# chapter1: De Broglie Matter Waves¶

## example 1.1;page no:10¶

In [3]:
# cal of de brogle wavelength of earth
#intiation of all variables
#given that
M = 6.*10**24 # Mass of earth in Kg
v = 3.*10**4 # Orbital velocity of earth in m/s
h = 6.625*10**-34 # Plank constant
print("Example 1.1,page no:10")
lamda=h/(M*v)
print("\n de Broglie wavelength of earth in metres is="),round(lamda,65)
Example 1.1,page no:10

de Broglie wavelength of earth in metres is= 3.68e-63

## example 1.2;page no:10¶

In [4]:
#cal of de Broglie wavelength of body
#intiation of all variables
#given that
M = 1 # Mass of object in Kg
v = 10 # velocity of object in m/s
h = 6.625*10**-34 # Plank constant
print("Example 1.2,page no:11");
lamda=h/(M*v)#calculation of de Broglie wavelength
print("\n de Broglie wavelength of body in metres is="),round(lamda,38)
Example 1.2,page no:11

de Broglie wavelength of body in metres is= 6.625e-35

## example 1.3;page no:11¶

In [5]:
#cal of de brogle wavelength of body
#intiation of all variables
# Given that
m = 1e-30 # Mass of any object in Kg
v = 1e5 # velocity of object in m/s
h = 6.625e-34 # Plank constant
print("Example 1.3,page no:11")
lamda=h/(m*v) # calculation of de Broglie wavelength in metres
print("\n de Broglie wavelength of body in metres is="),round(lamda,12)
Example 1.3,page no:11

de Broglie wavelength of body in metres is= 6.625e-09

## example 1.4;page no:15¶

In [6]:
#cal of velocity,momenteum and wave lenght of electron
#intiation of all variables
# Given that
import math
KE = 4.55e-25 # Kinetic energy of an electron in Joule
m = 9.1e-31 # Mass of any object in Kg
h = 6.62e-34 # Plank constant
print("Example 1.4,page no:15")
v = math.sqrt(2*KE/m) # Calculation of velocity of moving electron
p = m*v #Calculation of momentum of moving electron
lamda= h/p # calculation of de Broglie wavelength
print("velocity of electron in m/s:"),round(v)
print("momentum of electron in Kgm/s:"),round(p,29)
print("de Broglie wavelength of electron is:"),round(lamda,9)
print("Note:The value given in the book for lamda is wrong hence corrected above")
Example 1.4,page no:15
velocity of electron in m/s: 1000.0
momentum of electron in Kgm/s: 9.1e-28
de Broglie wavelength of electron is: 7.27e-07
Note:The value given in the book for lamda is wrong hence corrected above

## example 1.5;page no:16¶

In [7]:
#cal of de brogle wavelength of proton
#intiation of all variables
#Given that
c = 3e8 # speed of light in m/s
v = c/20 # Speed of proton in m/s
m = 1.67e-27 # Mass of proton in Kg
h = 6.625e-34 # Plank constant
print("Example 1.5,page no:16")
lamda= h/(m*v) # calculation of de Broglie wavelength
print("de Broglie wavelength of proton is:"),round(lamda,17)
# Answer in book is 6.645e-14m which is a calculation mistake
Example 1.5,page no:16
de Broglie wavelength of proton is: 2.645e-14

## example 1.6;page no:16¶

In [1]:
#cal of de brogle wavelength of neutron
#intiation of all variables
#given that
import math
e = 12.8 # Energy of neutron in MeV
c = 3.e8 # speed of light in m/s
m = 1.675e-27 # Mass of neutron in Kg
h = 6.62e-34 # Plank constant
print("Example 1.6,page no:16")
rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV
if e/rest_e < 0.015:
E = e
else:
E = rest_e +e
lamda = h/(math.sqrt(2*m*e*1e6*1.6e-19)) # calculation of de Broglie wavelength
print("\n de Broglie wavelength of neutron in angstrom="),round(lamda*1e10,7)
# Answer in book is 8.04e-5 angstrom which is misprinted
Example 1.6,page no:16

de Broglie wavelength of neutron in angstrom= 7.99e-05

## example 1.7;page no:17¶

In [7]:
#cal of de brogle wavelength of neutron
#intiation of all variables
#Given that
import math
e = 1.602e-19 # charge on electron in coulomb
V = 50. # Applied voltage in volts
m = 9.1e-31 # Mass of electron in Kg
h = 6.62e-34 # Plank constant
print("Example 1.7,page no:17")
lamda= h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength
print("\n de Broglie wavelength of neutron in angstrom="),round(lamda*1e10,3)
Example 1.7,page no:17

de Broglie wavelength of neutron in angstrom= 1.734

## example 1.9;page no:18¶

In [8]:
#cal of de brogle wavelength associated with the electron
#intiation of all variables
#Given that
import math
e = 1.6e-19 # charge on electron in coulomb
V = 54 # Applied voltage in volts
m = 9.1e-31 # Mass of electron in Kg
h = 6.63e-34 # Plank constant
print("Example 1.9,page no:18")
lamda = h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength
print("de Broglie wavelength associated with the electron in angstrom="),round(lamda*1e10,2)
Example 1.9,page no:18
de Broglie wavelength associated with the electron in angstrom= 1.67

## example 1.10;page no:19¶

In [9]:
#cal of velocity of electron,momentum of electron,de Broglie wavelength of electron
#intiation of all variables
#Given that
import math
E = 10. # Energy of electron in KeV
me = 9.1e-31 # Mass of electron in Kg
h = 6.63e-34 # Plank constant
print("Example 1.10,page no:19")
v = math.sqrt(2*E*1.6e-16/me) # Calculation of velocity of moving electron
p = me*v #Calculation of momentum of moving electron
lamda = h/p # calculation of de Broglie wavelength
print("velocity of electron in m/s:"),round(v,2)
print("momentum of electron in Kgm/s:"),round(p,27)
print("de Broglie wavelength of electron in angstrom="),round(lamda*1e10,3)
# Answers in book are v = 5.93e6 m/s, p = 5.397e-24 kgm/s, lambda = 1.23 angstrom
# Which is due to wrong calculation
Example 1.10,page no:19
velocity of electron in m/s: 59299945.33
momentum of electron in Kgm/s: 5.3963e-23
de Broglie wavelength of electron in angstrom= 0.123

## example 1.11;page no:20¶

In [10]:
#cal of velocity and kinetic energy of neutron
#intiation of all variables
#Given that
lamda= 1 # de Broglie wavelength of neutron in angstrom
m = 1.67e-27 # Mass of electron in Kg
h = 6.62e-34 # Plank constant
print("Example 1.11,page no:20")
v = h/(m*lamda*1e-10) # Calculation of velocity of moving neutron
print(" velocity of neutron in m/s:"),round(v,3)
E = 1./2.*m*v**2 # Calculation of kinetic energy of moving neutron
print(" Kinetic energy of neutron in eV="),round(E/1.6e-19,3)
Example 1.11,page no:20
velocity of neutron in m/s: 3964.072
Kinetic energy of neutron in eV= 0.082

## example 1.12;page no:20¶

In [12]:
#cal of de brogle wavelength of electron
#intiation of all variables
#given that
import math
E = 2 # Energy of accelerated electron in KeV
m = 9.1e-31 # Mass of electron in Kg
h = 6.62e-34 # Plank constant
print("Example 1.12,page no:20")
lamda = h/math.sqrt(2*m*E*1e3*1.6e-19) # Calculation of velocity of moving electron
print("Wavelength of electron in  metres="),round(lamda,13)
# Answer in book is 2.74e-12m
Example 1.12,page no:20
Wavelength of electron in  metres= 2.74e-11

## example 1.13;page no:21¶

In [13]:
#cal of de brogle wavelength of matter wave
#intiation of all variables
#given that
import math
v = 2e8 # speed of moving proton in m/s
c = 3e8 # speed of light in m/s
m = 1.67e-27 # Mass of proton in Kg
h = 6.62e-34 # Plank constant
print("Example 1.13,page no:21")
lamda = h/(m*v/math.sqrt(1-(v/c)**2)) # Calculation of velocity of moving electron
print("Wavelength of matter wave in  angstrom="),round(lamda*1e10,7)
Example 1.13,page no:21
Wavelength of matter wave in  angstrom= 1.48e-05

## example 1.14;page no:22¶

In [31]:
#cal of momentum,total energy and ratio of kinetic energy of photon
#intiation of all variables
#given that
lamda = 1# wavelength in m/s
m_e = 9.1e-31 # Mass of electron in Kg
m_p = 1.67e-27 # Mass of proton in kg
c = 3e8 # speed of light in m/s
h = 6.63e-34 # Plank constant
print("Example 1.14,page no:22")
p_e = h/(lamda*1e-10) # Momentum of electron
p_p = h/(lamda*1e-10) # Momentum of photon
print("Momentum of photon in  Kgm/s while Momentum of electron in Kgm/s which are equal:"),round(p_p,26),round(p_e,26)
E_e = p_e**2/(2*m_e) +m_e*c**2 # Total energy of electron
E_e1=(2.42*10**-17)+(m_e*c**2/1.6*10**-19)
E_p = h*c/(lamda*1e-10) # Total energy of photon
print("Total Energy of photon in  joule while Total Energy of electron in  MeV:"),round(E_p,18),E_e1
K_e = p_e**2/(2*m_e) # Kinetic energy of electron
K_p = h*c/(lamda*1e-10)# Kinetic energy of photon
r_K = K_e/K_p # Ratio of kinetic energies
print("Ratio of kinetic energies in:"),round(r_K,4)
Example 1.14,page no:22
Momentum of photon in  Kgm/s while Momentum of electron in Kgm/s which are equal: 6.63e-24 6.63e-24
Total Energy of photon in  joule while Total Energy of electron in  MeV: 1.989e-15 2.42e-17
Ratio of kinetic energies in: 0.0121

## example 1.15;page no:24¶

In [2]:
#cal of de brogle wavelength of neutron
#intiation of all variables
#given that
import math
e = 25 # Energy of neutron in eV
c = 3e8 # speed of light in m/s
m = 1.67e-27 # Mass of neutron in Kg
h = 6.62e-34 # Plank constant
print("Example 1.15,page no:24")
rest_e = m*c**2/(1e6*1.6e-19)# rest mass energy of neutron in MeV
if e/rest_e < 0.015:
E = e;
else:
E = rest_e +e;
lamda = h/(math.sqrt(2*m*e*1.6e-19)) # calculation of de Broglie wavelength
print("de Broglie wavelength of neutron in angstrom:"),round(lamda*1e10,4)
Example 1.15,page no:24
de Broglie wavelength of neutron in angstrom: 0.0573

## example 1.16;page no:24¶

In [14]:
#cal of de brogle wavelength of neutron
#intiation of all variables
#given that
import math
e = 2*1.6e-19 # charge on alpha particle in coulomb
V = 200 # Applied voltage in volts
m = 4*1.67e-27 # Mass of alpha particle in Kg
h = 6.63e-34 # Plank constant
print("Example 1.16,page no:24")
lamda=h/(math.sqrt(2*e*V*m)) # calculation of de Broglie wavelength
print("de Broglie wavelength of neutron in angstrom:"),round(lamda*1e10,5)
Example 1.16,page no:24
de Broglie wavelength of neutron in angstrom: 0.00717

## example 1.17;page no:25¶

In [18]:
#cal of de brogle wavelength of ball and electron
#intiation of all variables
#given that
M = 20 # Mass of ball in Kg
V = 5 # velocity of of ball in m/s
m = 9.1e-31 #Mass of electron in Kg
v = 1e6 # velocity of of electron in m/s
h = 6.62e-34 # Plank constant
print("Example 1.17,page no:25")
lambda_b = h/(M*V) # calculation of de Broglie wavelength for ball
lambda_e = h/(m*v) # calculation of de Broglie wavelength electron
print("de Broglie wavelength of ball in angstrom:"),round(lambda_b*1e10,34)
print("de Broglie wavelength of electron in angstrom:"),round(lambda_e*1e10,2)
# answer in book is 6.62e-22 angstrom for ball
Example 1.17,page no:25
de Broglie wavelength of ball in angstrom: 6.62e-26
de Broglie wavelength of electron in angstrom: 7.27

## example 1.18;page no:26¶

In [15]:
#cal of de brogle wavelength of neutron
#intiation of all variables
#given that
import math
E = 1 # Energy of neutron in eV
m = 1.67e-27 # Mass of neutron in Kg
h = 6.62e-34 # Plank constant
print("Example 1.18,page no:26")
lamda = h/math.sqrt(2*m*E*1.6e-19) # Calculation of velocity of moving electron
print("Wavelength of neutron in angstrom:"),round(lamda*1e10,3)
# Answer in book is 6.62e-22 angstrom
Example 1.18,page no:26
Wavelength of neutron in angstrom: 0.286

## example 1.19;page no:27¶

In [20]:
#cal of Applied voltage on electron
#intiation of all variables
#given that
lamda = 0.5# wavelength of electron in angstrom
m = 9.1e-31 # Mass of electron in Kg
h = 6.62e-34 # Plank constant
q = 1.6e-19 # charge on electron in coulomb
print("Example 1.19,page no:27")
V = h**2/(2*m*q*(lamda*1e-10)**2) # Calculation of velocity of moving electron
print("Applied voltage on electron in V:"),round(V,1)
# Answer in book is 601.6 Volt
Example 1.19,page no:27
Applied voltage on electron in V: 602.0

## example 1.21;page no:29¶

In [21]:
#cal of  wavelength of neutron
#intiation of all variables
#given that
import math
k = 8.6e-5 # Boltzmann constant
t = 37 # Temperature in degree Celsius
h = 6.62e-34 # Plank constant
m = 1.67e-27 # Mass of neutron
print("Example 1.21,page no:29")
lamda = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength
print("Wavelength of neutron at degree Celsius in angstrom:"),round(lamda*1e10,2)
Example 1.21,page no:29
Wavelength of neutron at degree Celsius in angstrom: 1.43

## example 1.22;page no:29¶

In [22]:
#cal of wavelength of helium
#intiation of all variables
#given that
import math
k = 8.6e-5 # Boltzmann constant
t = 27 # Temperature in degree Celsius
h = 6.62e-34 # Plank constant
m = 6.7e-27 # Mass of helium atom
print("Example 1.22,page no:29")
lamda = h/math.sqrt(3*m*(k*1.6e-19)*(t+273))# Calculation of wavelength
print("Wavelength of helium at degree Celsius in angstrom:"),round(lamda*1e10,3)
Example 1.22,page no:29
Wavelength of helium at degree Celsius in angstrom: 0.727

## example 1.23;page no:30¶

In [17]:
#cal of Interatomic spacing of crystal
#intiation of all variables
#given that
import math
E = 200. # energy of electrons in eV
x = 20. # distance of screen in cm
D = 2. # diameter of ring in cm
h = 6.62e-34 # Plank constant
m = 9.1e-31 # Mass of electron in kg
print("Example 1.23,page no:30")
lamda= h/math.sqrt(2*m*E*1.6e-19) # Calculation of wavelength
print("lamda="),round(lamda,13)
print("D/2*x="),D/(2*x)
p=D/(2*x)
print("tan(theta)="),p
d = lamda/(2*p)# calculation of interatomic spacing of crystal
print("Interatomic spacing of crystal in angstrom:"),round(d*1e10,2)
Example 1.23,page no:30
lamda= 8.67e-11
D/2*x= 0.05
tan(theta)= 0.05
Interatomic spacing of crystal in angstrom: 8.67

## example 1.24;page no:31¶

In [21]:
#cal of velocity of electron
#intiation of all variables
#given that
r = 0.5 # Bohr radius of hydrogen in angstrom
m = 9.1e-31 # Mass of neutron in Kg
h = 6.6e-34 # Plank constant
print("Example 1.24,page no:31")
v = h/(2*3.14*r*1e-10*m) # velocity of electron in ground state
print("Velocity of electron in ground state in M/s="),round(v/10**6,2)
# Answer in book is 2.31e6 m/s
Example 1.24,page no:31
Velocity of electron in ground state in M/s= 2.31

## example 1.25;page no:32¶

In [25]:
#cal of Velocity of electron in ground state
#intiation of all variables
#given that
lamda = 5890 # wavelength of yellow radiation in angstrom
m = 9.1e-31 # Mass of neutron in Kg
h = 6.63e-34 # Plank constant
print("Example 1.25,page no:32")
v = h/(lamda*1e-10*m) # velocity of electron in ground state
print("Velocity of electron in ground state in m/s:"),round(v,1)
# Answer in book is 1.24e3 m/s
Example 1.25,page no:32
Velocity of electron in ground state in m/s: 1237.0

## example 1.26;page no:33¶

In [26]:
#cal of Velocity and kinetic energy of neutron
#intiation of all variables
#given that
lamda = 2 # wavelength of neutron in angstrom
m = 1.67e-27 # Mass of neutron in Kg
h = 6.63e-34 # Plank constant
print("Example 1.26,page no:33")
v = h/(lamda*1e-10*m) # velocity of neutron
k = 0.5*m*v**2 # Kinetic energy of neutron
print("Velocity of neutron in m/s:"),round(v,1)
print("Kinetic energy of neutron in eV:"),round(k/1.6e-19,3)
# Answer in book is 0.021eV
Example 1.26,page no:33
Velocity of neutron in m/s: 1985.0
Kinetic energy of neutron in eV: 0.021

## example 1.29;page no:36¶

In [22]:
#cal of theta and theta1
#intiation of all variables
#given that
import math
v1 = 50 # Previous applied voltage
v2 = 65 # final applied voltage
k = 12.28
d = 0.91 # Spacing in a crystal in angstrom
print("Example 1.29,page no:36")
lamda = k/math.sqrt(v1)
theta= math.asin(lamda/(2*d))# Angel for initial applied voltage
lamda1 = k/math.sqrt(v2)# wavelength for final applied voltage
theta1 = math.asin(lamda1/(2*d))# Angel for final applied voltage
#print("lamda1/1.82="),math.asin(lamda1/1.82)
print("theta"),round(theta*180/3.14,1)
print("theta1="),round(theta1*180/3.14,2)
print("For first order, sin(theta) in For second order sin(theta) must be which is not possible for any value of angle.So no maxima occur for higher orders:"),round(2*math.sin(theta),2)
#print("Angle of diffraction for first order of beam  is  degree at  Volts:"),round((math.theta1*180/math.pi),2)
# Answer in book is 57.14 degree
Example 1.29,page no:36
theta 72.6
theta1= 56.84
For first order, sin(theta) in For second order sin(theta) must be which is not possible for any value of angle.So no maxima occur for higher orders: 1.91

## example 1.30;page no:45¶

In [28]:
#cal of Group velocity of seawater waves
#intiation of all variables
#given that
import math
lamda = 680 # Wavelength in m
g = 9.8 #Acceleration due to gravity
print("Example 1.30,page no:45")
v_g = 0.5*math.sqrt(g*lamda/(2*3.14)) # Calculation of group velocity
print("Group velocity of seawater waves in m/s:"),round(v_g,2)
# Answer in book is 16.29 m/s
Example 1.30,page no:45
Group velocity of seawater waves in m/s: 16.29

## example 1.32;page no:47¶

In [29]:
#cal of group and phase velocity of de brogle waves
#intiation of all variables
#given that
import math
lamda = 2e-13 # de Broglie wavelength of an electron in m
c = 3e8 # Speed of light in m/s
m = 9.1e-31 # Mass of electron in Kg
h = 6.63e-34 # Plank constant
print("Example 1.32,page no:47")
E = h*c/(lamda*1.6e-19)
E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy
E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV
v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity
v_p  = c**2/v_g # Phase velocity
print("Group velocity of de Broglie waves is c  :"),round(v_g/c,4)
print(" phase velocity of de Broglie waves is c"),round(v_p/c,4)
Example 1.32,page no:47
Group velocity of de Broglie waves is c  : 0.9966
phase velocity of de Broglie waves is c 1.0034

## example 1.33;page no:48¶

In [23]:
#cal of Kinetic energy of electron,group velocity and phase velocity of de Broglie waves
#intiation of all variables
#given that
import math
lamda = 2.e-12 # de Broglie wavelength of an electron in m
c = 3.e8 # Speed of light in m/s
m = 9.1e-31 # Mass of electron in Kg
h = 6.63e-34 # Plank constant
print("Example 1.33,page no:48")
E = h*c/(lamda*1.6e-19) # Energy due to momentum
E_rest = m*c**2/(1.6e-19) # Calculation of rest mass energy
E_total = math.sqrt(E**2+E_rest**2) # Total energy in eV
KE = E_total - E_rest # Kinetic energy
v_g = c*math.sqrt(1-(E_rest/E_total)**2) # Group velocity
v_p  = c**2/v_g # Phase velocity
print("Kinetic energy of electron in KeV:"),round(KE/1000,2)
print("Group velocity of de Broglie waves is c in m/s:"),round(v_g/c,4)
print("phase velocity of de Broglie waves is c in m/s:"),round(v_p/c,3)
# Answer in book is v_g = 0.6035c & v_p = 1.657c
Example 1.33,page no:48
Kinetic energy of electron in KeV: 293.33
Group velocity of de Broglie waves is c in m/s: 0.7719
phase velocity of de Broglie waves is c in m/s: 1.295