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My question isn't how they receive the energy to jump, but why. When someone views an element's emission spectrum, we see a line spectrum which proves that they don't exist outside of their orbitals (else we would see a continuous spectrum). Electrons can be released in the form of beta decay, thus proving that they are capable of traveling outside of orbitals contrary to the statement my teacher said that they stay within orbitals. Then, to add to the confusion, the older model of rings floating around a nucleus has, from what I can tell, been outdated, which would support this model. My teacher's explanation was that the electrons made a quantum jump of some kind. How do electrons move between orbitals or do we know how they jump, excluding the reason that energy causes them to jump, and why are positrons formed sometimes instead of electrons in Beta decay? When I'm asking "how do electrons jump" I would like to know how an electron can jump between each orbital such as how it moves and how it knows where to jump since it appears to be a jump where the electron doesn't slow into a orbital position. Specifically how they jump what is this Atomic electron transition, I understand that they jump and that they do this through absorbing and releasing energy but what is this Atomic electron transition other than what is already on the wikipedia article http://en.wikipedia.org/wiki/Atomic_electron_transition.

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"the older model of rings floating around a nucleus has from what i can tell been outdated" Hmmm ... yes. Outdated for roughly 70 years. Basically your instructor may be working for a number of misconceptions about the nature of election orbitals. Mind you, even in that framework the decay electrons are generally unbound and so would not lie on any of the rings in the first place. –  dmckee Dec 14 '12 at 3:12
Beta decay is a nuclear process (specifically a weak process), rather than a chemical one. This said, orbital transitions can occur in inverse beta decay (which is to say, electron capture) however this is only a side effect of an unoccupied core orbital being generated. –  Richard Terrett Dec 14 '12 at 4:21
@RichardTerrett The other thing about inverse beta decay that is interesting here is that it's rate is only as high as it is because s-orbitals have non-zero probability at $r=0$, but that starts to take us far afield. –  dmckee Dec 14 '12 at 5:20
sorry about the miss type in my bounty at the beginning i sad "how would" when i meant *"I would" and i said "there appears to be linear motion" when i meant "there appears to be no linear motion" –  Bored915 Dec 16 '12 at 23:29
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4 Answers

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Imagine an electron a great distance from an atom, with nothing else around. The electron doesn't "know" about the atom. We declare it to have zero energy. Nothing interesting is going on. This is our reference point.

If the electron is moving, but still far from the atom, it has kinetic energy. This is always positive. The electron, still not interacting with the atom, may move as it pleases. It has positive energy, and in any amount possible. Its wave function is a simple running plane wave, or some linear combination of them to make, for example, a spherical wave. Its wavelength, relating to the kinetic energy, may be any value.

When the electron is close to the atom, opposite charges attract, and the electron is said to be stuck in a potential well. It is moving, so has positive (always) kinetic energy, but the Coulomb potential energy is negative and in a greater amount. The electron must slow down if it moves away from the atom, to maintain a constant total energy for the system. It reaches zero velocity (zero kinetic energy) at some finite distance away, although quantum mechanics allows a bit of cheating with an exponentially decreasing wavefunction beyond that distance.

The electron is confined to a small space, a spherical region around the nucleus. That being so, the wavelength of its wavefunction must in a sense "fit" into that space - exactly one, or two, or three, or n, nodes must fit radially and circumferentially. We use the familiar quantum number n,l,m. There are discrete energy levels and distinct wavefunctions for each quantum state.

Note that the free positive-energy electron has all of space to roam about in, and therefore does not need to fit any particular number of wavelengths into anything, so has a continuous spectrum of energy levels and three real numbers (the wavevector) to describe its state.

When the atom absorbs a photon, the electron jumps from let's say for example from the 2s to a 3p orbital, the electron is not in any orbital during that time. Its wave function can be written as a time-varying mix of the normal orbitals. A long time before the absorption, which for an atom is a few femtoseconds or so, this mix is 100% of the 2s state, and a few femtoseconds or so after the absorption, it's 100% the 3p state. Between, during the absorption process, it's a mix of many orbitals with wildly changing coefficients. There was a paper in Physical Review A back around 1980 or 1981, iirc, that shows some plots and pictures and went into this in some detail. Maybe it was Reviews of Modern Physics. Anyway, keep in mind that this mixture is just a mathematical description. What we really have is a wavefunction changing from a steady 2s, to a wildly boinging-about wobblemess, settling to a steady 3p.

A more energetic photon can kick the electron out of the atom, from one of its discrete-state negative energy orbital states, to a free-running positive state - generally an expanding spherical wave - it's the same as before, but instead of settling to a steady 3p, the electron wavefunction ends as a spherical expanding shell.

I wish I could show some pictures, but that would take time to find or make...

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Here I will address some misconceptions in the question, not addressed by the answer of DarenW.

My question isn't how they receive the energy to jump, but why. When someone views an element's emission spectrum, we see a line spectrum which proves that they don't exist outside of their orbitals (else we would see a continuous spectrum).

These emission and absorption spectra




emission spectrum



Come from the atomic orbitals, as explained in DarenW's answer. That is, the nucleus with its positive charge, say Helium with charge +2, has around it two electrons "orbiting" in allowed by the solutions of the quantum mechanical problem "orbits". Where "orbits" means a spatial location in 3 dimensional space where the probability of finding electrons is high, of spherical shape about the nucleus with very specific quantum numbers.

Electrons can be released in the form of beta decay, thus proving that they are capable of traveling outside of orbitals contrary to the statement my teacher said that they stay within orbitals. T

This is a misconception. Beta decays happen when a neutron turns into a proton and an electron, and they are phenomena pertaining to the nucleus, not the atom. The atom is described well by electromagnetic interactions, the nucleus is described by strong interactions and weak interactions. Beta decays are a weak interaction. Thus the electron of the beta decay is a free electron once it materializes and is ejected from the nucleus, particularly if all free electron orbital locations are filled. The nucleus then changes into an Z+1 charge isotope nucleus.

Here is how the neutron decay is currently visualized

beta decay

How do electrons move between orbitals, excluding energy added to excite electrons,

You have to add energy to excite the electrons to higher orbitals, and usually it is with the kick of a photon of the energy of the gap between orbitals.

and why are positrons formed sometimes instead of electrons in Beta decay?

From wikipedia on electron capture

In all the cases where β+ decay is allowed energetically, the electron capture process, when an atomic electron is captured by a nucleus with the emission of a neutrino, is also allowed

It means that a proton in the nucleus turns into a neutron a positron and a neutrino. This lowers the nuclear Z by one unit, and will induce a cascade of higher orbital electrons falling in the hole left by the captured one.

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Of course electrons CAN travel between orbitals, although they do this in not conventional (classical) way.

The question of traveling electrons between orbitals is the subject or relativistic quantum mechanics, or as it is called another way, of quantum field theory or quantum electrodynamics.

By words I can describe the situation in following way.

The orbitals are not PLACES, they are EIGEN STATES of energy operator. Electron can exist in any state, but this any state is representable by superposition of eigenstates.

So, an electron traveling from orbital $\psi_1$ to orbital $\psi_2$ is described by the state $a \psi_1 + b \psi_2$ where $a$ and $b$ are complex weights of the components of superposition. They are changing over time, having $a=1; b=0$ at the beginning of the process and $a=0; b=1$.

Also, you know that $|a|^2 + |b|^2=1$ at any instant.

The law of this changing is exponential, i.e. $a(t) \sim e^{-\lambda t}$.

The parameters of this exponent are depending on state lifetime. The shorter lifetime, the more exponent slope. Also lifetime is also related with state uncertainty. The wider the state, the shorter it's lifetime.

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The answers so far seem pretty good, but I'd like to try a slightly different angle.

Before I get to atomic orbitals, what does it mean for an electron to "be" somewhere? Suppose I look at an electron, and see where it is (suppose I have a very sophisticated/sensitive/precise microscope). This sounds straightforward, but what did I do when I 'looked' at the electron? I must have observed some photon that had just interacted with that electron. If I want to get an idea of the motion of the electron (no just its instantaneous momentum, but its position as a function of time), I need to observe it for a period of time. This is a problem, though, because I can only observe the electron every time it interacts with a photon that I can observe. It's actually impossible for me to observe the electron continuously, I can only get snapshots of its position.

So what does the electron do between observations? I don't think anyone can answer that question. All we can say is that at one time the electron was observed at point A, and at a later time it was observed at point B. It got from A to B... somehow. This leads to a different way of thinking about where an electron (or other particle) is.

If I know some of the properties of the electron, I can predict that I'm more likely to observe an electron in some locations than in others. Atomic orbitals are a great example of this. An orbital is described by 4 quantum numbers, which I'll call $n$, $l$, $m$, $s$ (there are several notations; I think this one is reasonably common). $n$ is a description of how much energy the electron has, $l$ describes its total angular momentum, $m$ carries some information about the orientation of its angular momentum and $s$ characterizes its spin (spin is a whole topic on its own, for now let's just say that it's a property that the electron has). If I know these 4 properties of an electron that is bound to an atom, then I can predict where I am most likely to observe the electron. For some combinations of $(n,l,m,s)$ the distribution is simple (e.g. spherically symmetric), but often it can be quite complicated (with lobes or rings where I'm more likely to find the electron). There's always a chance I could observe the electron ANYWHERE, but it's MUCH MORE LIKELY that I'll find it in some particular region. This is usually called the probability distribution for the position of the electron. Illustrations like these are misleading because they draw a hard edge on the probability distribution; what's actually shown is the region where the electron will be found some high percentage of the time.

So the answer to how an electron "jumps" between orbitals is actually the same as how it moves around within a single orbital; it just "does". The difference is that to change orbitals, some property of the electron (one of the ones described by $(n,l,m,s)$) has to change. This is always accompanied by emission or absorption of a photon (even a spin flip involves a (very low energy) photon).

Another way of thinking about this is that the electron doesn't have a precise position but instead occupies all space, and observations of the electron position are just manifestations of the more fundamental "wave function" whose properties dictate, amongst other things, the probability distribution for observations of position.

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I think im going to reward you with the bounty and one of the other answers as the answer since covered most of it before you though you answered the original question which was the bounty question. Then to add to it you understood exactly what i was trying to get at that i failed at translating as my main question. –  Bored915 Dec 18 '12 at 19:27
Glad I could help. I agree the other answers do a good job of covering some of the formal description of the question in the framework of QM, but unless you know QM (based on what it sounds like you're covering in class, I'm guessing you have at most an intro) it can be a bit hard to follow all the detail. –  Kyle Dec 18 '12 at 19:40
I finished freshmen physics and am taking IB physics now, and science has been a passion of mine so I've been trying to learn ahead with the motto the more annoying the science the more fun it is. –  Bored915 Dec 18 '12 at 20:23
Ah the IB... I remember that physics course. Only the SL version was offered at my school when I took it. I remember reading up on all the optional units I couldn't take. Good luck with your studies :) –  Kyle Dec 18 '12 at 21:00
Thanks i got exams this week(no f's please crossing fingers) –  Bored915 Dec 19 '12 at 0:03
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