2
$\begingroup$

The Wikipedia article about the H$\alpha$ spectral line states

"it takes nearly as much energy to excite the hydrogen atom's electron from $n = 1$ to $n = 3$ (12.1 eV, via the Rydberg formula) as it does to ionize the hydrogen atom (13.6 eV), ionization is far more probable than excitation to the $ n = 3$ level"

Since 12.1 eV < 13.6 eV, more energy is required to ionize the atom then to excite it to the n=3 level. Hence shouldn't excitation to the n = 3 level be more probable?

$\endgroup$

1 Answer 1

2
$\begingroup$

The article is referring to hydrogen atoms in nebulae within a galaxy. The hydrogen in these clouds is excited by uv light from young, massive stars recently formed in the nebulae. These stars have an approximately black body spectrum, so the light they emit spans a wide range of wavelengths. In particular the intensity at an energy of 13.6eV is comparable to the intensity at 12.1eV, and this means a star hot enough to excite the $n=1$ to $n=3$ transition also emits light energetic enough to fully ionise the hydrogen atom. The result is that hydrogen atoms near such a star are likely to end up full ionised instead of just excited to the $n=3$ state.

The $n=3$ population arises when the ionised hydrogen atoms recombine with electrons. The recombination produces neutral atoms in a range of excited states, and a fraction of these will be in the $n=3$ state. The decay of these atoms then creates the $\mathrm H\alpha$ emission.

$\endgroup$
1
  • $\begingroup$ Is there any significant difference in the cross-section for $n = 1 \to 3$ excitation vs. the cross-section for ionization starting from the $n = 1$ state? $\endgroup$ Commented Dec 28, 2020 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.