# Tag Info

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Do electrons change orbitals as per QM instantaneously? In every reasonable interpretation of this question, the answer is no. But there are historical and sociological reasons why a lot of people say the answer is yes. Consider an electron in a hydrogen atom which falls from the $2p$ state to the $1s$ state. The quantum state of the electron over time ...

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Electrons do not move inside atoms. If an electron is in a given energy level $E$, the wavefunction is given by $\psi(x,y) = \phi(x)_{n\ell m} \,\mathrm{e}^{-\mathrm{i}E t/\hbar}$. The time dependence is a pure phase factor, hence the real-space probability density of the electron is $|\psi(x)|^2 = |\phi(x)|^2 \neq f(t)$, not a function of time. These are ...

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An animation is worth a million words:

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In hydrogen: It incorrectly predicts the number of states with given energy. This number can be seen through Zeeman splitting. In particular, it doesn't have the right angular momentum quantum numbers for each energy levels. Most obvious is the ground state, with has $\ell=0$ in Schrodinger's theory but $\ell=1$ in Bohr's theory. It doesn't hold well ...

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There are a few reasons why the particle produced needs to be a photon. Aside from conserving energy, we also need to conserve momentum, charge and spin, for example. So you would need to ask what other particle, instead of a photon, could be emitted while satisfying all those conservation requirements. If you just consider energy and spin conservation, ...

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The probability of finding the electron in some volume $V$ is given by: $$P = \int_V \psi^*\psi\,dV \tag{1}$$ That is we construct the function called the probability density: $$F(\mathbf x, t) = \psi^*\psi$$ and integrate it over our volume $V$, where as the notation suggests the probability density is generally a function of position and sometimes ...

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They are "fuzzy" in position space because they have well defined energies (and therefore give rise to sharp distributions of photon energy when they change states). Foundation Quantum states (the class to which orbitals belong) are elements in a Hilbert Space (which physicist sometime call "infinite dimensional vector spaces" just to watch the ...

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The Sommerfeld model, and the Bohr model from which it is derived, are toy models developed in an attempt to describe spectral lines in the era before modern quantum mechanics. You might be interested to look at the question Is it possible to recover the old Bohr-Sommerfeld model from the QM description of the atom by turning off some parameters? for more on ...

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Polyelectronic atoms don't have atomic orbitals - though they are a very useful approximation for describing the properties of polyelectronic atoms. The 1s, 2s, etc orbitals are solutions for a central potential, and for any smooth monotonic central potential we'll get solutions of this form. The radial part of the orbitals will be different for different ...

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Any answer based on analogies rather than mathematics is going to be misleading, so please bear this in mind when you read this. Most of us will have discovered that if you tie one end of a rope to a wall and wave the other you can get standing waves on it like this: Depending on how fast you wave the end of the rope you can get half a wave (A), one wave (...

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Orbitals are solutions to time-independent quantum wave equations. That is, there is no time-dependence. There is no little ball in there moving around, the electron has a quantum characteristic and exists with neither a well defined position nor a well defined momentum.

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There is always energy loss due to recoil. Considering a single atom that is in free space, the total momentum is conserved, and the total reaction energy is constrained by the transition energy $E_0$ (ignoring the natural line width due to the energy-time uncertainty), so we get in the centre of mass frame: $$0 = \hbar k + mv$$ $$E_0 = \frac 1 2 mv^2 + \... 24 There are no final models in science, there's always room for improvement. And major paradigm shifts cannot be totally ruled out. However, we can be quite confident in our current model of the electronic structure of the atom, which is based on quantum electrodynamics (QED), which has been validated to very high precision. Wikipedia has numerous orbital ... 21 The answer is thermodynamics, and the assumption that you're working in a colder environment than the temperature corresponding to a Planck distribution where your photons would be "on average" fairly present. In other words, inside a star, where it is hotter, the atoms are NOT in their ground state most of the time - in fact, if it is hot enough, they are ... 21 This is an interesting and non-trivial problem. Basically the Coulomb potential assumes a point particle but, if the proton is modelled as a solid sphere of finite radius, part of the electron wave function would be "inside" the proton, where the assumption of point charge no longer holds. To account for this one must modify the Coulomb potential from 1/... 19 A few years ago the XUV physics group at the AMOLF Institute in Amsterdam were (to my knowledge the first to be) able to directly image the orbitals of excited hydrogen atoms using photoionization microscopy. For more details see the paper, Hydrogen Atoms under Magnification: Direct Observation of the Nodal Structure of Stark States. A.S. Stolodna et al. ... 19 Generally speaking, atomic and molecular orbitals are not physical quantities, and generally they cannot be connected directly to any physical observable. (Indirect connections, however, do exist, and they do permit a window that helps validate much of the geometry we use.) There are several reasons for this. Some of them are relatively fuzzy: they present ... 19 A "shell" is the term for all states with the same principal quantum number n, but in each shell there are also possible different values for the angular momentum quantum number 0\leq \ell \leq n, the magnetic quantum number -\ell \leq m_\ell \leq \ell and the spin quantum number m_s\in\{-1/2,1/2\}. So for n>1, 6 electrons in a shell do not ... 18 First of all, note that different authors disagree on what should be the Coulomb potential V in d spatial^1 dimensions. We will assume that it satisfies Gauss's law, i.e.$$\tag{1} V(r)~\propto~\left\{\begin{array}{rcl} r^{2-d} &\text{for}& d~\neq~ 2, \\ \ln(r)&\text{for}& d~=~2. \end{array}\right.$$We will here only discuss the ... 17 We describe the whole system with a state, this state is a combination of the single particle states (orbitals). Each orbital we define in terms of an orbital momentum shell. A full shell has zero total angular momentum, therefore multiple full shells still have zero total angular momentum. Finally a full shell combined with a few valence electrons in higher ... 16 Hints: Heavier isotopes have higher reduced mass \mu. For a hydrogen-like atom, the energy levels E are proportional (& the radius r is inversely proportional) to the reduced mass \mu. More generally, it can be deduced from dimensional analysis alone that the conclusions of pt. 2 hold for any spinless non-relativistic multi-electron atom ... 16 This is an example of the "correspondence principle" in the broadest sense, that new theories should explain why old ones got some things right. The linked article discusses the Bohr model, but leaves some of your sub-questions unanswered. Going beyond that, how does an "electrons are somewhere specific" approximation lead to useful models of sharing and ... 15 The first images of hydrogen s orbitals were obtained in 2013 by physicists in the Netherlands. 15 I want to add about spontaneous emission. Excited states of atoms are not stationary states because of atoms are not isolated QM systems. There always is interaction with electromagnetic field. Schrodinger equation for atoms in simplest form takes into account only Coulomb interaction between electrons and nucleus. In this simplified approach excited states ... 14 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 ... 14 You're right on a lot of counts. The wavefunction of the system is indeed a function of the form$$ \Psi=\Psi(\mathbf r_1,\mathbf r_2), $$and there's no separating the two, because of the cross term in the Schrödinger equation. This means that it is fundamentally impossible to ask for things like "the probability amplitude for electron 1", because that ... 14 This will be a purely mathematical treatment. It needs to be combined with some practical playing around to really "get" it. Traveling wave Let's start with the description of a harmonic traveling wave in one-dimension. Here "harmonic" just means the mathematical form of the wave is sinusiodal in both time and space. For concreteness we'll using talk ... 14 The parallel between the Bohr picture and the Lewis diagrams isn't that great if you consider that the electron is moving in the Bohr model, while the electrons are static in a Lewis diagram. If a Bohr electron was "at rest" outside a nucleus, as it is in a Lewis diagram or one of your organic-chemistry diagrams, it would immediately accelerate towards ... 13 Well, the wave function of the electron in the ground state of a hydrogen atom (and very similarly in other atoms) behaves like$$ R(r) \sim \exp(-r / a)  where $a$ is the Bohr radius, effectively the radius of the atom. The exponential is in principle nonzero for an arbitrarily large $r$, so the electron may be found arbitrarily far from the nucleus at a ...

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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 ...

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