CHEM 442 Problem Set 4a

There are two sides to this problem set!!! ! Useful Integrals

a n⇡x m⇡x a a n⇡x n⇡x sin sin dx = sin cos dx =0 a a 2 n,m a a Z0 Z0 a n⇡x a2 a n⇡x a3 a3 x sin2 dx = x2 sin2 dx = a 4 a 6 4⇡2n2 Z0 Z0

Problems

1. The particle in a one-dimensional box of length a can be described by the wavefunction:

n⇡x (x)=sin n a

where n(x) = 0 outside the box. Normalize this wavefunction.

2. Being certain to obey all boundary conditions, sketch out the general shape of the wavefunc- tions and of the probability density of the first five states of a particle in this box.

3. As n gets very very large (approaches the classical limit), what happens to the probability distribution?

4. Show that x has the same value for all states of a particle in a one-dimensional box. Does h i the value make sense?

5. Show that p has the same value for all states of a particle in a one-dimensional box. Does h i this value make sense?

6. Calculate the following for the n = 2 state of the particle in a one-dimensional box of length a, and then use your results to verify that this system obeys the .

(a) x , x2 , h i h i x

(b) p , p2 , h i h i p 7. Remembering that the Hamiltonian for motion in one dimension is given by:

¯h2 d2 Hˆ = + V (x) 2m dx2

use your normalized wavefunction and the above Hamiltonian to derive a general expression for the levels of the particle in a one-dimensional box of length a.

8. Sketch out the relative of the first five states of a .

9. As m gets very very large (approaches the classical limit), what happens to the energies?

10. Assume an (m =9.109 10 31 kg) is contained in a box of length a = 100 nm. This e ⇥ is a realistic situation for an electron confined to a long carbon nanotube. What wavelength of light would the electron need to absorb to move between the n = 1 and n = 2 states in this box?

2 CHEM 442 Problem Set 4b

There are multiple pages to this problem set!!! ! In fact, this first page is almost entirely review! ! Generally Useful Properties

The simple harmonic oscillator can be described by the wavefunction in the potential listed below:

1 2 y2/2 V = kx (x)=H (y)e 2 v v where Hv(y) is a Hermite Polynomial:

vHv(y) 01 12y 24y2 2 ...... 1/4 ¯h2 and y = x/↵ where ↵ = . mk ✓ ◆ The Hermite Polynomials can be generated from the recursion relation:

Hv+1 2yHv +2vHv 1 =0 The energy levels of the simple harmonic oscillator are given by:

1 E = v + ¯h! v 2 ✓ ◆ where ! = k/m, keeping in mind that k here is the force constant, not Boltzmann’s constant. p Often, we wish to solve the harmonic oscillator for a diatomic molecule, and in that case, the mass we use is the reduced mass, given by: m m µ = 1 2 m1 + m2 The normalization constant for the harmonic oscillator is ugly, and is dependent on the quantum number v, unlike the particle in a one-dimensional box, whose normalization had no dependence on its quantum number n. The harmonic oscillator normalization constant is given by:

1 1/2 v Nv = ↵⇡ 2 v! ⇣p ⌘ Problems

1. Verify that using N0 normalizes 0.

1 ax2 ⇡ Helpful Integral: e dx = a Z1 r 2. Calculate the energies for the first five states of the simple harmonic oscillator. How does the spacing of these levels compare to the one-dimensional particle in a box? 3. Generate the Hermite Polynomials for v = 3 and v = 4. 4. Given that the frequency of the fundamental vibrational transition of carbon monoxide (CO) 1 1 2 is at 2143 cm , calculate the force constant in units of N m . A Newton (N) is a kg m s . 5. A function is said to be even if it satisfies:

f(x)=f( x)

and is said to be odd if it satisfies:

f(x)= f( x)

Using this, and the Hermite Polynomials for v = 0 through v = 4, show that when v is even, Hv is even, and when v is odd, Hv is odd. In this case, it will suce to show it for v =0 through v = 4, and we will generalize to the rest of them. 6. If f(x) is an odd function, it can be shown that:

A f(x)dx =0 A Z because the areas from ( A to 0) and from (0 to +A) will be equal in magnitude but opposite in sign, and will thus cancel to zero.

It can also be shown (but we don’t have to, because a very helpful mathematician already did so), that derivative of an even function is an odd function, and that the derivative of an odd function is an even function.

Finally, when multiplying odd and even functions, it works just like multiplying odd and even integers. (odd) (odd) = (even), (odd) (even) = (odd), etc. ⇥ ⇥

Using these relationships, and the definitions of even and odd functions from Problem 5, show that x and p both evaluate to zero for any value of v. You need only show enough work h i h i to prove that the integrand for each is odd, and thus evaluates to zero. Do these values make sense?

2 7. I’ve constructed a new wavefunction from a linear combination of the orthonormal set of harmonic oscillator wavefunctions, and it is given by:

(x)=3 0(x)+4 1(x).

(a) I was far too lazy to be bothered normalizing my new wavefunction - so please do that for me.

(b) I also don’t want to be bothered to do more than one measurement on this system. If I measure the energy just once, what are the possible values I’ll get, and with what probabilities?

3 CHEM 442 Problem Set 4c

There are multiple pages to this problem set!!! ! Probably Useful Junk:

The energy level expression for the solutions to the particle in a ring problem look like:

m2¯h2 E = l ml 2mr2 while the normalized wavefunction is:

1 iml✓ ml (✓)= e r2⇡

1. The wavefunction for a particle in a 3D box with sides of a, b, and c along the x, y, and z axes, respectively, is given by:

8 nx⇡x ny⇡y nz⇡z (x, y, z)nx,ny,nz = sin sin sin rabc a b c What is the average position in 3D space of the particle in this box?

2. The momentum operator in three dimensions is given by:

@ @ @ pˆ = i¯h + + @x @y @z ✓ ◆ What is p for the 3D particle in a box of lengths a, b, and c? h i

3. Let’s go back to a one-dimensional box for just a moment to simplify the math here. Assume our particle is trapped in a box between 0 and a, and we want to make measurements of the 2 energy. What is the variance (E) we expect in these measurements? Does the value you get back make sense?

4. What are the degeneracies of the first five energy levels for the particle in a 3D box with a = b =1.5c? 5. The structure of the porphyrin molecule is shown below.

N

NH HN

N

10 Å

(a) What would we predict is the wavelength of light corresponding to the lowest energy absorption transition for exciting a ⇡-electron in porphyrin if we assume that the elec- trons are simply confined to a 2D box with length of 10A?˚

(b) The experimental value is 17,000 cm 1, likely quite a bit away from the answer you ⇠ arrived at in part b. Speculate on why our model did a poorer job of predicting this value than the example with 1,3-butadiene in the lecture?

6. Consider the systems we have studied so far (the particle in a box, the particle in a harmonic potential, the particle in a ring). Which of these systems has a ground state energy value that is not zero?

7. Which of these three systems has an allowed energy level with a quantum number of zero? For those where a quantum number of zero is not allowed, explain why.

8. Can the particle in a ring ever stop moving (have zero energy)?

9. What is the qualitative di↵erence between the ml = +9 and ml = -9 energy levels of the particle in a ring?

10. If I wanted to adapt the energy level expression for the particle on a ring to be workable for a pair of particles, rigidly attached to each other, and rotating about a central point, what would I need to change (qualitatively)? (Hint: you might want to consider comparing to what was used for the particle in a harmonic potential).

2 11. Here’s the structure of benzene:

1.4 Å

(a) What wavelength of light would need to be absorbed to excite a ⇡-electron in this sys- tem? Use the best system from our three models to do the calculation.

(b) The actual experimental value is 260 nm. Did we do a better job with this representation than we did using a box for porphyrin?

(c) Yer darned right we did. So try modeling porphyrin as a ring and re-calculate that transition frequency. How’d we do this time?

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