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Dr. Pradeep Samantaroy
Department of Chemistry
Rayagada Autonomous College, Rayagada
pksroy82@gmail.com; pksroy82@yahoo.in
9444078968
Understanding the Case
Classical Approach
0 X L
Assume there is no air resistance? What do you expect?
Prepared by Dr. Pradeep Samantaroy
Understanding the Case
Quantum Approach
0 X L
V = 0
V = ∞ V = ∞
Zone I Zone II Zone III
V(x)=0 for L >x >0
V(x)=∞ for x ≥ L, x ≤0
Prepared by Dr. Pradeep Samantaroy
Let’s solve the case…!


ExV
dx
xd
m


)(
)(
2 2
22

The Schrodinger’s equation for the electron in one dimensional box
can be represented as
Applying boundary conditions to Zone I


E
dx
xd
m


*
)(
2 2
22

0
2

Hence, the probability of finding electron in the Zone I is zero.
The same logic can be applied to zone III.
Prepared by Dr. Pradeep Samantaroy


E
dx
xd
m


2
22
)(
2


 2
2
2
)(
k
dx
xd



E
m
dx
xd
22
2
2)(


This is similar to the general differential equation:
kxBkxA cossin 
So we can start applying boundary conditions:
x=0 ψ=0 kBkA 0cos0sin0  0B
x=L ψ=0 0AkLAsin0  nkL 
where n is any integer
Solution to the Zone II
And the general solution to the differential equation is:
L
xn
AII

 sinNow the wave function becomes
Prepared by Dr. Pradeep Samantaroy
How do you find the value of A?
Normalizing wave function
1)sin(
0
2

L
dxkxA
1
4
2sin
2 0
2







L
k
kxx
A 1
4
2sin
2
2












L
n
L
L
n
L
A


Since n is any integer 1
2
2





 L
A
L
A
2

Hence normalized wave function is:
L
xn
L
II

 sin
2

Prepared by Dr. Pradeep Samantaroy
2
2 2

mE
k 
m
k
E
2
22


2
22
42 m
hk
E 
2
2
2
22
42 

m
h
L
n
E 
2
22
8mL
hn
E 
Calculation of Energy Levels
Can n be zero??
Prepared by Dr. Pradeep Samantaroy
E1 = h2/ 8mL2
E2 = 4h2/ 8mL2
E3 = 9h2/ 8mL2
E4 = 16h2/ 8mL2
Ψ |Ψ2| E
The Important Graphs
Prepared by Dr. Pradeep Samantaroy
Let’s solve numericals
An electron is confined in a one-dimensional box of length 2 A°.
Calculate the ground state energy in electron volts.
Is quantization of energy level observable?
Prepared by Dr. Pradeep Samantaroy
Let’s solve numericals
Calculate the average value of the energy of a particle of mass m
confined to move in a one-dimensional box of width a and
infinite height with potential energy zero inside the box. The
normalized wave function of the particle is
Ψn(x) = (2/a)1/2 sin (nπx/a) where n = 1,2,3…
Prepared by Dr. Pradeep Samantaroy

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Dr Pradeep Samantaroy's Quantum Approach to 1D Box

  • 1. Dr. Pradeep Samantaroy Department of Chemistry Rayagada Autonomous College, Rayagada pksroy82@gmail.com; pksroy82@yahoo.in 9444078968
  • 2. Understanding the Case Classical Approach 0 X L Assume there is no air resistance? What do you expect? Prepared by Dr. Pradeep Samantaroy
  • 3. Understanding the Case Quantum Approach 0 X L V = 0 V = ∞ V = ∞ Zone I Zone II Zone III V(x)=0 for L >x >0 V(x)=∞ for x ≥ L, x ≤0 Prepared by Dr. Pradeep Samantaroy
  • 4. Let’s solve the case…!   ExV dx xd m   )( )( 2 2 22  The Schrodinger’s equation for the electron in one dimensional box can be represented as Applying boundary conditions to Zone I   E dx xd m   * )( 2 2 22  0 2  Hence, the probability of finding electron in the Zone I is zero. The same logic can be applied to zone III. Prepared by Dr. Pradeep Samantaroy
  • 5.   E dx xd m   2 22 )( 2    2 2 2 )( k dx xd    E m dx xd 22 2 2)(   This is similar to the general differential equation: kxBkxA cossin  So we can start applying boundary conditions: x=0 ψ=0 kBkA 0cos0sin0  0B x=L ψ=0 0AkLAsin0  nkL  where n is any integer Solution to the Zone II And the general solution to the differential equation is: L xn AII   sinNow the wave function becomes Prepared by Dr. Pradeep Samantaroy
  • 6. How do you find the value of A? Normalizing wave function 1)sin( 0 2  L dxkxA 1 4 2sin 2 0 2        L k kxx A 1 4 2sin 2 2             L n L L n L A   Since n is any integer 1 2 2       L A L A 2  Hence normalized wave function is: L xn L II   sin 2  Prepared by Dr. Pradeep Samantaroy
  • 7. 2 2 2  mE k  m k E 2 22   2 22 42 m hk E  2 2 2 22 42   m h L n E  2 22 8mL hn E  Calculation of Energy Levels Can n be zero?? Prepared by Dr. Pradeep Samantaroy
  • 8. E1 = h2/ 8mL2 E2 = 4h2/ 8mL2 E3 = 9h2/ 8mL2 E4 = 16h2/ 8mL2 Ψ |Ψ2| E The Important Graphs Prepared by Dr. Pradeep Samantaroy
  • 9. Let’s solve numericals An electron is confined in a one-dimensional box of length 2 A°. Calculate the ground state energy in electron volts. Is quantization of energy level observable? Prepared by Dr. Pradeep Samantaroy
  • 10. Let’s solve numericals Calculate the average value of the energy of a particle of mass m confined to move in a one-dimensional box of width a and infinite height with potential energy zero inside the box. The normalized wave function of the particle is Ψn(x) = (2/a)1/2 sin (nπx/a) where n = 1,2,3… Prepared by Dr. Pradeep Samantaroy