QM: Excited Vibrational States of HCl Molecules and Solutions Manual Discrepancy

  • Thread starter Thread starter jinksys
  • Start date Start date
  • Tags Tags
    Qm
Click For Summary

Homework Help Overview

The discussion revolves around the vibrational states of HCl molecules, specifically whether an HCl molecule is likely to be in its first excited vibrational state at room temperature. The problem involves comparing energy calculations from different sources, including a solutions manual and an online resource.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the energy expressions for vibrational states, questioning the use of different formulas for calculating energy levels. There is discussion about the significance of the ground state energy and the energy difference required to transition to the first excited state.

Discussion Status

The discussion is active, with participants examining the discrepancies between different sources and clarifying the concepts of vibrational energy levels. Some guidance has been offered regarding the interpretation of energy states, but multiple interpretations are still being explored.

Contextual Notes

Participants are considering the implications of temperature on vibrational states and the arbitrary nature of potential energy references in the context of the problem. There is uncertainty regarding the relationship between kT and the ground state energy.

jinksys
Messages
122
Reaction score
0

Homework Statement



Q:The bond between the hydrogen and chlorine atoms in an HCL molecule has a force constant of 516 N/m. Is it likely that a HCl molecule will be in its first excited vibrational state at room temp?

When doing my HW, I always check my answers against the solutions manual. However, I noticed that the solutions manual (Beiser, Concepts of Modern Physics) has a different solution that cramster.com. Beiser says that the Energy for vibrational energy level 1 is:\hbar \sqrt{k/m'}.

Shouldn't it be 3/2 \hbar\sqrt{k/m'}?
 
Physics news on Phys.org
Why would it be either?
 
vela said:
Why would it be either?

The energy for a vibrational energy level is E=(v+0.5)\hbar\sqrt{k/m'},
so for the first excited energy level, v=1, E={3/2}\hbar\sqrt{k/m'}.

Correct?
 
Yup.
 
I think that if you want to know the probability that the molecule will be in the excited state rather than the ground state, what you want to know is the difference in energies between the two. The energy of the ground state is arbitrary, really, in the same way that the zero of potential energy is arbitrary. I think that is why the book uses the number it is.
 
That doesn't make any sense to me. Why does Beiser use a "v" of 0.5?
 
That's not what Beiser's doing. When it's too cold to excite the vibrational state of the molecule, the molecule still has vibrational energy because the vibrational ground state has an energy of \hbar\omega/2. So to get to the first excited state, which has an energy of (3/2)\hbar\omega, it needs only an additional energy of \hbar\omega.
 
Both Beiser and cramster calculate kT for 300 K and then compare it to the energy of the first vibrational energy level. I'm confused now, is kT equal to the ground state energy?
 

Similar threads

Replies
5
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
5K
  • · Replies 3 ·
Replies
3
Views
11K
Replies
6
Views
4K