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CHEM106: Assignment 5
Particle in a Box
1. For a one-dimensional particle in a box system, the solution of the Schrödinger
n 2h 2
equation leads to the quantized energy En 
. What is the zero-point energy for
8mL2
the system?
The zero-point energy (ZPE) for a quantum mechanical system is defined as the
lowest possible energy allowed. In the case of the harmonic oscillator, the energy
depends on the quantum number n, and n = 1, 2, 3, ….
12 h 2
h2
Thus,
.
ZPE  E1 

8mL2 8mL2
2. The -electron of a conjugated polyene molecule can be modeled as a particle in a
box. Assuming a length of 0.6 nm for the one-dimensional box,
A. calculate the energy gap between the first excited state and the ground state for
the electron.
For the one-dimensional particle in a box, the energy is given by
n 2h 2
.
En 
8mL2
The energy gap between the first excited state (n = 2) and the ground state
(n = 1) is
DE = E2 - E1
=
2 2 h2
12 h2
8meL2 8meL2
=
3h2
8meL2
3´(6.626 ´10 -34 Js)2
=
8 ´9.1093897 ´10 -31 kg´(6 ´10 -10 m)2
= 5.02 ´10 -19 J.
B. what is the corresponding frequency of the photon absorbed as the electron
makes a transition from the ground state to the first excited state?
 E  h
   E / h  5.02  10 19 J / 6.626  10 34 Js  7.58  1014 Hz
3. For a one-dimensional particle in a box system, the solution of the Schrödinger
2
n
equation generates a wave function n ( x ) 
sin(
x ) for the state n. Verify that
L
L
n (x ) satisfies the Schrödinger equation.
For the one-dimensional particle in a box system, the Hamiltonian operator can be
defined as
2 d 2
2 d 2
ˆ
ˆ
ˆ
.
H  T V  
0
2m dx 2
2m dx 2
When we operate the Hamiltonian on the wave function, we obtain
2
d2 2
np
Ĥ Y n (x) = sin( x)
2
2m dx L
L
=
np 2 2
np
)
sin( x)
2m L
L
L
2
(
n2 h2 2
np
sin( x)
2
8mL L
L
= En Y n (x).
Thus, the given wave function n (x ) satisfies the Schrödinger equation.
=
4. For the ground state of the particle in a box system 1 ( x ) 
2
x
, calculate the
sin
L
L
probability of finding the particle between x = 0 and x = L/2.
[Useful integral:  sin 2 (x )d x 
1
1
x
sin( 2x ) ]
2
4
The probability of finding the particle between x = 0 and x = L/2 is
L/l
L/2
L/2
2



p   * ( x )( x )d x   sin( x ) sin( x )d x   sin 2 ( x )d x .
L 0
L
L
L
0
0
According to the given above,  sin 2 (x )d x 
we have
1
1
x
sin( 2x ) ,
2
4
2 1
L
2p
L/2
[ xsin(
x)] 0
L 2
4p
L
2 L L
2p L
= { [sin(
) - sin(0)]}
L 4 4p
L 2
2L 1
=
= .
L4 2
P=
5. Find out the degeneracies of the lowest four energy levels for the three- dimensional
particle in a box system with the dimension Lx = Ly = Lz.
The energy for the three-dimensional particle in a box system is given by
E  Ex  E y  Ez 
n x2 h 2
8mL2x

n 2y h 2
8mL2y

n z2 h 2
8mL2z
.
If Lx = Ly = Lz, the above expression reduces to
h2
E 
(n x2  n 2y  n z2 ) .
2
8mLx
The degeneracy of an energy level equals to the number of quantum states for the
given energy. The lowest four energy levels and the corresponding degeneracy is
listed below:
Energy Levels
h2
(in units of
)
8mL2x
3
6
9
12
Possible States
nx ny nz
Degeneracy
111
1 2 1, 2 1 1, 1 1 2
2 1 2, 2 2 1, 1 2 2
222
1
3
3
1