# More quantum mechanics nightmare

1. Dec 17, 2004

### jsundberg

More quantum mechanics nightmare :(

A new set of problems to solve, that's mostly problems. Makes me cry. :D
All I want is for christmas to come, but this has to be done.

A:
Calculate the energy eigenvalues for a electron in a one-dimensional box with the length 0.1nm. Express the result in kJ/mol. Extend your calculations to first a proton, then an argon atom (looked upon as a point particle without inner structure).

B: Starting from the guess that the right function for a energy eigenfunction for a harmonic vibration in a potential V(x) = 1/2kx^2 is psi(x) = xexp(-ax^2) verify the guess and calculate the parameter a and the corresponding energy eigenvalue En. The particle mass is m. What grade of excitation does this eigenfunction have? Normalise the wavefunction and caluclate the energyeigenvalue in kJ/mol if the particle is a proton and k = 200N/m.

Hints, pointers, everything appreciated. I need to understand this.

2. Dec 17, 2004

### theFuture

A: What is the general form the wave function must have for this potential? What are the boundary conditions (i.e. what does psi look like at the edge of the potential)? Think about wavelength in relation to the length of the box. So now that you have psi go back to schroedingers time independent, plug it in and out the eigenstates should come.

B: This is the same thing, but they already gave you the form so plug it in and show it works. Then the calulations should come easily.

3. Dec 17, 2004

### jsundberg

Thanks for your time & help.
Still stuck, I think I'm stuck on more basic principles actually.
I think I've got the solution for the eigen values: En = n^2h^2/8mL^2

But how should I formulate this? Especially in kJ/mol.
I know h, m, and L (and those are constant).
should I just write En = n^2*constant1/constant2?
but that seems strange.

4. Dec 17, 2004

### dextercioby

That energy level (labeled through "n") corresponds to a particle of mass "m".If "m" is in Kg,then 6.023*10^{23} particles of mass "m" make up a mol of particles.So compute E_{1} and multiply it with Avodadro's number(the one stated before) to find the energy per mol.Then do the same for "n=2" aso.

Daniel.

5. Dec 17, 2004

### jsundberg

Thanks again!

Yes, but then I only get the energy eigenvalues for 2 quantum numbers. Shouldn't I find a general formula for n (unspecified) levels?

6. Dec 17, 2004

### dextercioby

If u computed the energy per mole for "n=1"(call it $E_{1}$),then it's easy to write for arbitrary "n":
$$E_{n}=E_{1}n^{2}$$
,where the units are J/mol,and "n" is a natural number different from zero.

Daniel.

7. Dec 17, 2004

### jsundberg

Thank you!
You saved my day. :D
Physics is not my thing really.

8. Dec 17, 2004

### dextercioby

Then what are doin' solving QM problems???? You should be doin what u like,and not what causes u headaches and and frustration??
Physics student???Maybe it wasn't such a smart move.

Daniel.

9. Dec 17, 2004

### jsundberg

:D
No, I'm a chemistry undergrad, going for a major in organic chemistry. I need this course in physical chemistry though to be able to take my master. The thermodynamic part was fun, and not quite this challening. I'll just have to try to make it work, it's over in a few more weeks. :)

10. Dec 18, 2004

### jsundberg

I have come so far to calculate the energy eigenvalues for the
electron, proton and argon atom:

En = n^2h^2/8mL^2
E1 = h^2/8mL^2
En = E1n^2

For the electron I got:
E1 = 3629kJ/mol
Proton:
E1 = 2kJ/mol
Argon:
E1 = 0.06kJ/mol

Now I'm going to compare the distance in energy between the ground
state and first excitated state between those three particles with RT
at the temperature 300K. The last part here confuses me a lot, RT? What do they mean? My textbook doesn't really help a lot :(

11. Dec 18, 2004

### theFuture

R is the ideal gas constant, T is the temperature. RT has units of energy/mol.

12. Dec 18, 2004

### Nylex

R = 8.314 J K^-1 mol^-1.

13. Dec 18, 2004

### Gokul43201

Staff Emeritus
Okay, you must know temperature is actually a measure of the KE of the molecules. For an ideal gas, the absolute temperature (in Kelvin) is directly proportional to the KE.

So, for a single atom/molecule $$<\frac {1}{2} mv^2>~~ \alpha~~ T$$

the proportionality constant $k_B$ is known as the Boltzmann Constant, and is accompanied by a prefactor of 3/2, which comes from the Equipartition Theorem.

So, $$<\frac {1}{2} mv^2> = \frac{3}{2} k_BT$$

Thus, $k_BT$ gives you an estimate of the average KE of a molecule at temperature T. To extrapolate this equation to a mole of gas, you simply have to multiply by Avogadro's Number, $N_{av}= 6.023*10^{23}$ molecules/mole.

So, the average KE of a mole of gas, is roughly $$<\frac {1}{2} Mv^2> = \frac {3}{2}N_{av}k_BT = \frac{3}{2}RT$$
where [tex]R = N_{av}k_B [/itex]

To get an order of magnitude estimate, the 3/2 prefactor is often skipped. In fact, this prefactor is different for different kinds of molecules (see Equipartition Theorem).

14. Dec 19, 2004

### jsundberg

That's a bit embarassing. Don't know what I thought of, of course I know what
RT is. Thanks for all the help, again. You guys saved my day once again. :D

Gokul: Really good explanation! Thanks.