Particle in 1 D box

4,668 views 10 slides Nov 30, 2020
Slide 1
Slide 1 of 10
Slide 1
1
Slide 2
2
Slide 3
3
Slide 4
4
Slide 5
5
Slide 6
6
Slide 7
7
Slide 8
8
Slide 9
9
Slide 10
10

About This Presentation

Particle in 1D Box, useful for MSc and BSc Chemistry students


Slide Content

Dr. Pradeep Samantaroy
Department of Chemistry
Rayagada Autonomous College, Rayagada
[email protected]; [email protected]
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A kLAsin0 nkL
where n is any integer

Solution to the Zone II
And the general solution to the differential equation is: L
xn
A
II

 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
22

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

E
1 = h
2
/ 8mL
2

E
2 = 4h
2
/ 8mL
2

E
3 = 9h
2
/ 8mL
2

E
4 = 16h
2
/ 8mL
2

Ψ |Ψ
2
| 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
Ψ
n(x) = (2/a)
1/2
sin (nπx/a) where n = 1,2,3…
Prepared by Dr. Pradeep Samantaroy
Tags