Calculating Rotational & Vibrational Energy of Hydrogen Molecule

[nicky04]

Homework Statement

The potential energy V between the two hydrogen atoms ( mH = 1.672*10^(-24) g) in a
hydrogen molecule is given by the empirical expression  V = V0 [ exp(-2a(r-r0)) - 2exp(-a(r-r0))] where
V0 = 7*10^(-12) erg, a = 2*10^8 cm-1, r0 = 8*10^-9 cm.
(a) Estimate the temperatures at which rotation and vibration begin to contribute to the specific
heat of hydrogen gas.
(b) Give the approximate values of Cv (in terms of the gas constant R for the following
temperatures: 25 K, 250 K, 2500 K, 10000 K.

Homework Equations

I am not sure since I am not sure about my starting point.

The Attempt at a Solution

Basically I can think of two ways to do this. Either I could calculate the rotational temperature given by
T_rot = h_bar^2 / 2 k I , with I = M*r0^2 and the reduced mass M = mh / 2.
The vibrational temperature is given by
T_vib = h_bar omega / k.
What is confusing me is the fact that the given energy potential is not being used. I thought, maybe I need it to calculate omega? I started solving the equation of motion for a harmonic oscillator with and additional potential, but with this potential it is not that easy to solve and I have a feeling that it's wrong anyway. Is there another way to calculate omega from V? I also tried it by saying
omega = sqrt( k / m) and
F=-kx.
With the given potential I could calculate F but then it depends on r and I am not sure my finaly solution should depend on r?

Another way would be to look at the partition functions for the rotation and the vibration. The first question there is, can I assume the atoms are in the ground state? Because the functions depend on l and n. Further I am not sure how to include the potential again, should I just add it in the exponent? Once I had the partition functions I could derive the specific heats and set them equal to the known specific heats depending on R.
Does any of this make sense or should I rethink it all over again?

 

[vela]

Express V as a Taylor series about its minimum and look at the coefficient of the second-order term.

 

[nicky04]

ok so the minimum would be at r=r0. then i have the taylor expansion as
V = V0 + V0 (-2a + 2a)(r-r0) + V0 (4a^2 - 2a^2)(r-r0)^2 + ...
= V0 + 2 V0 a^2 (r-r0)^2
correct?
what do i do with this now though? should i go th first way that i described or the second? or something else?

 

[vela]

Your expansion isn't quite correct. The sign of the first term is wrong, and it looks like you forgot to divide by n!.

Once you get the expansion right, the coefficient of (r-r₀)² is equal to 1/2 μω², where μ is the reduced mass.

 

[nicky04]

yes you are right, i forgot that in the expansion. i calculated the temperatures now, they seem reasonable. thank you so much!

 

[Mindscrape]

Out of curiosity, as a passer-by, what would be the approach after the potential is taylor expanded? I would think to use kinetic energy, as nicky suggested.

 

[vela]

That's the correct approach. You're looking for when the kinetic energy of the molecules becomes comparable to the rotational and vibrational energies. At cooler temperatures, those degrees of freedom are frozen out.