# Chapter 5:Principles of Quantum Mechanics¶

## Example 5.1 , Page no:102¶

In [1]:
import math
from __future__ import division

#given
lambda1=1; #in Angstrom (wavelength)
m=1.67E-27; #in Kg (mass of neutron)
h=6.625E-34; #in J-s (Planck's constant)
e=1.6E-19; #in C (charge of electron)

#calculate
lambda2=lambda1*1E-10; #since lambda is in Angstrom
#Since lambda=h/(m*v)
#Therefore we have
v=h/(m*lambda2); #calculation of velocity
K=m*v**2/2; #calculation of kinetic energy
K1=K/e; #changing unit fro J to eV

#result
print"The velocity is v=",round(v,3),"m/s";
print"The kinetic energy is K=",'%.3E'%K,"J";
print"\t\t       =",'%.3E'%K1,"eV";

The velocity is v= 3967.066 m/s
The kinetic energy is K= 1.314E-20 J
= 8.213E-02 eV


## Example 5.2 , Page no:103¶

In [2]:
import math
from __future__ import division

#given
K=50; #in eV (Kinetic energy)
m0=9.1E-31; #in Kg (mass of electron)
h=6.625E-34; #in J-s (Planck's constant)
e=1.6E-19; #in C (charge of electron)

#calculate
K=K*e; #changing unit from eV to J
#Since K=m*v^2/2
#Therefore v=sqrt(2*K/m)
#Since lambda=h/(m*v)
#Therefore we have
lambda1=h/math.sqrt(2*m0*K); #calculation of wavelength
lambda2=lambda1*1E10; #changing unit from m to Angstrom

#result
print"The wavelength is =",'%.3E'%lambda1,"m";
print"\t\t  =",round(lambda2,3),"Angstrom";

The wavelength is = 1.736E-10 m
= 1.736 Angstrom


## Example 5.3 , Page no:104¶

In [3]:
import math
from __future__ import division

#given
E=2000; #in eV (Kinetic energy)
m=9.1E-31; #in Kg (mass of electron)
h=6.625E-34; #in J-s (Planck's constant)
e=1.6E-19; #in C (charge of electron)

#calculate
E=E*e; #changing unit from eV to J
#Since E=m*v^2/2
#Therefore v=sqrt(2*E/m)
#Since lambda=h/(m*v)
#Therefore we have
lambda1=h/math.sqrt(2*m*E); #calculation of wavelength
lambda2=lambda1*1E9; #changing unit from m to nanometer

#result
print"The wavelength is =",'%.3E'%lambda1,"m";
print"\t\t  =",round(lambda2,4),"nm";

The wavelength is = 2.745E-11 m
= 0.0275 nm


## Example 5.4 , Page no:104¶

In [4]:
import math
from __future__ import division

#given
m_e=9.1E-31; #in Kg (mass of electron)
m_n=1.676E-27; #in Kg (mass of neutron)
h=6.625E-34; #in J-s (Planck's constant)
c=3E8; #in m/s (velocity of light)

#calculate
E_e=m_e*c**2; #rest mass energy of electron
E_n=2*E_e; #given (kinetic energy of neutron)
#Since K=m*v^2/2
#Therefore v=sqrt(2*K/m)
#Since lambda=h/(m*v)
#Therefore we have
lambda1=h/math.sqrt(2*m_n*E_n); #calculation of wavelength
lambda2=lambda1*1E10; #changing unit from m to Angstrom

#result
print"The wavelength is =",'%.3E'%lambda1,"m";
print"\t\t  =",'%.3E'%lambda2,"Angstrom";

The wavelength is = 2.827E-14 m
= 2.827E-04 Angstrom


## Example 5.5 , Page no:104¶

In [5]:
import math
from __future__ import division

#given
V=1600; #in V (Potential)

#calculate
lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom

#result
print"The wavelength is =",round(lambda1,3),"Angstrom";
print "Note : The answer in the book is wrong "

The wavelength is = 0.307 Angstrom
Note : The answer in the book is wrong


## Example 5.6 , Page no:105¶

In [6]:
import math
from __future__ import division

#given
E=1000; #in eV (Kinetic energy of photon)
K=1000; #in eV (Kinetic energy of electron)
m0=9.1E-31; #in Kg (mass of electron)
h=6.6E-34; #in J-s (Planck's constant)
c=3E8; #in m/s (velocity of light)
e=1.6E-19; #in C (charge on electron)

#calculate
E=E*e; #changing unit from eV to J
lambda_p=h*c/E; #For photon E=hc/lambda
lambda_p1=lambda_p*1E10; #changing unit from m to Angstrom
#Since K=m*v^2/2
#Therefore v=sqrt(2*K/m)
#Since lambda=h/(m*v)
#Therefore we have
K=K*e; #changing unit from eV to J
lambda_e=h/math.sqrt(2*m0*K); #calculation of wavelength
lambda_e1=lambda_e*1E10; #changing unit from m to Angstrom

#result
print"For photon,the wavelength is =",'%.3E'%lambda_p,"m";
print"\t\t\t     =",round(lambda_p1,1),"Angstrom";
print"For electron,the wavelength is =",'%.3E'%lambda_e,"m";
print"\t\t\t       =",round(lambda_e1,2),"Angstrom";

For photon,the wavelength is = 1.238E-09 m
= 12.4 Angstrom
For electron,the wavelength is = 3.868E-11 m
= 0.39 Angstrom


## Example 5.7 , Page no:105¶

In [7]:
import math
from __future__ import division

#given
lambda1=1.66*10**-10; #in m (wavelength)
m=9.1*10**-31; #in Kg (mass of electron)
h=6.626*10**-34; #in J-s (Planck's constant)
e=1.6E-19; #in C (charge on electron)

#calculate
#Since lambda=h/(m*v)
#Therefore we have
v=h/(m*lambda1); #calculation of velocity
K=m*v**2/2; #calculation of kinetic energy
K1=K/e; #changing unit from J to eV

#result
print"The velocity of electron is v=",'%.2E'%v,"m/s";
print"The kinetic energy is K=",'%.2E'%K,"J";
print"\t\t       =",round(K1,3),"eV";
print "Note: The answer in the book for kinetic energy is wrong "

The velocity of electron is v= 4.39E+06 m/s
The kinetic energy is K= 8.75E-18 J
= 54.714 eV
Note: The answer in the book for kinetic energy is wrong


## Example 5.8 , Page no:106¶

In [8]:
import math
from __future__ import division

#given
T=400; #in K (temperature)
m=6.7E-27; #in Kg (mass of He-atom)
h=6.6E-34; #in J-s (Planck's constant)
k=1.376E-23; #in J/degree (Boltzmann constant)

#calculate
#Since lambda=h/(m*v)
#E=mv^2/2;
#Therefore lambda=h/sqrt(2*m*E)
#E=kT
#Therefore lambda=h/sqrt(2*m*k*T)
lambda1=h/math.sqrt(2*m*k*T)
lambda2=lambda1*1E10; #changing unit from m to Angstrom

#result
print"The de-Broglie wavelength of He-atom is =",'%.2E'%lambda1,"m";
print"\t\t\t\t        =",round(lambda2,4),"Angstrom";

The de-Broglie wavelength of He-atom is = 7.69E-11 m
= 0.7685 Angstrom


## Example 5.9 , Page no:106¶

In [9]:
import math
from __future__ import division

#given
m_e=9.1E-31; #in Kg (mass of electron)
m_p=1.6E-27; #in Kg (mass of proton)
h=6.626E-34; #in J-s (Planck's constant)
c=3E8; #in m/s (velocity of light)

#calculate
E=m_e*c**2; #in J (rest energy of electron)
#Since lambda=h/(m*v)
#E=mv^2/2;
#herefore lambda=h/sqrt(2*m*E)
#Also E=m_e*c^2;
#therefore lambda=h/sqrt(2*m_p*m_e*c^2)
lambda1=h/math.sqrt(2*m_p*m_e*c**2); #calculation of wavelength
lambda2=lambda1*1*10**10; #changing unit from m to Angstrom

#result
print"The de-Broglie wavelength of proton is =",'%.2E'%lambda1,"m";
print"\t\t\t\t       =",'%.2E'%lambda2,"Angstrom";

The de-Broglie wavelength of proton is = 4.09E-14 m
= 4.09E-04 Angstrom


## Example 5.10 , Page no:106¶

In [10]:
import math
from __future__ import division

#given
V=10000; #in V (Potential)

#calculate
lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom

#result
print"The wavelength is =",round(lambda1,3),"Angstrom";

The wavelength is = 0.123 Angstrom


## Example 5.11 , Page no:107¶

In [11]:
import math
from __future__ import division

#given
V=100; #in V (Potential)
n=1; #order of diffraction
d=2.15; #in Angstrom (lattice spacing)

#calculate
lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom
#Since 2*d*sind(theta)=n*lambda
#therefore we have
theta=math.asin(n*lambda1/(2*d)); #calculation of glancing angle
theta1=theta*180/3.14;

#result
print"The wavelength is =",round(lambda1,3),"Angstrom";
print"The glancing angle is =",round(theta1,1),"degree";
print "Note: In question V=100 eV but the solution is using V=100V "

The wavelength is = 1.227 Angstrom
The glancing angle is = 16.6 degree
Note: In question V=100 eV but the solution is using V=100V


## Example 5.12 , Page no:107¶

In [12]:
import math
from __future__ import division

#given
V=344; #in V (Potential)
n=1; #order of diffraction
theta=60; #in degree (glancing angle)

#calculate
lambda1=12.27/math.sqrt(V); #calculation of wavelength in Angstrom
#Since 2*d*sind(theta)=n*lambda
#therefore we have

#result
print"The wavelength is =",round(lambda1,3),"Angstrom";
print"The spacing of the crystal is =",round(d,4),"Angstrom";

The wavelength is = 0.662 Angstrom
The spacing of the crystal is = 0.3819 Angstrom


## Example 5.13 , Page no:107¶

In [13]:
import math
from __future__ import division

#given
r=0.53*10**-10; #in m (radius of first Bohr orbit)
h=6.6*10**-34; #in J-s (Planck's constant)
m=9.1*10**-31; #in Kg (mass of electron)
n=1; #First Bohr orbit
pi=3.14; #value of pi used in the solution

#calculate
#Since 2*pi*r=n*lambda  and  lambda=h/(m*v)
#Threfore we have v=h*n/(2*pi*r*m)
v=h*n/(2*pi*r*m); #calculation of velocity

#result
print"The velocity of electron is =",'%.3E'%v,"m/s";

The velocity of electron is = 2.179E+06 m/s


## Example 5.14 , Page no:108¶

In [14]:
import math
from __future__ import division

#given
dx=0.2; #in Angstrom (uncertainty in the position)
h=6.6E-34; #in J-s (Planck's constant)
m0=9.1E-31; #in Kg (mass of electron)
pi=3.14; #value of pi used in the solution

#calculate
dx=dx*1E-10; #since dx is in Angstrom
#Since dx*dp=h/4*pi  (uncertainty relation)
dp=h/(4*pi*dx); #calculation of uncertainty in the momentum
#since dp=m*dv
dv=dp/m0; #calculation of uncertainty in the velocity

#result
print"The uncertainty in the momentum is dp=",'%.3E'%dp,"Kg-m/";
print"The uncertainty in the velocity is dv=",'%.3E'%dv,"m/s";

The uncertainty in the momentum is dp= 2.627E-24 Kg-m/
The uncertainty in the velocity is dv= 2.887E+06 m/s


## Example 5.15 , Page no:108¶

In [15]:
import math
from __future__ import division

#given
m_e=9.1E-31; #in Kg (mass of electron)
m_p=1.67E-27; #in Kg (mass of proton)
dx_p=1; #in nanometer (uncertainty in position of electron)
dx_n=1; #in nanometer (uncertainty in position of proton)

#calculate
#since dp=h/(4*pi*dx)
#since h/(4*pi) is constant and dx is same for electron and proton
#therefor both electron and proton have same uncertainty in the momentum
#since dv=dp/m  and  dp is same for both
#therefore dv_e/dv_p=m_p/m_e
#therefore
K=m_p/m_e; #ratio of uncertainty in the velocity of electron and proton

#result
print"The ratio of uncertainty in the velocity of electron to that of proton is =",round(K);

The ratio of uncertainty in the velocity of electron to that of proton is = 1835.0


## Example 5.16 , Page no:108¶

In [16]:
import math
from __future__ import division

#given
dx=5*10**-15; #in m (radius of nucleus or uncertainty in the position)
h=6.6*10**-34; #in J-s (Planck's constant)
m=1.67E-27; #in Kg (mass of proton)
pi=3.14; #value of pi used in the solution
e=1.6E-19; #in C (charge of electron)

#calculate
#Since dx*dp=h/4*pi  (uncertainty relation)
dp=h/(4*pi*dx); #calculation of uncertainty in the momentum
p=dp; #minimum value of momentum to calculate mimimum kinetic energy
K=(p**2)/(2*m); #calculation of minimum kinetic energy of proton
K1=K/e; #changing unit from J to eV
K2=K1/1E6; #changing unit from eV to MeV

#result
print"The minimum uncertainty in the momentum of proton is dp =",'%.3E'%dp,"Kg-m/s";
print"The minimum kinetic energy of proton is K=",'%.3E'%K,"J";
print"\t\t\t\t \t =",'%.3E'%K1,"eV";
print"\t\t\t\t\t =",round(K2,1),"MeV";

The minimum uncertainty in the momentum of proton is dp = 1.051E-20 Kg-m/s
The minimum kinetic energy of proton is K= 3.307E-14 J
= 2.067E+05 eV
= 0.2 MeV


## Example 5.17 , Page no:109¶

In [17]:
import math
from __future__ import division

#given
K=1; #in KeV (kinetic energy of electron)
dx=1; #in Angstrom (uncertainty in the position)
h=6.63E-34; #in J-s (Planck's constant)
m=9.1E-31; #in Kg (mass of electron)
pi=3.14; #value of pi used in the solution
e=1.6E-19; #in C (charge of electron)

#calculate
dx=dx*1E-10; #since dx is in Angstrom
#Since dx*dp=h/4*pi  (uncertainty relation)
dp=h/(4*pi*dx); #calculation of uncertainty in the momentum
K=K*1E3*1.6E-19; #changing unit from KeV to J
p=math.sqrt(2*m*K); #calculation of momentum
poc=(dp/p)*100; #calculation of percentage of uncertainty

#result
print"The uncertainty in the momentum of electron is dp=",'%.3E'%dp,"Kg-m/s";
print"The momentum of electron is p=",'%.3E'%p,"Kg-m/s";
print"The percentage of uncertainty in the momentum  is =",round(poc,1);

The uncertainty in the momentum of electron is dp= 5.279E-25 Kg-m/s
The momentum of electron is p= 1.706E-23 Kg-m/s
The percentage of uncertainty in the momentum  is = 3.1


## Example 5.18 , Page no:109¶

In [18]:
import math
from __future__ import division

#given
v=6.6E4; #m/s (speed of electron)
poc=0.01; #percentage of uncertainty
h=6.63E-34; #in J-s (Planck's constant)
m=9E-31; #in Kg (mass of electron)
pi=3.14; #value of pi used in the solution

#calculate
p=m*v; #calculation of momentum
dp=(poc/100)*p; #calculation of uncertainty in the momentum
#Since dx*dp=h/4*pi  (uncertainty relation)
dx=h/(4*pi*dp); #calculation of uncertainty in the position

#result
print"The momentum of electron is p=",'%.3E'%p,"Kg-m/s";
print"The uncertainty in the momentum of electron is dp=",'%.3E'%dp,"Kg-m/s";
print"The uncertainty in the position of electron is dx=",'%.3E'%dx,"Kg-m/s";

The momentum of electron is p= 5.940E-26 Kg-m/s
The uncertainty in the momentum of electron is dp= 5.940E-30 Kg-m/s
The uncertainty in the position of electron is dx= 8.887E-06 Kg-m/s


## Example 5.19 , Page no:111¶

In [19]:
import math
from __future__ import division

#given
lambda1=1; #in Angstrom (wavelength)
pi=3.14; #value of pi used in the solution
dlambda=1E-6; #uncertainty in wavelength

#calculate
lambda2=lambda1*1E-10; #sinc lambda is in Angstrom
#By uncertainty principle, dx*dp>=h/(4*pi)  --(1)
#since p=h/lambda  -----(2)
#Or p*lambda=h
#diffrentiting this equation
#p*dlambda+lambda*dp=0
#dp=-p*dlambda/lambda     ----(3)
#from (2) and (3)  dp=-h*dlambda/lambda^2   ---(4)
#from (1) and(4)  dx*dlambda>=lambda^2/4*pi
#Or dx=lambda^2/(4*pi*dlambda)
dx=lambda2**2/(4*pi*dlambda); #calculation of uncertainty in the position

#result
print"The uncertainty in the position of X-ray photon is dx=",'%.3E'%dx,"m";
print "Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6 "
print " ANSWEER IS WRONG IN THE TEXTBOOK"

The uncertainty in the position of X-ray photon is dx= 7.962E-16 m
Note: wavelength accuracy is given as 1 in 1E8 but in book solution has used 1 in 1E6
ANSWEER IS WRONG IN THE TEXTBOOK


## Example 5.20 , Page no:111¶

In [20]:
import math
from __future__ import division

#given
dt=1*10**-8; #in sec (average life time)
pi=3.14; #value of pi used in the solution

#calculate
#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)
#and E=h*v      v is the frequency
#therefore we have dv>=1/(4*pi*dt)
dv=1/(4*pi*dt); #calculation of minimum uncertainty in the frequency

#result
print"The minimum uncertainty in the frequency of the photon is dv=",'%.3E'%dv,"sec^-1";

The minimum uncertainty in the frequency of the photon is dv= 7.962E+06 sec^-1


## Example 5.21 , Page no:111¶

In [21]:
import math
from __future__ import division

#given
dt=1E-12; #in sec (average life time)
h=6.63E-34; #in J-s (Planck'c constant)
pi=3.14; #value of pi used in the solution
e=1.6*1E-19; #in C (charge of electron)

#calculate
#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)
dE=h/(4*pi*dt); #calculation of minimum uncertainty in the energy
dE1=dE/e; #changing unit from J to eV

#result
print"The uncertainty in the energy of the photon is dE=",'%.3E'%dE,"J";
print"\t\t\t\t\t         =",'%.3E'%dE1,"eV";

The uncertainty in the energy of the photon is dE= 5.279E-23 J
= 3.299E-04 eV


## Example 5.22 , Page no:111¶

In [22]:
import math
from __future__ import division

#given
dT=2.5E-14; #in sec (average life time)
h=6.63E-34; #in J-s (Planck'c constant)
pi=3.14; #value of pi used in the solution
e=1.6*1E-19; #in C (charge of electron)

#calculate
#Since dE*dt>=h/(4*pi)  (uncertainty relation for energy)
dE=h/(4*pi*dT); #calculation of minimum uncertainty in the energy
dE1=dE/e; #changing unit from J to eV

#result
print"The uncertainty in the energy of the photon is dE=",'%.3E'%dE,"J";
print"\t\t\t\t\t         =",'%.3E'%dE1,"eV";

The uncertainty in the energy of the photon is dE= 2.111E-21 J
= 1.320E-02 eV


## Example 5.23 , Page no:112¶

In [23]:
import math
from __future__ import division

#given
a=2; #in Angstrom (length of the box)
m=9.1E-31; #in Kg (mass of electron)
h=6.626E-34; #in J-s (Planck'c constant)
n2=2; n4=4; #two quantum states
e=1.6*1E-19; #in C (charge of electron)

#calculate
a=a*1E-10; #since a is in Angstrom
#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)
E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd quantum state
E21=E2/e; #changing unit from J to eV
E4=n4**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 4nd quantum state
E41=E4/e; #changing unit from J to eV

#result
print"The energy corresponding to the 2nd quantum state is E2=",'%.3E'%E2,"J";
print"\t\t\t\t\t\t       =",'%.3E'%E21,"eV";
print"The energy corresponding to the 4nd quantum state is E4=",'%.3E'%E4,"J";
print"\t\t\t\t\t\t       =",'%.3E'%E41,"eV";

The energy corresponding to the 2nd quantum state is E2= 6.031E-18 J
= 3.769E+01 eV
The energy corresponding to the 4nd quantum state is E4= 2.412E-17 J
= 1.508E+02 eV


## Example 5.24 , Page no:113¶

In [24]:
import math
from __future__ import division

#given
a=1E-10; #in m (width of the well)
m=9.1E-31; #in Kg (mass of electron)
h=6.626E-34; #in J-s (Planck'c constant)
n1=1; n2=2; n3=3; #ground and first two excited states
e=1.6*1E-19; #in C (charge of electron)

#calculate
#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)
E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the Ground state
E11=E1/e; #changing unit from J to eV
E2=n2**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 1st excited state
E21=E2/e; #changing unit from J to eV
E3=n3**2*h**2/(8*m*a**2); #calculation of energy corresponding to the 2nd excited state
E31=E3/e; #changing unit from J to eV

#result
print"The energy corresponding to the ground state is E1=",'%.3E'%E1,"J";
print"\t\t\t\t\t =",'%.3E'%E11,"eV";
print"The energy corresponding to the 1st excited state is E2=",'%.3E'%E2,"";
print"\t\t\t\t\t =",'%.3E'%E21,"eV";
print"The energy corresponding to the 2nd excited state is E3=",'%.3E'%E3,"J";
print"\t\t\t\t\t =",'%.3E'%E31,"eV";
print " There is slight variation in the answer due to round off error"

The energy corresponding to the ground state is E1= 6.031E-18 J
= 3.769E+01 eV
The energy corresponding to the 1st excited state is E2= 2.412E-17
= 1.508E+02 eV
The energy corresponding to the 2nd excited state is E3= 5.428E-17 J
= 3.392E+02 eV
There is slight variation in the answer due to round off error


## Example 5.25 , Page no:113¶

In [25]:
import math
from __future__ import division

#given
dx=1E-8; #in m (length of box or uncertainty in the position)
h=6.626E-34; #in J-s (Planck'c constant)
m=9.1E-31; #in Kg (mass of electron)

#calculate
#From uncertainty principle dx*dp=h and dp=m*dv
#therefore we have
dv=h/(m*dx); #calculation of minimum uncertainty in the velocity
dv1=dv*1E-3; #changing unit from m/s to Km/s

#result
print"The minimum uncertainty in the velocity of electron i dv=",'%.3E'%dv,"m/s";
print"\t\t\t\t\t\t        =",round(dv1,2),"Km/s";
print "Note: There is slight variation in the answer due to round off error"

The minimum uncertainty in the velocity of electron i dv= 7.281E+04 m/s
= 72.81 Km/s
Note: There is slight variation in the answer due to round off error


## Example 5.26 , Page no:113¶

In [26]:
import math
from __future__ import division

#given
a=4E-10; #in m (length of the box)
m=9.1E-31; #in Kg (mass of electron)
h=6.626E-34; #in J-s (Planck'c constant)
n1=1; #ground state
e=1.6*1E-19; #in C (charge of electron)

#calculate
#Since  E_n=n^2*h^2/(8*m*a^2)   (Energy corresponding to nth quantum state)
E1=n1**2*h**2/(8*m*a**2); #calculation of energy corresponding to the ground state
E11=E1/e; #changing unit from J to eV

#result
print"The minimum energy of electron is E1=",'%.3E'%E1,"J (Note: The answer in the book  is wrong due to printing error)";
print"\t\t\t\t    =",round(E11,3),"eV";

The minimum energy of electron is E1= 3.769E-19 J (Note: The answer in the book  is wrong due to printing error)
= 2.356 eV