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Diatomic excitations

Page history last edited by Joe Redish 7 years, 1 month ago

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

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