# Chapter 1:Origin of Quantum Concepts¶

## Example no:1,Page no:58¶

In [87]:
import math
#Variable declaration
c1=0.01
c2=0.1
c3=1
c4=10
b=2.898*10**-3                                        #Wien's constant (meter-kelvin)
h=(6.625*10**-34)/(2*math.pi)                         #Planck's constant (joule-second)
c=3*10**8                                             #speed of light (meter/second)
k=1.38*10**-23                                        #Boltzmann constant (joule/kelvin)
T=3000                                                #Temperature of black body (kelvin)
Delta_lembda=1*10**-9                                 #wavelength interval (meter)

#Calculation
from sympy import *
kT=Symbol('kT')
#(a)Average energy of Planck's oscillator:
E1=round(c1/(math.exp(c1)-1))*kT                           #Average energy of Planck's oscillator
E2=round(c2/(math.exp(c2)-1),2)*kT                           #Average energy of Planck's oscillator
E3=round(c3/(math.exp(c3)-1),2)*kT                           #Average energy of Planck's oscillator
E4=round(c4/(math.exp(c4)-1),5)*kT                           #Average energy of Planck's oscillator
#(b)Power radiated by a unit area of a black body
P=(4*(math.pi**2)*h*(c**2)*(T**5)*Delta_lembda)/((b**5)*((math.exp((2*math.pi*h*c)/(b*k)))-1.0))           #The power radiated per unit area (watt/meter**2)

#Result
print"(a)The average energy of Plancks oscillator:"
print"(i) Energy =",E1
print"Energy =",E2
print"Energy = ",E3
print"Energy =",E4
print"(b) The power radiated per unit area =%.f"%P,"W/m**2"
print"NOTE:Approximate values are used in book,that's why different answer"

(a)The average energy of Plancks oscillator:
(i) Energy = 1.0*kT
Energy = 0.95*kT
Energy =  0.58*kT
Energy = 0.00045*kT
(b) The power radiated per unit area =3115 W/m**2
NOTE:Approximate values are used in book,that's why different answer


## Example no:2,Page no:59¶

In [8]:
#Variable declaration
v=2*10**-2                                                                       #side of the cube (meter)
lembda=5000*10**-10                                                              #wavelength (meter)
delta_lembda=10*10**-10                                                          #range of wavelength (meter)
k=1.38*10**-23                                                                   #Boltzmann constant (joule/kelvin)
T=1500                                                                          #Temperature of the cavity (kelvin)

#Calculation
import math
#(i)Number of modes:
N=(8*math.pi*v**3*delta_lembda)/lembda**4                                             #number of modes
#(ii)Total radiant energy in the cavity:
U=N*k*T                                                                         #energy density (joule)

#Result
print"\n(a) Number of modes =%.3e"%N,"joule\n(b) Energy density =%.2e"%U ,"J"
print"Note: In book the answers of both the parts are WRONG by one order of magnitude in powers"

(a) Number of modes =3.217e+12 joule
(b) Energy density =6.66e-08 J
Note: In book the answers of both the parts are WRONG by one order of magnitude in powers


## Example no:3,Page no:59¶

In [2]:
import math

#Variable declaration
m=0.1                                                    #mass of a spring-mass system (kg)
k=10                                                     #spring constant of a spring-mass system (newton/meter)
A=0.1                                                    #Amplitude of system oscillation (meter)
h=(6.625*10**-34)/(2*math.pi)                            #Planck's constant (joule-second)
delta_n=1                                                #change in quantum number

#Calculation
#(a) Quantum number n associated with the energy of the oscillator
f=(k/m)**(1.0/2.0)                                          #frequency of oscillator (radian/second)
E=0.5*f*A**2                                            #Energy of oscillator (joule)
n=E/(h*f)                                               #Quantum number of the oscillator
#(b) Fractional change in energy
change_E=delta_n/n                                      #fractional change in energy

#Result
print"\n(a) Quantum number of the oscillator =%.e"%n
print"(b) Fractional change in energy =%.g"%change_E
print"(c) This example illustrates that the energy levels of macroscopic oscillators are so close together that even most delicate instruments cannot reveal the quantized nature of energy levels. All this is due to smallness of Planck’s constant h. In the limit h->0, the energy levels become continuous."
print"WRONG ANSWER NOTE:The answer given in the book for quantum number is just the order of it as it is a very large number. But the answer generated by the code is the exact value of it."

(a) Quantum number of the oscillator =5e+31
(b) Fractional change in energy =2e-32
(c) This example illustrates that the energy levels of macroscopic oscillators are so close together that even most delicate instruments cannot reveal the quantized nature of energy levels. All this is due to smallness of Planck’s constant h. In the limit h->0, the energy levels become continuous.
WRONG ANSWER NOTE:The answer given in the book for quantum number is just the order of it as it is a very large number. But the answer generated by the code is the exact value of it.


## Example no:4,Page no:63¶

In [4]:
import math
#Variable declaration
e=1.6*10**-19                                 #Charge of electron (coulombs)
h=(6.625*10**-34)/(2*math.pi)                 #Planck's constant (joule-second)
c=3*10**8                                     #Speed of light (meter/second)

#Calculation
ch=(2*math.pi*h*c*10**9)/e                    #Value of ch (eV nm)

#Result
print"\nch =",round(ch),"eV nm"
print"WRONG ANSWER:Approximate answer is given in book"

ch = 1242.0 eV nm
WRONG ANSWER:Approximate answer is given in book


## Example no:5,Page no:63¶

In [6]:
import math
#Variable declaration
h=(6.625*10**-34)/(2*math.pi)                          #Planck's constant (joule-second)
c=3*10**8                                              #speed of light (meter/second)
lembda=2000                                            #Wavelength of the light (Å)
phi=4.2                                                #work function of aluminium surface (eV)
ch=12400                                               #constant (eV Å)

#Calculation
#(a) maximum kinetic energy of photoelectrons
Tmax=(ch/lembda)-phi                                   #maximum kinetic energy of photoelectrons (eV)

#(b) minimum kinetic energy of photoelectrons
Tmin=0

#(c) cut-off wavelength
lembda_cut=ch/phi                                      # cut-off wavelength (Å)

#(d) stopping potential
v=2                                                    #stopping potential (volt)

#Result
print"\n(a) Maximum kinetic energy of photoelectrons =",round(Tmax),"eV"
print"(b) Minimum kinetic energy of photoelectrons =",Tmin
print"(c) Cut-off(Threshold) wavelength =",round(lembda_cut),"Å\n"
print"(d) Stopping potential =",v,"volt"
print"Note: In book answer of cut_off wavelength is wrong"

(a) Maximum kinetic energy of photoelectrons = 2.0 eV
(b) Minimum kinetic energy of photoelectrons = 0
(c) Cut-off(Threshold) wavelength = 2952.0 Å

(d) Stopping potential = 2 volt
Note: In book answer of cut_off wavelength is wrong


## Example no:6,Page no:64¶

In [11]:
#Variable declaration
lembda1=4000.0                                  #wavelength of light (Å)
V1=2.0                                          #stopping potential (volt)
lembda2=6000.0                                  #wavelength of light (Å)
V2=1.0                                          #stopping potential (volt)
e=1.6*10**-19                                 #Charge of electron (coulombs)
c=3.0*10**8                                     #speed of light (meter/second)
ch=12400.0                                      #constant (eV Å)

#Calculation
#(i) Planck's constant
h=(e*(V1-V2)*lembda1*10**-10*lembda2*10**-10)/(c*((lembda2*10**-10)-(lembda1*10**-10)))   #Planck's constant (joule-second)
#(ii) Work function
phi=(ch/lembda1)-V1                           #work function of the material (eV)

#Result
print" Plancks constant = h =",h,"J-s\n"
print" Work function of the material =",phi,"eV"

 Plancks constant = h = 6.4e-34 J-s

Work function of the material = 1.1 eV


## Example no:7,Page no:64¶

In [15]:
#Variable declaration
ch=12400                               #constant (eV Å)
phi_Tantalum=4.2                       #work function of Tantalum (eV)
phi_Tungsten=4.5                       #work function of Tungsten (eV)
phi_Aluminium=4.2                      #work function of Aluminium (eV)
phi_Barium=2.5                         #work function of Barium (eV)
phi_Lithium=2.3                        #work function of Lithium (eV)

#Calculation
lembda_Tantalum=ch/phi_Tantalum         #Threshold wavelength of Tantalum (Å)
lembda_Tungsten=ch/phi_Tungsten         #Threshold wavelength of Tungsten (Å)
lembda_Aluminium=ch/phi_Aluminium       #Threshold wavelength of Aluminium (Å)
lembda_Barium=ch/phi_Barium             #Threshold wavelength of Barium (Å)
lembda_Lithium=ch/phi_Lithium           #Threshold wavelength of Lithium (Å)

#Result
print"Tantalum %d"%lembda_Tantalum,"Å"
print"Tungsten %d"%lembda_Tungsten,"Å"
print"Aluminium %d"%lembda_Aluminium,"Å"
print"Barium %d"%lembda_Barium,"Å"
print"Lithium %d"%lembda_Lithium,"Å"
if(lembda_Tantalum<8000 and lembda_Tantalum>4000):
print"Tantalum can be used for designing photocell"

if(lembda_Tungsten<8000 and lembda_Tungsten>4000):             #Checking whether Threshold wavelength of Tungsten lies in visible range or not
print"Tungsten can be used for designing photocell"

if(lembda_Aluminium<8000 and lembda_Aluminium>4000):           #Checking whether Threshold wavelength of Aluminium lies in visible range or not
print"Aluminium can be used for designing photocell"

if(lembda_Barium<8000 and lembda_Barium>4000):                #Checking whether Threshold wavelength of Barium lies in visible range or not
print"Barium can be used for designing photocell"

if(lembda_Lithium<8000 and lembda_Lithium>4000):              #Checking whether Threshold wavelength of Lithium lies in visible range or not
print"Lithium can be used for designing photocell"

Tantalum 2952 Å
Tungsten 2755 Å
Aluminium 2952 Å
Barium 4960 Å
Lithium 5391 Å
Barium can be used for designing photocell
Lithium can be used for designing photocell


## Example no:8,Page no:68¶

In [16]:
import math
#Variable declaration
lembda_c=0.024                                                                  #Compton wavelength of electron (Å)
lembda=1                                                                        #Wavelength of X-rays (Å)
ch=12400                                                                        #constant (eV Å)

#Calculation
#(a) Compton shift
shift1=lembda_c*(1-math.cos(Theta1))                                                 #Compton shift (Å)
shift2=lembda_c*(1-math.cos(Theta2))                                                 #Compton shift (Å)
shift3=lembda_c*(1-math.cos(Theta3))                                                 #Compton shift (Å)
#(b) Kinetic energy imparted to the recoil electron
T1=(ch*shift1)/(lembda*(lembda+shift1))                                         #Kinetic energy imparted to the electron (eV)
T2=(ch*shift2)/(lembda*(lembda+shift2))                                         #Kinetic energy imparted to the electron (eV)
T3=(ch*shift3)/(lembda*(lembda+shift3))                                         #Kinetic energy imparted to the electron (eV)

#Result
print"\n(a) Compton shift:\n\t (i)",shift1," Å\n\t (ii)",shift2," Å\n\t (iii)",shift3," Å\n"
print"(b) Kinetic energy imparted to the recoil electron:\n\t (i)",round(T1)," eV\n\t (ii) %d"%T2," eV\n\t (iii)",round(T3)," eV"

(a) Compton shift:
(i) 0.012  Å
(ii) 0.024  Å
(iii) 0.048  Å

(b) Kinetic energy imparted to the recoil electron:
(i) 147.0  eV
(ii) 290  eV
(iii) 568.0  eV


## Example no:9,Page no:69¶

In [17]:
import math
#Variable declaration
lembda_c=0.024                                                                  #Compton wavelength of electron (Å)
Theta=(45*math.pi)/180                                                              #Scattering angle (radian)

#Calculation
lembda=lembda_c*(1-math.cos(Theta))                                                  #Wavelength of incident photon (Å)

#Result
print"\n(a) Wavelength of incident photon = ",round(lembda,4)," Å (gamma ray)"
print"(b) Photon lies in the gamma ray spectrum"

(a) Wavelength of incident photon =  0.007  Å (gamma ray)
(b) Photon lies in the gamma ray spectrum


## Example no:10,Page no:69¶

In [1]:
#Variable declaration
E=1                                      #Energy of photon (MeV)
eta=0.25                                 #Relative change in photon's wavelength

#Calculation
T=(E*eta)/(1+eta)                       #Kinetic energy of recoil electron (MeV)

#Result
print"\nThe kinetic energy of recoil electron =",T,"MeV"

The kinetic energy of recoil electron = 0.2 MeV


## Example no:11,Page no:69¶

In [18]:
import math
#Variable declaration
E=0.25                                         #Energy of photon (MeV)
Theta=(120*math.pi)/180                        #Scattering angle of photon (radian)
a=0.51                                         #Value of m0*c**2 (Mev)

#Calculation
E_das=E/(1+(E/a)*(1-math.cos(Theta)))          #Energy of the scattered photon (MeV)

#Result
print"\nEnergy of the scattered photon =",round(E_das,3),"Mev"

Energy of the scattered photon = 0.144 Mev


## Example no:12,Page no:69¶

In [46]:
#Variable declaration
p=1.02                                                 #momentum of the photon (MeV/c)
p_dash=0.255                                           #momentum of the photon after scattering (MeV/c)
a=0.51                                                 #Value of m0*c**2 (Mev)

#Calculation
Theta=2*math.degrees(math.asin(((0.5*a*(p-p_dash))/(p*p_dash))**(1.0/2.0)));

#Result
print"\nAngle of the photon after scattering =",Theta,"degree"

Angle of the photon after scattering = 120.0 degree


## Example no:13,Page no:70¶

In [75]:
#Variable declaration
Theta=120                  #Scattering angle of photon (degree)
T=0.45                    #Kinetic energy of electron (MeV)
a=0.51                    #Value of m0*c**2 (Mev)
import math

#Calculation

E=(T/2.0)*(1.0+math.sqrt(1.0+(2.0*a)/(T*((math.sin(math.radians(Theta/2.0)))**2))))  #Energy of the incident photon (MeV)

#Result
print"\nEnergy of the incident photon =%.3g"%E,"Mev"

Energy of the incident photon =0.676 Mev


## Example no:14,Page no:74¶

In [80]:
#Variable declaration
lembda0=2536*10**-10                    #wavelength of exciting line (meter)
lembda=2612*10**-10                     #wavelength of Raman line (meter)

#Calculation
v0=1.0/lembda0                              #wave number of exciting line (1/meter)
v=1.0/lembda                                #wave number of Raman line (1/meter)
shift=v0-v                                #the Raman shift (1/meter)

#Result
print"\nThe Raman shift =",shift,"m-1"
print"Note: v0 and v values in the book are VERY LESS PRECISE,Therefore an approximate answer"

The Raman shift = 114733.745248 m-1
Note: v0 and v values in the book are VERY LESS PRECISE,Therefore an approximate answer


## Example no:15,Page no:75¶

In [86]:
#Variable declaration
lembda0=5000*10**-10                     #Wavelength of radiation (meter)
lembda=5050.5*10**-10                    #Wavelength of Raman line (meter)

#Calculation
#(a) Raman frequency
v0=1.0/lembda0                             #Wave number of radiation (1/meter)
v=1.0/lembda                               #Wave number of Raman line (1/meter)
shift=v0-v                               #Raman shift (1/meter)
va=v0+shift                              #Frequency of antistoke's line (1/meter)
#(b) Position of the antistokes' line
lembdaa=(10.0**10.0)/va                        #Wavelength of antistoke's line (Å)

#Result
print"(a) Raman frequency =%.2e"%va,"m**-1"
print"(b) Wavelength of antistokes line =%.1f"%lembdaa,"Å (APPROX)"

(a) Raman frequency =2.02e+06 m**-1
(b) Wavelength of antistokes line =4950.5 Å (APPROX)