Chemistry 362
Dr. Jean M. Standard
Problem Set 4 Solutions
1.
The two-dimensional particle in a box model can be used to estimate the energy levels of the porphyrin
molecule shown below.
N
H
N
N
H
N
a
a
Assume that the two-dimensional box is a square of approximately 7 Å on each side. The conjugated
portion of the porphyrin molecule consists of 18 pi electrons. Calculate the energy of the highest occupied
molecular orbital (HOMO) and the lowest-unoccupied molecular orbital (LUMO) in Joules. Calculate
the wavelength in nanometers of a transition from the HOMO to the LUMO for the porphyrin molecule.
Treating the porphyrin molecule as a two-dimensional box gives energy levels of the form:
En x ,n y =
n y2 h 2
n x2 h 2
+
.
8mL2
8mL2
where L is the length of one side. The energy level diagram for this two-dimensional box would look
€
something like that shown below.
(3,4) (4,3)
(2,4) (4,2)
(3,3)
(1,4)
(4,1)
(2,3)
(3,2)
(1,3)
(3,1)
(2,2)
(1,2) (2,1)
(1,1)
(nx, ny)
2
1.
continued
With 18 pi electrons, the HOMO corresponds to the (1,4) or (4,1) state; these are degenerate in energy.
Substituting n x = 1 and n y = 4 (or n x = 4 and n y = 1 ) into the energy expression and using
L = 7.0 Å = 7.0 ×10 −10 m , the energy of the HOMO is
€
€
€
€
€
12 h 2
E HOMO =
8mL2
=
42h2
8mL2
+
(
)
17 6.62618 × 10 −34 Js
(
2
)(
)
8 9.10953 × 10 −31 kg 7.0 × 10 −10 m
2
E HOMO = 2.0902 × 10 −18 J .
From the energy diagram, the LUMO corresponds to the (3,3) level. Substituting n x = 3 and n y = 3 into the
€ using L = 7.0 Å = 7.0 ×10 −10 m , the energy of the LUMO is
energy expression and
ELUMO =
€
=
32 h 2
8mL2
32 h 2
8mL2
+
(
)
18 6.62618 × 10 −34 Js
(
€
€
2
)(
)
8 9.10953 × 10 −31 kg 7.0 × 10 −10 m
2
ELUMO = 2.2132 × 10 −18 J .
For a transition from the HOMO to the LUMO, the energy difference ΔE is ΔE = ELUMO − E HOMO . A photon
€
with an energy corresponding
to ΔE would have a frequency given by E photon = ΔE = hν . Since, for light,
λ
λν = c , we can substitute ν = , and solve for the wavelength to give
c
€
€
€
€
λ =
€
€
hc
hc
=
.
ΔE
ELUMO − E HOMO
Inserting numerical values leads to
€
λ =
(6.62618 × 10
(2.2132 × 10
)(
−34
Js 2.99793× 10 8 ms −1
−18
J − 2.0902 × 10 −18 J
)
)
λ = 1.62 ×10 −6 m or λ = 1620 nm.
€
This wavelength is actually too long to appear in a visible spectrum of porphyrin, which suggests either that the
€ to adequately represent€the system or that there is some other transition besides the
particle in a box model fails
HOMO → LUMO transition that corresponds to the visible spectrum.
3
2.
For the two-dimensional model of porphyrin described in problem 1, determine the wavelength in
nanometers of a transition from the HOMO to the LUMO+1 for the porphyrin molecule. Which
transition, the HOMO → LUMO or HOMO → LUMO+1 falls in the visible range of the spectrum?
The energy of the HOMO is the same as in problem 5, so the only energy that we need to calculate is the
LUMO+1 energy. From the energy diagram given in problem 1, the LUMO+1 corresponds to the (2,4) or (4,2)
−10
level. Substituting n x = 2 and n y = 4 into the energy expression and using L = 7.0 Å = 7.0 ×10 m , the
energy of the LUMO+1 is
€
€
ELUMO +1 =
=
22 h 2
8mL2
42 h2
8mL2
+
€
(
)
20 6.62618 × 10 −34 Js
(
2
)(
)
8 9.10953 × 10 −31 kg 7.0 × 10 −10 m
2
ELUMO +1 = 2.4591 × 10 −18 J .
For a transition from the HOMO to the LUMO+1, the energy difference ΔE is ΔE = ELUMO +1 − E HOMO . A
€ corresponding to ΔE would have a frequency given by E photon = ΔE = hν . Since, for
photon with an energy
λ
light, λν = c , we can substitute ν = , and solve for the wavelength to give
c
€
€
€
€
λ =
€
€
hc
hc
=
.
ΔE
ELUMO +1 − E HOMO
Inserting numerical values leads to
€
(6.62618 × 10
(2.4591× 10
−34
λ =
−18
)(
Js 2.99793× 10 8 ms −1
J − 2.0902 × 10
−18
J
)
)
λ = 5.39 ×10 −7 m or λ = 539 nm.
€
The wavelength of the HOMO → LUMO+1 transition does correspond to a transition in the visible region of
€
the spectrum. The HOMO€→ LUMO transition calculated
in problem 5 does not.
4
3.
Consider an electron trapped in a three-dimensional square box of length L on each side.
a.
By analogy with the two-dimensional case, give the equation for the quantized energy levels of the
particle in a 3D box. What is the form of the wavefunction?
By analogy with the 2D case, we will need three quantum numbers (one for each dimension, nx , ny , and
nz ) to describe the energy levels of the particle in a 3D box,
Enx ,ny ,nz = ny2 h 2
nz2 h 2
nx2 h 2
+ + ,
8mL2
8mL2
8mL2
where L is the length of one side. Similarly, the wavefunction will be represented as a product of three
one-dimensional terms,
!n πy$
!n πz$
! 2 $3/2 ! n π x $
ψnx ,ny ,nz (x, y, z) = # & sin # x & sin # y & sin # z & .
"L%
" L %
" L %
" L %
b.
What is the expression for the energy of the ground state of the electron in the 3D box?
From part (a), the energy levels of the particle in a 3D box are
Enx ,ny ,nz = ny2 h 2
nz2 h 2
nx2 h 2
+ + .
8mL2
8mL2
8mL2
The lowest quantum number in each direction is 1, so substituting nx = 1 , ny = 1 , and nz = 1 into the
expression yields the ground state energy, E111 ,
E111 = 12 h 2
12 h 2
12 h 2
+ + ,
8mL2
8mL2
8mL2
or E111 = c.
3h 2
.
8mL2
Give the quantum numbers for the first excited state of the electron in a 3D box. What is the
degeneracy of the first excited state?
The first excited state would correspond to increasing only one of the three quantum numbers from 1 to 2.
There are three possible ways to do this: ( nx , ny , nz ) = (1, 1, 2 ) or (1, 2, 1) or ( 2, 1, 1) . Thus, we say
that the degeneracy of the first excited state is three.
5
3.
continued
d.
What are the quantum numbers for the second and third excited states?
In order to determine the ordering of the higher excited states, you have to check the sum of the squares of
the quantum numbers to determine the total energies. The energy diagram is shown below. The second
and third excited states are both triply degenerate. For the second excited state, the quantum numbers are
9h 2
(nx , ny, nz ) = (2, 2, 1) or (2, 1, 2) or (1, 2, 2) , with an energy of E221 = 8mL2 . The third excited state
11h 2
has quantum numbers ( nx , ny , nz ) = (3, 1, 1) or (1, 3, 1) or (1, 1, 3) , and an energy of E311 = .
8mL2
(2,2,2)
(3,1,1) (1,3,1) (1,1,3)
(2,2,1) (2,1,2) (1,2,2)
(2,1,1) (1,2,1) (1,1,2)
(1,1,1)
(nx, ny, nz)
4.
pˆ x ] . Hint: to evaluate the commutator, it is easiest to consider the effect
of the commutator operating on a function, such as Aˆ , Bˆ f (x) .
Evaluate the commutator
[ xˆ ,
[
[ xˆ ,
€
]
pˆ x ] f (x) = ( xˆpˆ x − pˆ x xˆ ) f (x)
€= #%−i!x d + i! d x &( f (x)
$
dx
dx '
# d
d &
= − i! % x
−
x ( f (x)
$ dx
dx '
#
&
d
= − i! % x f )(x) −
( x f (x)) (
$
'
dx
= − i! ( x f )(x) − f (x) − x f )(x))
[ xˆ ,
pˆ x ] f (x) = i!f (x)
Therefore, we see that the commutator
€
[ xˆ ,
pˆ x
]
is given by
[ xˆ ,
€
€
pˆ x ] = i! .
6
5.
−ikx
A particle traveling in the negative x direction has a wavefunction given by ψ (x) = e .
a.
Show that this wavefunction is an eigenfunction of the momentum operator and thus that the
€
momentum is known exactly.
−ikx
We can show that the wavefunction ψ (x) = e
is an eigenfunction of the momentum operator by
applying the operator,
d −ikx
e
dx
= (−i!)(−ik)e −ikx
€
pˆ xψ (x) = − i!
= −!ke −ikx
pˆ xψ (x) = − !kψ (x) .
−ikx
Therefore, ψ (x) = e
is an eigenfunction of pˆ x and the eigenvalue is −!k . Thus, the momentum of the
system is the eigenvalue −!k . €
€
b.
€
€
€
What is the uncertainty
in the momentum, Δpx , of the particle?
−ikx
Since the wavefunction ψ (x) = e
is an eigenfunction of pˆ x , the momentum is the eigenvalue −!k
€
exactly. Therefore, there is no uncertainty
in the momentum – we know it exactly,
€
c.
Δp
€x = 0 .
€
€ from part (b), use the Heisenberg Uncertainty Principle to
Given the uncertainty in the momentum
determine the uncertainty in the position, Δx .
!
From the Heisenberg Uncertainty Principle, Δpx Δx ≥
. Solving for Δx ,
€
2
€
Δx ≥
!
.€
2Δpx
From part (b), we have Δpx = 0 . Substituting,
€
Δx ≥
€
!
, or Δx = ∞ .
0
This result tells us that if we know the momentum
€ exactly, then we know nothing about the position. It is
€
completely uncertain.
7
6.
Calculate the uncertainty in the position of a baseball of mass 100 grams traveling at 90 mph if the
velocity is known to 0.1 mph.
To calculate the momentum, we can use the equation px = mv . The velocity in m/s is
" 1hr %" 1609 m %
v = ( 90 mph)$
'$
'
€
# 3600s &# 1mi &
v = 40.2 m/s .
Then, the momentum can be calculated as
€
px = mv
= ( 0.1kg)( 40.2 m/s)
px = 4.02 kg m/s .
The uncertainty in the velocity is
€
Δv = 0.1mph
= 0.0447 m/s .
Assuming there is no uncertainty in m (or that the uncertainty is very small compared to the uncertainty in px ),
€
we get the relation
Δpx = mΔv .
€
Substituting the mass along with uncertainty in the velocity into this relation, we get the uncertainty in the
momentum,
€
Δpx = mΔv
= ( 0.1kg)( 0.0447 m/s)
Δpx = 0.0045 kg m/s .
Solving the Heisenberg Uncertainty Principle for Δx ,
€
Δx ≥
€
!
.
2Δpx
Substituting the uncertainty in momentum calculated above,
€
Δx ≥
(1.05459 × 10
−34
)
Js
2( 0.0045 kg m/s)
Δx ≥ 1.2 × 10 −32 m .
This is an extremely small uncertainty in the position. Thus, we see that for a macroscopic particle like a
€ Principle has negligible effect.
baseball, the Heisenberg Uncertainty
8
7.
Assume that an electron is confined inside a one-dimensional box of width 1 Å. Estimate the uncertainty
in position of the electron from the probability distribution for an electron in its ground state. Next,
determine the energy of the electron in its ground state and calculate the magnitude of the electron
p2
momentum using the relation E =
(this is valid because the potential energy is zero inside the box).
2m
Finally, use the Heisenberg Uncertainty Principle to obtain the uncertainty in the momentum.
€
We can estimate the uncertainty
in the position to be roughly half the width of the box,
Δx = 5.0 ×10 −11 m .
This is just an estimate; the same qualitative results would be obtained if the full width was used, or even some
€ idea is to get the correct order of magnitude.
other fraction of the box width. The
The ground state energy E1 for the particle in a box is
E1 =
€
h2
.
8mL2
Substituting, for an electron in a box with a width of 1 Å, the ground state energy is
€
E1 =
h2
8mL2
−34
=
2
(6.62618 × 10 Js)
8 ( 9.10953× 10 kg)(1.0 × 10
−31
−10
)
m
2
E1 = 6.025× 10 −18 J .
To determine the momentum of the particle, we use
€
E = T =
px2
,
2m
since V=0 inside the box for the particle. Solving for the momentum, we have
€
px =
2mE .
Substituting the ground state energy from above, the momentum of the particle is
€
px =
2mE
[(
)(
= 2 9.10953× 10 −31 kg 6.025× 10 −18 J
px = 3.31× 10 −24 kg m/s .
€
1/ 2
)]
9
7. continued
Finally, to estimate the uncertainty in the momentum, we have from the Heisenberg Uncertainty Principle,
!
2Δx
Δpx ≥
(1.05459 × 10 Js)
2 ( 5.0 × 10 m)
−34
≥
−11
Δpx ≥ 1.1× 10 −24 kg m/s
€
a.
What proportion is the uncertainty in the momentum relative to the magnitude of the momentum?
The proportion is given as
Δpx
1.1× 10 −24 kg m/s
=
px
3.3× 10 −24 kg m/s
Δpx
= 0.33.
px
Thus, we see that the uncertainty in the momentum is about 30% of the momentum. We took Δx to be
fairly large (50% of the width€of the box), so Δpx is smaller, only about 30% of the magnitude of the
momentum. If Δx were estimated to be a little smaller, then Δpx would be larger, and vice versa.
€
€
€
b.
€
How does the uncertainty in the momentum and the proportion relative to the magnitude of the
momentum change if the box size is increased to 10 Å?
If the box size is increased to 10 Å, then the estimate for the uncertainty in position becomes
Δx = 5.0 ×10 −10 m .
For a larger box, the ground state energy E1 is
€
h2
E1 =
€
8ma 2
−34
=
E1 = 6.025 × 10 −20 J.
€
2
(6.62618 × 10 Js)
8 ( 9.10953 × 10 kg)(1.0 × 10 m)
−31
−9
2
10
7
b.
continued
The momentum of the particle then becomes
px =
2mE
[(
)(
= 2 9.10953× 10 −31 kg 6.025× 10 −20 J
1/ 2
)]
px = 3.31× 10 −25 kg m/s .
Finally, to estimate the uncertainty in the momentum, we have from the Heisenberg Uncertainty Principle,
€
!
Δpx ≥
2Δx
(1.05459 × 10 Js)
2 ( 5.0 × 10 m)
−34
≥
−10
Δpx ≥ 1.1× 10 −25 kg m/s .
The proportion is given as
€
Δpx
1.1× 10 −25 kg m/s
=
px
3.3× 10 −25 kg m/s
Δpx
= 0.33.
px
Thus, we see that the uncertainty in the momentum is once again about 30% of the momentum for this
€
quantum mechanical particle constrained
to a relatively small region of space.