7.4.P9
Consider a diatomic molecule (like O2) in air. The bond between the oxygen atoms acts very much like a spring: the two atoms can vibrate back and forth along the line joining them.
But since atoms in a molecule are quantum systems, in fact, not every energy is permitted to the vibrating system. Only a discrete set of energies are allowed as indicated on the graph of a spring potential energy shown at the right. The allowed energies are equally spaced, having the form
E0, E1 = E0 + ε, E2 = E0 + 2ε, E3 = E0 + 3ε, … En = E0 + nε,
where E0 is the energy of the lowest allowed (ground) state, and ε is a constant determined by the strength of the interaction and the masses of the atoms.
|
|
A. At room temperature, kBT = 25 meV. Suppose for a particular diatomic molecule, ε = 100 meV. With these values of the parameters, you will almost always find the molecule in its ground state. Explain why using a calculation.
B. At what value of kBT would the probability of finding the molecule in its first excited state be equal to 20% of the probability of finding it in its ground state?
C. At the temperature found in B, how many states would have a probability of being found with a probability of greater than 1%?
D. At the temperature found in B, what is the probability to find the molecule in its ground state to two significant figures?
Ben Dreyfus and Joe Redish 3/24/16
Comments (0)
You don't have permission to comment on this page.