View Full Version : More quantum mechanics nightmare :(
jsundberg
Dec17-04, 04:41 AM
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.
theFuture
Dec17-04, 12:52 PM
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.
jsundberg
Dec17-04, 01:53 PM
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.
dextercioby
Dec17-04, 02:24 PM
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.
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.
jsundberg
Dec17-04, 02:31 PM
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?
dextercioby
Dec17-04, 02:46 PM
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?
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.
jsundberg
Dec17-04, 02:48 PM
Thank you!
You saved my day. :D
Physics is not my thing really.
dextercioby
Dec17-04, 03:05 PM
Thank you!
You saved my day. :D
Physics is not my thing really.
Then what are doin' solving QM problems???? :confused: 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. :rolleyes:
Daniel.
jsundberg
Dec17-04, 04:02 PM
: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. :)
jsundberg
Dec18-04, 07:14 AM
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 :(
theFuture
Dec18-04, 10:46 AM
R is the ideal gas constant, T is the temperature. RT has units of energy/mol.
Gokul43201
Dec18-04, 11:53 AM
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).
jsundberg
Dec19-04, 06:44 AM
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.
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.