In the numerical example in the text, I calculated only the ratio of the probabilities of a hydrogen atom being in two different states. At such a low temperature the absolute probability of being in a first excited state is essentially the same as the relative probability compared to the ground state. Proving this rigorously, however, is a bit problematic, because a hydrogen atom has infinitely many states. Estimate the partition function for a hydrogen atom at 5800 K, by adding the Boltzmann factors for all the states shown explicitly . (For simplicity you may wish to take the ground state energy to be zero, and shift the other energies accordingly.)
In the numerical example in the text, I calculated only the ratio of the probabilities of a hydrogen atom being in two different states. At such a low temperature the absolute probability of being in a first excited state is essentially the same as the relative probability compared to the ground state. Proving this rigorously, however, is a bit problematic, because a hydrogen atom has infinitely many states. Estimate the partition function for a hydrogen atom at 5800 K, by adding the Boltzmann factors for all the states shown explicitly . (For simplicity you may wish to take the ground state energy to be zero, and shift the other energies accordingly.)
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In the numerical example in the text, I calculated only the ratio of the probabilities of a hydrogen atom being in two different states. At such a low temperature the absolute probability of being in a first excited state is essentially the same as the relative probability compared to the ground state. Proving this rigorously, however, is a bit problematic, because a hydrogen atom has infinitely many states.
Estimate the partition function for a hydrogen atom at 5800 K, by adding the Boltzmann factors for all the states shown explicitly . (For simplicity you may wish to take the ground state energy to be zero, and shift the other energies accordingly.)
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