Chapter 5: Foundations of Quantum Mechanics¶

Example 5.1,Page number 5-23¶

In :
import math

#Given Data:
m=6.68*10**-27       #mass of alpha particle
V=30*10**3           #potential difference
e=1.6*10**-19        #charge of an electron
q=2*e                #Charge of alpha particle
h=6.63*10**-34       #Planck's constant

#Calculations:
lamda=h/math.sqrt(2*m*q*V)     #de Broglie wavelength
print"de Broglie wavelength associated with alpha particle is =" ,lamda,"m"
de Broglie wavelength associated with alpha particle is = 5.85429607723e-14 m

Example 5.2,Page number 5-23¶

In :
import math

#Given Data:
m=1                  #mass of given particle in kg
h=6.63*10**-34       #Planck's constant
v=1*10**3            #velocity of particle

#Calculations:
lamda=h/(m*v)         #de Broglie wavelength
print"de Broglie wavelength associated with particle is =",lamda,"m"
print"This wavelength is too small for any practical significance."
de Broglie wavelength associated with particle is = 6.63e-37 m
This wavelength is too small for any practical significance.

Example 5.3,Page number 5-24¶

In :
import math

#Given Data:
m1=40*10**-3         #mass of bullet in kg
m2=9.1*10**-31       #mass of electron in kg
h=6.63*10**-34       #Planck's constant
v=1100              #velocity of bullet and electron

#Calculations:
lamda1=h/(m1*v)        #de Broglie wavelength
print"de Broglie wavelength associated with bullet is =",lamda1,"m"

lamda2=h/(m2*v)        #de Broglie wavelength
print"de Broglie wavelength associated with electron is =",lamda2,"m"

print"Wavelength of bullet is too small.Hence it can not be measured with help of diffraction effect."
de Broglie wavelength associated with bullet is = 1.50681818182e-35 m
de Broglie wavelength associated with electron is = 6.62337662338e-07 m
Wavelength of bullet is too small.Hence it can not be measured with help of diffraction effect.

Example 5.4,Page number 5-24¶

In :
import math

#Given Data:
V=100            #potential difference
d=2.15*10**-10   #lattice spacing

#Calculations:
lamda=12.26*10**-10/(math.sqrt(V))          #wavelength associated with electron in meter

#using bragg's law for first order  lamda=2d sin(theta)
theta=math.degrees(math.asin(lamda/(2*d)))      #glancing angle in degrees
print"Glancing angle at which first reflection occurs is =",theta,"Degrees"
Glancing angle at which first reflection occurs is = 16.5657992687 Degrees

Example 5.5,Page number 5-25¶

In :
import math

#Given Data:
mn=1.674*10**-27     #mass of neutron
h=6.63*10**-34       #Planck's constant
lamda=1*10**-10      #wavelength of neutron

#Calculations:

#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength
E1=h**2/(2*mn*lamda**2)   #Energy of neutron in joules
E=E1/(1.6*10**-19)        #Energy of neutron in electron-Volts

print"Energy of neutron is =",E,"eV"
Energy of neutron is = 0.0820581317204 eV

Example 5.6,Page number 5-25¶

In :
import math

#Given Data:
mn=1.67*10**-27      #mass of neutron
h=6.6*10**-34        #Planck's constant
lamda=3*10**-10        #wavelength of neutron
d=3.036*10**-10      #lattice spacing

#Calculations:

#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength
E1=h**2/(2*mn*lamda**2)  #Energy of neutron in joules
E=E1/(1.6*10**-19)    # Energy of neutron in electron-Volts
print"Energy of neutron is =",E,"eV"

#using bragg's law for first order  lamda=2d sin(theta)
theta=math.degrees(math.asin(lamda/(2*d)))      #glancing angle in degrees
print" Glancing angle at which first orde reflection occurs is =",theta,"Degrees"
Energy of neutron is = 0.00905688622754 eV
Glancing angle at which first orde reflection occurs is = 29.6085193042 Degrees

Example 5.7,Page number 5-26¶

In :
import math

#Given Data:
m=9.108*10**-31       #mass of electron
h=6.625*10**-34       #Planck's constant
lamda=5*10**-7          #wavelength of electron

#Calculations:

#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength
E1=h**2/(2*m*lamda**2)     #Energy of electron in joules
E=E1/(1.6*10**-19)      #Energy of electron in electron-Volts
print"Energy of electron is =",E,"eV"
Energy of electron is = 6.02363650088e-06 eV

Example 5.8,Page number 5-27¶

In :
import math

#Given Data:
mn=1.676*10**-27        #mass of neutron
me=9.1*10**-31          #mass of electron
h=6.625*10**-34         #Planck's constant

#Calculations:
#Part 1:
En1=0.025              #Energy in eV of neutron
En=En1*(1.6*10**-19)   #Energy in joules

lamda1=h/math.sqrt(2*mn*En)     #wavelength of a beam of neutron
print"wavelength of a beam of neutron is =",lamda1,"m"

#Part 2:
lamda2=2*10**-10            #wavelength of electron and photon

#//we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength
Ee1=h**2/(2*me*lamda2**2)     #Energy of electron in joules
Ee=Ee1/(1.6*10**-19)          #Energy of electron in electron-Volts
print"Energy of electron is =",Ee,"eV"

p1=h/lamda2               #momentum of electron
print" Momentum of electron is =",p1,"kg.m/s"

C=3*10**8                 #Velocity of light
Ep=h*C/lamda2             #Energy of photon in joules
print"Energy of photon is =",Ep,"Joules"

p2=h/lamda2              #momentum of photon
print"Momentum of photon is =",p2,"kg.m/s"
wavelength of a beam of neutron is = 1.80927208246e-10 m
Energy of electron is = 37.6808250343 eV
Momentum of electron is = 3.3125e-24 kg.m/s
Energy of photon is = 9.9375e-16 Joules
Momentum of photon is = 3.3125e-24 kg.m/s

Example 5.9,Page number 5-28¶

In :
import math

#Given data:
#We have alpha particle,neutron,proton and electron.

#To find: shortest wavelength

print"We know, lamda=h/sqrt(2*m*E)     #de Broglie wavelength"

#Wavelength is inversely proportional to mass of particle for constant energy
print"i.e., Wavelength is inversely proportional to mass of particle for constant energy. "

print"We have alpha particle,neutron,proton and electron."

#AS,alpha particle has highest mass.Thus it will have shortest wavelength.
print"Out of above, alpha particle has highest mass."

print"Hence it will have shortest wavelength."
We know, lamda=h/sqrt(2*m*E)     #de Broglie wavelength
i.e., Wavelength is inversely proportional to mass of particle for constant energy.
We have alpha particle,neutron,proton and electron.
Out of above, alpha particle has highest mass.
Hence it will have shortest wavelength.

Example 5.10,Page number 5-28¶

In :
import math

#Given Data:
me=9.108*10**-31        # mass of electron
mp=1.66*10**-27         # bmass of proton
h=6.625*10**-34         # Planck's constant
lamda=1*10**-10         # wavelength of electron and proton

#Calculations:

#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength
Ee1=h**2/(2*me*lamda**2)           #Energy of electron in joules
Ee=Ee1/(1.6*10**-19)               #Energy of electron in electron-Volts
print"Energy of electron is =",Ee,"eV"

Ep1=h**2/(2*mp*lamda**2)      #Energy of photon in joules
Ep=Ep1/(1.6*10**-19)          #Energy of photon in electron-Volts
print"Energy of photon is =",Ep,"eV"
Energy of electron is = 150.590912522 eV
Energy of photon is = 0.0826254235693 eV

Example 5.11,Page number 5-29¶

In :
import math

#Given Data:
m1=50*10**-9         #mass of particle in kg
m2=9.1*10**-31       #mass of electron in kg
h=6.625*10**-34      #Planck's constant
v1=1                #velocity of particle
v2=3*10**6           #velocity of electron

#Calculations:
lamda1=h/(m1*v1)*10**10        #de Broglie wavelength
print"de Broglie wavelength associated with particle is =",lamda1,"Angstrom"

lamda2=h/(m2*v2)*10**10        #de Broglie wavelength
print"de Broglie wavelength associated with electron is =",lamda2,"Angstrom"

print"Wavelength of electron is measurable."
de Broglie wavelength associated with particle is = 1.325e-16 Angstrom
de Broglie wavelength associated with electron is = 2.42673992674 Angstrom
Wavelength of electron is measurable.

Example 5.12,Page number 5-29¶

In :
import math

#Given Data:
me=9.1*10**-31           #mass of electron in kg
h=6.63*10**-34          #Planck's constant

#Calculations:

E1=2*10**3              #Energy in eV of electron
E=E1*(1.6*10**-19)      #Energy in joules

lamda=h/math.sqrt(2*me*E)      #wavelength of electron
print"Wavelength of electron is =",lamda,"m"
Wavelength of electron is = 2.7472794985e-11 m

Example 5.13,Page number 5-30¶

In :
import math

#Given Data:
me=9.1*10**-31            #mass of electron
h=6.63*10**-34            #Planck's constant
lamda=2*10**-10           #wavelength of electron and photon

#Calculations:
p1=h/lamda                #momentum of electron
print"Momentum of electron is =",p1,"kg.m/s"

Ee=p1**2/(2*me)          #Energy of electron in joules
print"Energy of electron is =",Ee,"Joules"

p2=h/lamda                #momentum of photon
print"Momentum of photon is =",p2,"kg.m/s"

c=3*10**8                #Velocity of light
Ep=h*c/lamda             #Energy of photon in joules
print"Energy of photon is =",Ep,"Joules"
Momentum of electron is = 3.315e-24 kg.m/s
Energy of electron is = 6.03803571429e-18 Joules
Momentum of photon is = 3.315e-24 kg.m/s
Energy of photon is = 9.945e-16 Joules

Example 5.14,Page number 5-31¶

In :
import math

#Given Data:
m=1.676*10**-27            #mass of neutron
h=6.625*10**-34            #Planck's constant
lamda=1*10**-10            #wavelength of neutron

#Calculations:
C=3*10**8                 #Velocity of light
Ep1=h*C/lamda             #Energy of photon in joules
E1=Ep1/(1.6*10**-19)      #Energy of photon in electron-Volts
print"Energy of photon is =",E1,"eV"

#we know,  lamda=h/sqrt(2*m*E)     #de Broglie wavelength
En1=h**2/(2*m*lamda**2)     #Energy of neutron in joules
E2=En1/(1.6*10**-19)     #Energy of neutron in electron-Volts
print"Energy of neutron is =",E2,"eV"

R=E1/E2                 #Ratio of energies of proton to neutron
print"Ratio of energies of proton to neutron is =",R
Energy of photon is = 12421.875 eV
Energy of neutron is = 0.0818366367094 eV
Ratio of energies of proton to neutron is = 151788.679245

Example 5.14.1,Page number 5-36¶

In :
import math

#Given Data:
v=900                  #velocity of electron in m/s
delv=v*0.001/100       #uncertainity in velocity
h=6.63*10**-34         #Planck's constant
m=9.1*10**-31           #mass of an electron

#Calculations:
delp=m*delv            #uncertainity in the measured values of momentum

#using heisenberg's uncertainity formula
delx=h/(2*3.142*delp)    #uncertainity in its position
print"Uncertainity with which position of electron can be located is >=",delx,"m"
Uncertainity with which position of electron can be located is >= 0.0128823012337 m

Example 5.14.2,Page number 5-37¶

In :
import math

#Given Data:
m=1.6*10**-27        #mass of proton in kg
h=6.63*10**-34       #Planck's constant
v=3./20*10**8         #velocity of particle

#Calculations:
lamda=h/(m*v)        #de Broglie wavelength
print"de Broglie wavelength associated with proton is =",lamda,"m"
de Broglie wavelength associated with proton is = 2.7625e-14 m

Example 5.14.3,Page number 5-37¶

In :
import math

#Given Data:
m=1.676*10**-27        #mass of neutron
h=6.634*10**-34        #Planck's constant

#Calculations:
E1=0.025              #Energy in eV of neutron
E=E1*(1.6*10**-19)    #Energy in joules
#As E=m*v**2/2
v=math.sqrt(2*E/m)        #Velocity of neutron beam

lamda=h/(m*v)          #wavelength of a beam of neutron
print"wavelength of a beam of neutron is =",lamda,"m"
wavelength of a beam of neutron is = 1.81172996152e-10 m

Example 5.14.4,Page number 5-37¶

In :
import math

#Given Data:
delx=10*10**-9          #uncertainity in position of electron
h=6.63*10**-34          #Planck's constant
m=9.1*10**-31           #mass of an electron
E=10**3*1.6*10**-19      #Energy of electron in joules

#Calculations:
p=math.sqrt(2*m*E)          #momentum of electron
#using heisenberg's uncertainity formula
delp=h/(2*math.pi*delx)    #uncertainity in the momentum

P=delp/p*100           #percentage of uncertainity in momentum
print"Percentage of uncertainity in momentum of electron is =",P,"percent"
Percentage of uncertainity in momentum of electron is = 0.0618355139385 percent

Example 5.15,Page number 5-31¶

In :
import math

#Given Data:
m=1.676*10**-27          #mass of neutron
h=6.63*10**-34           #Planck's constant
lamda=2*10**-12            #wavelength of neutron
c=3*10**8                #Velocity of light

#Calculations:
p=h/lamda                 #momentum of neutron
KE=p**2/(2*m)            #Kinetic Energy of neutron in joules
print"Kinetic Energy of electron is =",KE,"Joules"

#velocity of particle is same as group velocity. Thus,
vg=p/m                  #group velocity
print"group velocity of neutron is =",vg,"m/s"

#using, vg*vp=c**2
vp=c**2/vg               #phase velocity
print" phase velocity of neutron is =",vp,"m/s"
Kinetic Energy of electron is = 3.27840841289e-17 Joules
group velocity of neutron is = 197792.362768 m/s
phase velocity of neutron is = 4.55022624434e+11 m/s

Example 5.16,Page number 5-32¶

In :
import math

#Given Data:
m=1.157*10**-30         #mass of particle in kg
h=6.63*10**-34          #Planck's constant
c=3*10**8               #Velocity of light

#Calculations:
E1=80                  #Energy in eV of particle
E=E1*(1.6*10**-19)     #Energy in joules

lamda=h/math.sqrt(2*m*E)     #wavelength of particle
print"Wavelength of particle is =",lamda,"m"

#Now,
vg=h/(lamda*m)          #group velocity
print"Group velocity of particle is =",vg,"m/s"

#using, vg*vp=c**2
vp=c**2/vg              #phase velocity
print"Phase velocity of particle is =",vp,"m/s"
Wavelength of particle is = 1.21822320075e-10 m
Group velocity of particle is = 4703848.2563 m/s
Phase velocity of particle is = 19133270270.7 m/s

Example 5.17,Page number 5-33¶

In :
import math

#Given Data:
v=400                #velocity of electron in m/s
delv=0.01/100       #uncertainity in velocity
h=6.63*10**-34       #Planck's constant
m=9.11*10**-31       #mass of an electron

#Calculations:
p=m*v                #momentum of an electron
delp=p*delv          #uncertainity in the measured values of momentum

#using heisenberg's uncertainity formula
delx=h/(2*math.pi*delp)    #accuracy in its position
print"Accuracy in its position is >=",delx,"m"
Accuracy in its position is >= 0.00289571150576 m

Example 5.18,Page number 5-33¶

In :
import math
#Given Data:
delx=10**-8             #maximum uncertainity in position of electron
h=6.63*10**-34          #Planck's constant
m=9.1*10**-31           #mass of an electron

#Calculations:
#using heisenberg's uncertainity formula
delp=h/(2*math.pi*delx)    #minimum uncertainity in the measured values of momentum

delv=delp/m            #minimum uncertainity in the velocity of an electron
print"Minimum uncertainity in the velocity of an electron is =",delv,"m/s"
Minimum uncertainity in the velocity of an electron is = 11595.5744253 m/s

Example 5.19,Page number 5-34¶

In :
import math

#Given Data:
delv=2*10**4            #uncertainity in velocity
h=6.63*10**-34          #Planck's constant
m=9.1*10**-31           #mass of an electron

#Calculations:
delp=m*delv             #uncertainity in the measured values of momentum

#using heisenberg's uncertainity formula
delx=h/(2*math.pi*delp)    #accuracy in its position
print"Minimum space required by electron to be confined in an atom is >=",delx,"m"
Minimum space required by electron to be confined in an atom is >= 5.79778721263e-09 m

Example 5.20,Page number 5-34¶

In :
import math

#Given Data:
delt=1.4*10**-10        #uncertainity in time spent by nucleus in excited state
h=6.63*10**-34          #Planck's constant

#Calculations:

#using, delE*delt>= h/(2*math.pi)
delE1= h/(2*math.pi*delt)       #uncertaininty in its energy in excited state in joules
delE=delE1/(1.6*10**-19)     #uncertaininty in its energy in excited state in eV
print"Uncertaininty in its energy in excited state is >=",delE,"eV"
Uncertaininty in its energy in excited state is >= 4.71070211026e-06 eV

Example 5.21,Page number 5-35¶

In :
import math

#Given Data:
a=2*10**-10             #width of potential well in m
h=6.63*10**-34          #Planck's constant
m=9.1*10**-31           #mass of an electron

#Calculations:
#we know equation for energy of an electron
n0=1
E01=n0**2*h**2/(8*m*a**2)   #Energy in ground state
E0=E01/(1.6*10**-19)      #Energy in eV
print"Energy of an electron in ground state is=",E0,"eV"

n1=2
E11=n1**2*h**2/(8*m*a**2)  #Energy in first excited state
E1=E11/(1.6*10**-19)      #Energy in eV
print" Energy of an electron in first excited state is=",E1,"eV"

n2=3
E21=n2**2*h**2/(8*m*a**2)   #Energy in second excited state
E2=E21/(1.6*10**-19)      #Energy in eV
print"Energy of an electron in second excited state is=",E2,"eV"
Energy of an electron in ground state is= 9.43443080357 eV
Energy of an electron in first excited state is= 37.7377232143 eV
Energy of an electron in second excited state is= 84.9098772321 eV

Example 5.22,Page number 5-36¶

In :
import math

#Given Data:
a=25*10**-10               #width of well
delx=5*10**-10             #uncertainity in position of particle
n=1                        #ground state

#calculation:
x1=a/2
psi1=math.sqrt(2/a)*math.sin(n*math.pi/a*x1)
P1=(psi1**2)*delx            #Probability of finding particle at distance of x1
print"Probability of finding particle at a distance of x1 is =",P1

x2=a/3
psi2=math.sqrt(2/a)*math.sin(n*math.pi/a*x2)
P2=(psi2**2)*delx             #Probability of finding particle at distance of x2
print"Probability of finding particle at a distance of x2 is =",P2
print"(There is print mistake in book)."

x3=a
psi3=math.sqrt(2/a)*math.sin(n*math.pi/a*x3)
P3=(psi3**2)*delx             #Probability of finding particle at distance of x3
print"Probability of finding particle at a distance of x3 is =",P3
Probability of finding particle at a distance of x1 is = 0.4
Probability of finding particle at a distance of x2 is = 0.3
(There is print mistake in book).
Probability of finding particle at a distance of x3 is = 5.99903913065e-33
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