Question about the quantum ladder

[melech]

I understand that as electrons in orbit around an atom reach higher and higher energy levels the energy differences get smaller. Does this mean that at high enough energy levels, or when the electron is not in confined orbit, there is no quantum jump at all and that the energy emission and absorption of the electron is absolutely continuous, as in classical physics? Planck's constant does not have a limiting effect in the energy flow?

[Andrew D]

No, it means that after an electron is excited enough, it doesn't take much more energy to completely separate it from the atom. The energy still comes in discrete amounts, but think of it this way: any number is an integer multiple of some infinitely small number. The process by which the electron leaves the atom is called 'ionization'.

[korjik]

I understand that as electrons in orbit around an atom reach higher and higher energy levels the energy differences get smaller. Does this mean that at high enough energy levels, or when the electron is not in confined orbit, there is no quantum jump at all and that the energy emission and absorption of the electron is absolutely continuous, as in classical physics? Planck's constant does not have a limiting effect in the energy flow?

Once the electron is no longer in a confined orbit, it is no longer part of the atom. Usual term is that the electron is now 'free'. A free electron dosent really absorb photons, it more refracts and scatters them (thompson scattering? I can never remember which is which). Technically, the spectrum of photons it can scatter is continous, but plank's constant still shows up in the photon energy.

[Ken G]

But the OP is onto something interesting about atomic energy levels. In statistical mechanics, when you have an atom in equilibrium with its environment at some temperature T, the relative likelihood of finding the atom in a level of energy E_n, compared to a level of energy E_m, is e^{(E_m - E_n)/kT}. For an idealized hydrogen atom, E_n ~ -1/n², using the convention where the minimum energy "free" state has zero energy. The problem with allowing this idealized hydrogen atom to have an infinite number of levels (unbounded n values) is that you get an infinite number of levels that all need to have a finite probability of being excited, by the above formula. That would force all the atoms to be excited to arbitrarily large n values, no matter what T is, so all atoms would ionize when in equilibrium at any temperature, no matter how low!

The solution of this obvious mistake is that the atom cannot actually have access to an infinite number of levels. We either say that the highest levels never participate in the thermal equilibrium, or we simply recognize that the hydrogen atom can never be completely isolated, so there is some maximum-n value where we can really consider the atom to be in a bound state-- higher n would get ionized by the environment before they could reach thermal equilibrium. That allows us to truncate the sum at some high n, and when we do that, at low T an interesting thing happens-- we get the opposite result from not truncating the sum, and the majority of the atoms are found in the ground state n=1.

But the bottom line is, a completely idealized atom does indeed in some sense go "non-quantum-mechanical" at arbitrarily large n, but our treatment of the atom must ignore those levels anyway, to avoid unphysical results.

[melech]

Ken G, Is this what is referred to as the ultra-violet catastrophe?

[korjik]

Ken G, Is this what is referred to as the ultra-violet catastrophe?

No. Ultraviolet catastrophe is a completely classical effect. IIRC, the catastrophe is more cause power per unit wavelength goes to infinity in the UV range if you use Maxwells equations and assume photons are not quantized. It is a similar math problem that comes about for a different physical reason.

[Ken G]

You're both right-- the UV catastrophe is a classical effect that is a bit different from the high-n hydrogen atom problem, but it also shares its characteristics. The key difference is that for the H atom, we know we have just one electron, and we want to know its expected energy. The UV catastrophe occurs for light, and again we want to know the expected energy, but now it is the energy of many photons-- photons can just "pop into existence." The similarity is that in the UV catastrophe, in thermal equilibrium, you would classically expect to get kT of energy in every frequency mode of the radiation field, and there are an infinite number of "high-n" modes, just like in the H atom. So again the problem is all those high-n modes, and you need some way to make them "not important." In the H atom, we make them not important by just truncating at some n, where we say higher n don't participate in the thermal equilibrium. In the UV catastrophe, we say that the high-n modes have a hard time getting excited because each one requires much more than kT of energy, and since we require a quantum of excitation (the process of excitation is a quantum process, it either happens or it doesn't in each interaction), it is very unlikely to ever get excited even in thermal equilibrium.