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

[jinksys]

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 $$3/2 \hbar\sqrt{k/m'}$$?

 

[vela]

Why would it be either?

 

[jinksys]

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?

 

[vela]

Yup.

 

[The_Duck]

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.

 

[jinksys]

That doesn't make any sense to me. Why does Beiser use a "v" of 0.5?

 

[vela]

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 $\hbar\omega$.

 

[jinksys]

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?