Question

The energy required to ionise a hydrogen atom is 13.6 eV, so you might expect that the number of ionised hydrogen atoms in the sun's atmosphere would be even less than the number in the first excited state. Yet at the end of Chapter 5 I showed that the fraction of ionised hydrogen is much larger, nearly one atom in 10,000. Explain why this result is not a contradiction, and why it would be incorrect to try to calculate the fraction of ionised hydrogen using the methods of this section.

Short answer

It makes estimating the probabilities unknown, it does make the ionisation state more likely than using solely Boltzmann factors.

Step-by-step solution

Given information

The energy required to ionise a hydrogen atom is 13.6 eV, so you might expect that the number of ionised hydrogen atoms in the sun's atmosphere would be even less than the number in the first excited state. Yet at the end of Chapter 5 I showed that the fraction of ionised hydrogen is much larger, nearly one atom in 10,000.

Explanation

The probability of ionised atoms is equal to Boltzmann factors $e^{-I/kT}$where I is the ionisation energy and T is the temperature, plus the Boltzmann factors multiplied by the degeneracy of the ionised state, which is essentially limitless. Although this makes estimating the probabilities unknown, it does make the ionisation state more likely than using solely Boltzmann factors. The likelihood does, in fact, depend on the number density of electrons in the surroundings.