Download Test #1 solutions

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Chem 5314 test #1 solutions, out of 40 marks
I want complete, detailed answers to the questions. Show all your work to get full credit.
Data:
Z
2π
sin(nx) cos(nx) dx = 0
(1)
0
The Schrödinger time dependent equation in one dimension is
~2 ∂ 2 ψ(x, t)
∂ψ(x, t)
=−
i~
+ V (x)ψ(x, t)
∂t
2m ∂x2
(2)
The Schrödinger time independent equation in one dimension is
−
~2 d2 ψn (x)
+ V (x)ψn (x) = En ψn (x)
2m dx2
(3)
The particle in a one dimensional box (of size L) energy levels are
En =
n2 π 2 ~2
2mL2
n = 1, 2, 3, ...
(4)
and the wavefunctions for the particle in a one dimensional box are
ψ(x) =
p
2/L sin(nπx/L) x ∈ (0, L)
The particle in a three-dimensional box (of side lengths a, b, c) energy levels are
π 2 ~2 n2x n2y n2z
+ 2 + 2
E=
nx , ny , nz = 1, 2, 3, ...
2m a2
b
c
(5)
(6)
energy levels of the particle on a ring:
En =
n2 ~2
2mR2
n = 0, 1, 2, ...
Speed of light = c = 3.0 × 108 m/s
electron mass = 9.11 × 10−31 kg
mass conversion from amu to kg: 1.66 × 10−27 kg/amu
Avogadro’s constant = NA = 6.022 × 1023 mol−1
Planck constants: h = 6.6262 × 10−34 Js and ~ = 1.05459 × 10−34 Js
π = 3.14159
1 eV = 1.6022×10−19 J (electron volt to joule conversion)
Relationship between the wavelength of a photon and its energy: λ = hc/E
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(7)
Problem 1 – 8 marks
For a quantum mechanical particle in free space (V = 0), the angular frequency ω and
the wave number k of its associated wavefunction are related by
~ω =
~2 k 2
2m
(8)
a) – 5 marks Verify that, if a monochromatic wave of the form ψ = ei(kx−ωt) is substituted
into the Schrödinger time dependent equation (also called the Schrödinger wave equation),
the above relation is reproduced.
Solution:
Taking partial derivatives gives
∂ψ
= −iωψ
∂t
(9)
∂ 2ψ
= −k 2 ψ
2
∂x
(10)
and
Substituting into Schoedinger’s Equation and canceling ψ from both sides and rearranging
gives the desired relationship between k and ω.
b) – 3 marks Show that ψ = cos(kx−ωt) fails to satisfy the Schrödinger time dependent
equation.
Solution:
In this case,
∂ψ
= ω sin(kx − ωt)
∂t
(11)
∂ 2ψ
= −k 2 ψ = −k 2 cos(kx − ωt)
2
∂x
(12)
and
If we substitute into the Schrödinger equation, we see that the two sides are not equal
(for an arbitrary choice of the independent variables x and t).
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Problem 2 – 16 marks
a) – 2 marks For a quantum mechanical particle on a ring, the lowest energy stationary
state is E = 0. What is the potential energy of a particle occupying this state? What is the
kinetic energy of a particle occupying this state?
Solution: Both the potential energy and the kinetic energy of this particle are zero. The
ring is defined as having zero potential energy, and since E = 0 the kinetic energy must also
be zero.
b) – 4 marks What is the normalized wavefunction for this state?
Solution: The wavefunction is a constant (independent of θ), and if we call the constant
A, normalization required that
Z
ψ=A⇒1=
2π
√
A2 dθ ⇒ A = 1/ 2π
(13)
0
c) – 2 marks Draw (make a graph of) the probability density of a particle occupying
this state.
Solution:
d) – 2 marks Repeat part (c) for a classical particle with E = 0. For example, you
might think of the classical particle as a bead on a necklace.
Solution:
The probably density is localized at one point (one θ value) since the particle is at rest.
e) – 2 marks Imagine you prepare a classical particle with E = 0 in the following way:
you give a classical particle a high kinetic energy, and then after some randomly determined
time, you remove all its energy. Now imagine you do this over and over again (millions of
times), and each time you plot the probability density just as you did in part (d). If you
make all of these plots on the same graph, what does it look like?
Solution:
It looks like part (c). Each time we remove the particle’s energy, it localizes to one point
(one θ value). Since this value is random, the contributions from different values combine
to give a uniform probability density.
f ) – 4 marks Compare your answers to (c), (d), and (e). Can you explain the quantum
mechanical behavior by using this approach of comparing to the classical behavior? Compare
and contrast the quantum mechanical and classical behaviors for E = 0.
Solution: The quantum particle is delocalized. We have an equal chance to find the
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particle anywhere on the ring. This is a manifestation of its wavelike nature. If we measure
its location it will collapse to a single value (as in part d) but we can’t predict which value we
will obtain. This is a manifestation of the inherent uncertainty in quantum mechanics, and
the interpretation of the wavefunction in terms of probabilities. If we take a classical particle
with a high kinetic energy, and then after some randomly determined time we remove all
its energy, it will localize (also) to a single value (as in part d), and (also) we can’t predict
which value we will obtain. So after a measurement event both the quantum and the classical
particle behave identically. However, before a measurement event (which is what you were
really asked about) the quantum and classical behavior is fundamentally different. In the
classical case the motion of the particle is deterministic, and if we selected a specific time
at which to remove the particle’s energy (and we knew the initial position and the initial
kinetic energy) we would be able to follow the entire trajectory of the particle as it moves
around the ring, and then stops. So, in that sense, the “randomly determined time” of
part e introduces uncertainty into the problem in a way that is not fundamental (we added
it). But in the quantum case we didn’t “add” any uncertainty – the uncertainty (which is
reflected in the probability density) is inherent to the particle and there is nothing we can
do to remove it. [Note: this answer goes into more detail than I was expecting because it
touches on some topics we didn’t discuss.]
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5
Problem 3 – 10 marks
For a quantum mechanical particle on a ring, consider the wavefunction
ψ(θ) = A sin(θ/2), where A is a real number.
a) – 2 marks Sketch (make a graph of) this wavefunction.
Solution: see the figure
b) – 4 marks Does ψ(θ) solve the time independent Schrödinger equation? Show your
detailed mathematical reasoning.
Solution: The TISE is
−
~2 d2 ψn (x)
+ V (x)ψn (x) = En ψn (x)
2m dx2
(14)
and then we change variables using x = Rθ, and put V = 0, to obtain
~2 d2 ψn (θ)
= En ψn (θ)
2mR2 dx2
(15)
A
d2 A sin(θ/2)
= − sin(θ/2)
2
dx
4
(16)
−
Now, we have that
so that
−
~2 d2 ψn (θ)
~2
=
ψ(θ)
2mR2 dx2
8mR2
(17)
Thus it seems that ψ solves the TISE with
~2
En =
8mR2
(18)
If we compare to the known energy values of
En =
n2 ~ 2
2mR2
(19)
we have to conclude that n = 1/2 which is not an integer. So there is a problem after all.
c) – 4 marks Does ψ(θ) satisfy the boundary conditions? Show your work.
Solution:
There is a cusp at θ = 0 = 2π (see Fig. 1) and therefore the function ψ(θ) is nondifferentiable at θ = 0 = 2π. This violates a boundary condition.
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Problem 4 – 6 marks
Carotene is a substance which occurs in carrots and mangos. It is a carbohydrate with
molecular formula C40 H56 . A model of a carotene molecule is presented in figure 1. According
to the theory of molecular bonding there are two kinds of electron bonds in carotene. The
σ-electrons are localized at fixed places between the carbon atoms. The π-electrons can
move along the entire length of the molecule. The π-electrons can be described by a onedimensional particle-in-a-box model.
FIG. 1: Carotene molecule.
Carotene has 22 π-electrons per molecule. The first excitation requires an energy of
2.76 eV. Calculate the effective length (in nanometers) along which the π-electrons can
move according to this model.
Solution:
The one-dimensional particle-in-a-box model has non-degenerate states with the lowest
state having a quantum number of n = 1; hence the 22 π-electrons in carotene fill up to
n = 11. Therefore the HOMO to LUMO transition, which corresponds to the first excitation,
is from n = 11 to n = 12.
Hence
∆n2 π 2 ~2
2me ∆E
(122 − 112 )π 2 (1.05459 × 10−34 Js)2
1 eV
=
−31
(2)(9.11 × 10 kg)(2.76 eV ) 1.6022 × 10−19 J
= 3.13 × 10−18 m−2
L2 =
⇒ L = 1.77 nm
(20)
(21)
(22)
(23)
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