          # charge of electron in columbs

# Calculations
E       = (h**2 * n**2)/(8*m*l**2 *e )   # Energy of electron in eV

# Result
print 'Energy of an electron = %3.3f' %E,' eV';

Energy of an electron = 37.639  eV


## Example 14, Page No:4.63¶

In [11]:
import math;

# Variable declaration
l       = 0.5*10**-9;          # width of one dimensional box in m
h       = 6.625*10**-34        # plancks constant in Jsec
m       = 9.11*10**-31         # mass of electron in Kg
n       = 1;                   # for ground state
e       = 1.6*10**-19          # charge of electron in columbs

# Calculations
E       = (h**2 * n**2)/(8*m*l**2 *e )   # Energy of electron in eV

# Result
print 'Least Energy of an electron = %3.4f' %E,' eV';

Least Energy of an electron = 1.5056  eV


In [12]:
import math;

# variable declaration
h       = 6.625*10**-34      # plancks constant
c       = 3*10**8;           # vel. of light
lamda   = 5893*10**-10;      # wavelength in m
P       = 100                # power of sodium vapour lamp

# Calculations
E       = (h*c)/lamda;       # Energy in joules
N       =  P/E               # Number of photons emitted

# Result
print 'Number of Photons emitted = %3.4e' %N,' per second';

Number of Photons emitted = 2.9650e+20  per second


In [15]:
import math;

# Variable declaration
lamda1  = 0.022*10**-10;      # wavelength of scatterd X-rays in m
theta   = 45;                 # scatterring angle in degrees
h       = 6.625*10**-34       # plancks constant
mo      = 9.11*10**-31        # mass in Kg
c       = 3*10**8;            # vel. of light

# Calculatioms
# from Compton theory ,Compton shift is given by
# lamda' - lamda = (h/(mo*c))*(1-cosθ)

theta_r = theta*math.pi/180;     # degree to radian conversion
lamda  = lamda1-( (h/(mo*c))*(1-math.cos(theta_r))) # incident Wavelength

# Result
print 'Wavelength of incident beam = %3.4f' %(lamda*10**10),' Å';

Wavelength of incident beam = 0.0149  Å


## Addl_Example 3 , Page No:4.65¶

In [17]:
import math;

# Variable Declaration
Ei      = 1.02*10**6          # photon energy in eV
theta   = 90;                 # scattered angle in degrees
h       = 6.625*10**-34       # plancks constant
mo      = 9.1*10**-31         # mass of electron in Kg
e       = 1.6*10**-19         # charge of electron
c       = 3*10**8;            # vel. of light in m/s

# Calculations
# from Compton theory ,Compton shift is given by
# lamda' - lamda = (h/(mo*c))*(1-cosθ)
theta_r = theta*math.pi/180;     # degree to radian conversion
c_lamda = ( (h/(mo*c))*(1-math.cos(theta_r)))  #Change in wavelength in m
dv      = c/c_lamda;         # change in frequency of the scattered photon
dE      = (h*dv)/e           # change in energy of scattered photon in eV
# This change in energy is transferred as the KE of the recoil electron
Er      = dE;                # Energy of recoil electron
Es      = Ei - Er            # Energy of scattered photon

# Result
print 'Energy of the recoil electron = %3.4f' %(Er*10**-6),' MeV','\n','Energy of the Scattered photon = %3.4f' %(Es*10**-6),'MeV';

Energy of the recoil electron = 0.5119  MeV
Energy of the Scattered photon = 0.5081 MeV


In [20]:
import math;

# Variable Declaration
lamda   = 0.124*10**-10;     # wavelength of X-rays in  m
theta   = 180;               # Scattering angle in degrees
h       = 6.625*10**-34      # plancks constant
mo      = 9.11*10**-31       # mass in Kg
c       = 3*10**8;           # vel. of light

# Calculatioms
# from Compton theory ,Compton shift is given by
# lamda' - lamda = (h/(mo*c))*(1-cosθ)

theta_r = theta*math.pi/180;     # degree to radian conversion
lamda1  = lamda+( (h/(mo*c))*(1-math.cos(theta_r))) # wavelength of scattered X-rays

# Result
print 'Wavelength of Scattered X-rays = %3.4f' %(lamda1*10**10),' Å';

Wavelength of Scattered X-rays = 0.1725  Å


In [23]:
import math;

# Variable Declaration
h       = 6.625*10**-34        # plancks constant
m       = 9.11*10**-31         # mass of electron in Kg
e       = 1.6*10**-19          # charge of electron
V       = 2000;                # potential in volts

# Calculations

lamda   = h/(math.sqrt(2*m*e*V))  # de Broglie wavelength

# Result
print 'The de-Broglie wavelength of electron = %3.4f' %(lamda*10**10),' Å';

The de-Broglie wavelength of electron = 0.2744  Å


In [25]:
import math;

# variable Declaration
h       = 6.625*10**-34       # plancks constant
m       = 1.678*10**-27       # mass of proton in Kg
e       = 1.6*10**-19         # charge of electron
Kb      = 1.38*10**-23;       # boltzmann constant
T       = 300                 # Temperature in kelvin

#Calculations
lamda   = h/(math.sqrt(3*m*Kb*T))  #de Broglie wavelength

#Result
print 'The de-Broglie wavelength = %3.4f' %(lamda*10**10),' Å';

The de-Broglie wavelength = 1.4512  Å


In [26]:
import math;

# Variable Declaration
h       = 6.625*10**-34         # plancks constant
m       = 9.11*10**-31          # mass of electron in Kg
lamda   = 3*10**-2;             # wavelength of electron wave
e       = 1.6*10**-19;          # charge of electron

# Calculations
E       = (h**2)/(2*m*lamda**2);   # Energy in Joules
E1      = E/e;

# Result
print 'Energy of the electron E = %3.4e' %E1,'eV';
print 'Note: Calculation mistake in textbook'

Energy of the electron E = 1.6729e-15 eV
Note: Calculation mistake in textbook


In [1]:
import math;

# Variable declaration
h       = 6.625*10**-34       # plancks constant
m       = 9.11*10**-31        # mass of electron in Kg
c       = 3*10**8;            # velocity of light in m/s

# Calculations
ve      = 0.7071*c           # velocity of electron
lamda   = h/(m*ve*math.sqrt(1-(ve/c)**2))    #de Broglie wavelength

# we know Compton wavelength  ,lamda' - lamda = (h/(mo*c))*(1-cosθ)
# maximum shift θ = 180
theta    = 180
theta1   = theta*math.pi/180;
d_lamda  = (h/(m*c))*(1-math.cos(theta1))
print 'de Broglie wavelength = %e' %lamda,' m';
print 'compton wavelength    = %e' %d_lamda,'m';
print 'The de-Broglie wacelength is equal to the compton wavelength';

de Broglie wavelength = 4.848152e-12  m
compton wavelength    = 4.848152e-12 m
The de-Broglie wacelength is equal to the compton wavelength


In [36]:
import math;

# Variable Declaration
l       = 10**-10;            # side of one dimensional box
h       = 6.625*10**-34       # plancks constant in Jsec
m       = 9.11*10**-31        # mass of electron in Kg
n1      = 1;                  # for 1st eigen value
n2      = 2;                  # for 2nd eigen value
n3      = 3;                  # for 3rd eigen value
n4      = 4;                  # for 4th eigen value
e       = 1.6*10**-19         # charge of electron in columbs

# Calculations
E1      = (h**2 * n1**2)/(8*m*l**2 *e )   #first Eigen value
E2      = (h**2 * n2**2)/(8*m*l**2 *e )   # second Eigen value
E3      = (h**2 * n3**2)/(8*m*l**2 *e )     # third Eigen value
E4      = (h**2 * n4**2)/(8*m*l**2 *e )    # fourth Eigen value

# Result
print '1st Eigen value  = %3.1f' %E1,'eV';
print '2nd Eigen value  = %3.1f' %E2,'eV';
print '3rd Eigen value  = %3.1f' %E3,'eV';
print '4th Eigen value  = %3.1f' %E4,'eV';

1st Eigen value  = 37.6 eV
2nd Eigen value  = 150.6 eV
3rd Eigen value  = 338.8 eV
4th Eigen value  = 602.2 eV


## Addl_Example 10 , Page No:4.68¶

In [37]:
import math;

# Variable Declaration
l       = 10**-10   ;         # length of one dimensional box in m
h       = 6.625*10**-34       # plancks constant in Jsec
m       = 9.11*10**-31        # mass of electron in Kg
n       = 1;                  # for ground state
e       = 1.6*10**-19         # charge of electron in columbs

# Calculations
E       = 2*(h**2 * n**2)/(8*m*l**2 *e )   #Energy of system having two electrons

# Result
print 'Energy of the system having two electrons = %3.4f' %E,' eV';

Energy of the system having two electrons = 75.2789  eV


## Addl_Example 11 , Page No:4.69¶

In [38]:
import math;

# Variable Declaration
b       = 40;       # angle subtended by final images at eye in degrees
a       = 10        # angle subtended by the object at the eye kept at near point in degrees

# Calculations
b_r     = b*math.pi/180;    # degree to radian conversion
a_r     = a*math.pi/180;    # degree to radian conversion
M       = math.tan(b_r)/math.tan(a_r);    # magnifying power

#Result
print 'Magnifying power = %3.3f' %M;

Magnifying power = 4.759

In [ ]: