# Chapter 2:Wave Nature of Material Particles¶

## Example no:1,Page no:89¶

In [39]:
#Variable declaration
m=9.1*10**-31                                                                    #mass of electron (kg)
h=6.6*10**-34                                                                    #planck's constant (joule-second)
e=1.6*10**-19                                                                    #charge of electron (coulomb)

#Calculation
from sympy import *
V=Symbol('V')
a=(h*10**10)/(2.0*m*e*V)**(1/2.0)                                                               #wavelength of electron = h/(2*m*e*v)**(1/2) (Å)

#Result
print"\n Wavelength of electron accelerated through a potential difference V =",a,"Å"

 Wavelength of electron accelerated through a potential difference V = 12.2306137249082*V**(-0.5) Å


## Example no:3,Page no:90¶

In [105]:
#Variable declaration
m_e=9.1*10**-31                                   #mass of electron (kg)
m=100*10**-3                                      #mass of object (kg)
v=1000                                            #velocity of electron and object (meter/second)
h=6.63*10**-34                                    #planck's constant (joule-second)

#Calculation
#(i) de Broglie wavelength of electron
lembda_e=h/(m_e*v)                                #de Broglie wavelength of electron
#(ii) de Broglie wavelength of object
lembda=h/(m*v)                                  #de Broglie wavelength of object

#Result
print"\n(i) de Broglie wavelength of electron =",round(lembda_e*10**10),"A"
print"(ii) de Broglie wavelength of object =%.2e"%lembda,"m"
print"Note: In the book the answer of part(ii) is wrong"

(i) de Broglie wavelength of electron = 7286.0 A
(ii) de Broglie wavelength of object =6.63e-36 m
Note: In the book the answer of part(ii) is wrong


## Example no:4,Page no:90¶

In [106]:
#Variable declaration
e=1.6*10**-19                               #charge of electron (coulomb)
T=100*e                                     #kinetic energy (joule)
m_e=9.1*10**-31                             #mass of electron (kg)
m_p=1.67*10**-27                           #mass of proton (kg)
m_alpha=4*m_p                              #mass of alpha particle (kg)
h=6.63*10**-34                              #planck's constant (joule-second)

#Calculation
lembda_e=(h*10**10)/(2*m_e*T)**(1.0/2.0)                        #de Broglie wavelength of electron (Å)
lembda_p=(h*10**10)/(2*m_p*T)**(1.0/2.0)                        #de Broglie wavelength of proton (Å)
lembda_alpha=(h*10**10)/(2*m_alpha*T)**(1.0/2.0)                #de Broglie wavelength of alpha particle (Å)

#Result
print"De Broglie wavelength of electron =",round(lembda_e,2),"Å"
print"De Broglie wavelength of proton =",round(lembda_p,3),"Å"
print"De Broglie wavelength of alpha particle =",round(lembda_alpha,3),"Å"

De Broglie wavelength of electron = 1.23 Å
De Broglie wavelength of proton = 0.029 Å
De Broglie wavelength of alpha particle = 0.014 Å


## Example no:5,Page no:90¶

In [107]:
#Variable declaration
h=6.63*10**-34                                                                   #planck's constant (joule-second)
m=9.1*10**-31                                                                    #mass of electron (kg)
lembda=5896*10**-10                                                              #wavelength of yellow spectral line of sodium (meter)
e=1.6*10**-19                                                                    #charge of electron (coulomb)

#Calculation
T_j=h**2/(2*m*lembda**2)                                                          #kinetic energy of the electron (joule)
T_eV=T_j/e                                                                      #kinetic energy of the electron (eV)

#Result
print"Kinetic energy of electron =%.2e"%T_j,"J =%.1e"%T_eV,"eV"

Kinetic energy of electron =6.95e-25 J =4.3e-06 eV


## Example no:6,Page no:91¶

In [108]:
#Variable declaration
h=6.63*10**-34                                                                   #planck's constant (joule-second)
m_n=1.67*10**-27                                                                 #mass of neutron (kg)
T=300                                                                           #Temperature (kelvin)
k=1.38*10**-23                                                                   #Boltzmann constant (joule/kelvin)

#Calculation
E=(3*k*T)/2.0                                                                     #Kinetic energy of thermal neutron (joule)
lembda=(h*10**10)/(2*m_n*E)**(1/2.0)                                                #Wavelength of thermal neutron (Å)

#Result
print"The wavelength of thermal neutron =%.3g"%lembda,"Å"

The wavelength of thermal neutron =1.46 Å


## Example no:7,Page no:91¶

In [109]:
#Variable declaration
h=6.63*10**-34                                                                   #planck's constant (joule-second)
m_H2=2*1.67*10**-27                                                              #mass of hydrogen molecule (kg)
T=27+273                                                                        #room temperature (kelvin)
k=1.38*10**-23                                                                   #Boltzmann constant (joule/kelvin)

#Calculation
lembda=(h*10**10)/(2*m_H2*k*T)**(1/2.0)                                             #de Broglie wavelength of hydrogen molecule (Å)

#Result
print"The de Broglie wavelength of hydrogen molecules at their most probable speed =%.2f"%lembda,"Å"

The de Broglie wavelength of hydrogen molecules at their most probable speed =1.26 Å


## Example no:8,Page no:91¶

In [110]:
#Variable declaration
a=0.51                                           #Value of m0*c**2 (Mev)

#Calculation
import math
T=a*(math.sqrt(2.0)-1)                             #Kinetic energy (MeV)

print"Kinetic energy of electron =",round(T,2),"MeV"

Kinetic energy of electron = 0.21 MeV


## Example no:9,Page no:92¶

In [111]:
#Variable declaration
a=0.51                                  #Value of m0*c**2 (MeV)
b=0.0124                                #Value of h*c (MeV Å)
lembda_X=0.1                            #Short wavelength limit of continuous X-ray spectrum (Å)

#Calculation
lembda=lembda_X/(1+(2*a*lembda_X)/b)**(1/2.0)        #de Broglie wavelength of relativistic electrons

print"De Broglie wavelength of relativistic electrons =",round(lembda,3),"Å"

De Broglie wavelength of relativistic electrons = 0.033 Å


## Example no:10,Page no:92¶

In [112]:
#Variable declaration
r=0.53                                                                          #Radius of the first Bohr orbit in hydrogen atom (Å)

#Calculation
import math
lembda=2*math.pi*r                                                                  #de Broglie wavelength of electron in first Bohr orbit in hydrogen atom

#Result
print"\nde Broglie wavelength of electron in first Bohr orbit in hydrogen atom =%.1f"%lembda,"Å"

de Broglie wavelength of electron in first Bohr orbit in hydrogen atom =3.3 Å


## Example no:12,Page no:93¶

In [113]:
#Variable declaration
v=10000                                                                         #speed of object (meter/second)
accu_v=0.0001                                                                   #accuracy of speed of object
m_b=0.05                                                                        #mass of the bullet (kg)
h=1.054*10**-34                                                                  #planck's constant (joule-second)
m_e=9.1*10**-31                                                                  #mass of electron (kg)

#Calculation
#(a) fundamental accuracy of position for bullet
p_b=m_b*v                                                                       #momentum of bullet (kg m/s)
p_uncer_b=p_b*accu_v                                                            #uncertainty in momentum of bullet (kg m/s)
x_uncer_b=h/p_uncer_b                                                           #minimum uncertainty in position of bullet (meter)
#(b) fundamental accuracy of position for electron
p_e=m_e*v                                                                       #momentum of electron (kg m/s)
p_uncer_e=p_e*accu_v                                                            #uncertainty in momentum of electron (kg m/s)
x_uncer_e=h/p_uncer_e                                                           #uncertainty in position of electron (meter)

print"\n(a) Minimum uncertainty in position of bullet =%.1e"%x_uncer_b,"meter"
print"(b) uncertainty in position of electron =%.3g"%x_uncer_e,"meter"

print"Note:The answers given in the book are wrong. \nAlso in the solution they have used speed=1000 while in the question it is given to be equal to 10000."

(a) Minimum uncertainty in position of bullet =2.1e-33 meter
(b) uncertainty in position of electron =0.000116 meter
Note:The answers given in the book are wrong.
Also in the solution they have used speed=1000 while in the question it is given to be equal to 10000.


## Example no:13,Page no:94¶

In [114]:
#Variable declaration
h=1.054*10**-34                                                                  #planck's constant (joule-second)
m=9.1*10**-31                                                                    #mass of electron (kg)
x_uncer=1*10**-10                                                                #uncertainty in the position of elctrons (meter)
e=1.6*10**-19                                                                    #charge of electron (coulomb)

#Calculation
#(i) uncertainty in the momentum of electron
p_uncer=h/x_uncer                                                               #The uncertainty in the momentum of electron (kg m/s)
#(ii) kinetic energy of electron
T=p_uncer**2/(2*m*e)                                                             #kinetic energy of electron (eV)

#Result
print"\n(i) The uncertainty in the momentum of electron =",p_uncer,"kg m/s"
print"(ii) Kinetic energy of electron =%.1f"%T,"eV"


(i) The uncertainty in the momentum of electron = 1.054e-24 kg m/s
(ii) Kinetic energy of electron =3.8 eV


## Example no:14,Page no:94¶

In [115]:
#Variable declaration
h=1.054*10**-34                                                                  #planck's constant (joule-second)
x=10**-14                                                                        #dimension of the nucleus (meter)
c=3*10**8                                                                        #speed of light (meter/second)
e=1.6*10**-19                                                                    #charge of electron (coulomb)

#Calculation
#(i) Uncertainty in the momentum of electron
p_uncer=h/x                                                                     #The uncertainty in the momentum of electron (kg m/s)
#(ii) kinetic energy of electron
T=(p_uncer*c)/(e*10**6)                                                          #kinetic energy of electron (MeV)

#Result
print"\n(i) The uncertainty in the momentum of electron =",p_uncer,"kg m/s"
print"(ii) Kinetic energy of electron =",round(T)," MeV"

(i) The uncertainty in the momentum of electron = 1.054e-20 kg m/s
(ii) Kinetic energy of electron = 20.0  MeV


## Example no:15,Page no:94¶

In [116]:
#Variable declaration
h=1.054*10**-34                                                                  #planck's constant (joule-second)
x=10**-14                                                                        #dimension of the nucleus (meter)
e=1.6*10**-19                                                                    #charge of electron (coulomb)
m=1.67*10**-27                                                                   #mass of proton (kg)

#Calculation
#(i) Uncertainty in the momentum of electron
p_uncer=h/x                                                                     #The uncertainty in the momentum of electron (kg m/s)
#(ii) kinetic energy of proton
T=(p_uncer**2)/(2*m*e*10**6)                                                      #kinetic energy of proton (MeV)

#Result
print"\n(i) The uncertainty in the momentum of electron =",p_uncer,"kg m/s"
print"(ii) Kinetic energy of proton =",round(T,2),"MeV  (approx)"

(i) The uncertainty in the momentum of electron = 1.054e-20 kg m/s
(ii) Kinetic energy of proton = 0.21 MeV  (approx)


## Example no:17,Page no:95¶

In [117]:
#Variable declaration
h=1.054*10**-34                                                                  #planck's constant (joule-second)
delta_t=10**-12                                                                  #time for which nucleus remains in excited state (second)

#Calculation
delta_E=h/delta_t                                                               #uncertainty in the energy of the gamma ray photon (joule)

#Result
print"\nThe uncertainty in the energy of the gamma ray photon =",delta_E,"J"

The uncertainty in the energy of the gamma ray photon = 1.054e-22 J


## Example no:18,Page no:95¶

In [118]:
#Variable declaration
delta_t=10**-8                                                                   #life-time of the average excited atom (second)

#Calculation
delta_f=1/delta_t                                                               #minimum uncertainty in the frequency of photon (radian/second)

#Result
print"\nminimum uncertainty in the frequency of photon =%.g"%delta_f,"rad/s"

minimum uncertainty in the frequency of photon =1e+08 rad/s


## Example no:19,Page no:95¶

In [119]:
#Variable declaration
h=1.054*10**-34                                                                  #planck's constant (joule-second)
e=1.6*10**-19                                                                    #charge of electron (coulomb)
m=9.1*10**-31                                                                    #mass of electron (kg)
E0=8.8542*10**-12                                                                #permittivity of free space (C**2/N*m**2)

#Calculation
#(i) radius of ground state of hydrogen atom
r=(4*math.pi*E0*h**2)/(m*e**2)                                                        #radius of ground state of hydrogen atom (meter)
#(ii) Binding energy of electron in hydrogen atom in the ground state
E=(-0.5*m*e**4)/(4*math.pi*E0*h)**2                                                   #binding energy of electron in hydrogen atom in the ground state (joule)

#Result
print"\n(i) Radius of ground state of hydrogen atom =%.1e"%r,"m (in scientific notation) OR 0.53*10**-10"
print"(ii) Binding energy of electron in ground state of hydrogen atom =%.2e"%E,"J"

(i) Radius of ground state of hydrogen atom =5.3e-11 m (in scientific notation) OR 0.53*10**-10
(ii) Binding energy of electron in ground state of hydrogen atom =-2.17e-18 J