Consider a diatomic molecule (like O 2 ) in air. The bond between the oxygen ato

Question

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 + , E2 = E0 + 2, E3 = E0 + 3, … En = E0 + n,

where E0 is the energy of the lowest allowed (ground) state, and 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, = 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?

Explanation / Answer

A

The probability of finding the molecule in the first excited state compared to the ground state is

z = e(-/kBT ). Since each higher excited state just replaces with 2, 3, etc, the probability of the second, third, etc. states compared to the ground state is z2, z3, etc. In our case, since /kBT = (100 meV)/(25 meV) = 4, we have that z = e-4 = 0.018, about 2%. Higher powers of z become much smaller very quickly. So at this temp, our diatomic molecule will be in the ground state more than 97% of the time.

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B
Let's call /kBT= x. Then z = e-x. If we want z to be 0.2, the x = -ln(0.2) = 1.61. So /kBT = 1.61 and kBT = /1.61 = 62 meV. This is more than twice the Kelvin temperature at room temp so over 500K.

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C
Since if the probability of the first excited state is z, the second is z2, the third is z3, etc. So the probability relative to the ground state of the second is 0.04 (4%), the third is 0.008 (0.8%), of the fourth is 0.00016 (~0,02%) etc. So only the ground and first excited states will be greater than 1% even at this high temperature.

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D
The normalization factor for the probabilities is 1/(1+z+z2+z3...) = 1/(1 + 0.2 + 0.04 + 0.008) = 0.801 or 80% to two sig figs.

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