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92

Things are not empty space. Our classical intuition fails at the quantum level. Matter does not pass through other matter mainly due to the Pauli exclusion principle and due to the electromagnetic repulsion of the electrons. The closer you bring two atoms, i.e. the more the areas of non-zero expectation for their electrons overlap, the stronger will the ...


37

Why do most metals appear silver in color, with gold being an exception? It is hardly surprising that the answer to this question relies heavily on quantum theory, but most people will be surprised to hear that the full answer brings relativistic considerations into the picture. So we are talking quantum relativistic effects. The quantum bit of the story ...


15

That's a really good question! There are three cases, the third of which is the most fundamental and most interesting. The first case is incomplete absorption, such as a gamma ray knocking loose a few electrons as it passes. In that case the differences are taken care of locally and fairly trivially by allocating energy, momentum and spin appropriately ...


12

D electrons in metal allow optical transitions in the visible regime. Visible light can be absorbed by elements, having unbound valence electrons in d shell. So Chemistry: optical d->s$^2$ transition Iron [Ar] 3d$^6$ 4s$^2$ Tin [Kr] 4d$^{10}$ 5s$^2$ 5p (full d shell) Aluminium [Ne] 3s$^2$ 3p$^1$ (is a special case: no d valence electrons, but Aluminium ...


11

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 ...


10

For me the physical implication of zero angular momentum is that the electron's probability distribution is spherically symmetric. At the deepest level, the angular momentum property in quantum mechanics describes how something transforms under rotations (see Noether's theorem). Although this is quite an abstract property, in the case of electron orbitals it ...


10

There is no structural similarity between an atom and the universe. Atoms are usually described as bound states of a quantum system, while the universe is usually described by general relativity, two very different and currently even incompatible approaches.


10

In any proper quantum mechanical understanding of the atom, a bound electron does not have a position and follow a path (i.e. have a time-varying position) in the sense that it would have in a classical or semi-classical theory. Instead the electron "has a state" or "occupies an orbital" (an orbital not a orbit!), and because there is not a path there is ...


9

Here is an experimentalist's answer: You state: The probability of a photon having just the right amount of energy for an atomic transition is 0. You must be aware that the statement falls just by the existence of lasers, so your question should have a how is it possible to have lasers. 1) An individual photon cannot be labeled as continuous. It has ...


9

The Darwin term can be obtained from the low energy approximation ($|p|^2/m^2<<1$) of the Dirac equation of the electron in a central field. An elegant way to perform this task is by means of the Foldy-Wouthuysen Transformation. The same approximation leads also to the other terms detected in the Hydrogen atom fine structure including the spin-orbit ...


9

It most certainly exist outside secret labs :) Like Gerben wrote, the fields are called molecular dynamics (MD) and quantum chemistry which, as computers grow faster, will be essential tools of nanotechnology and medicine. Molecular Dynamics is currently implemented by making certain approximations in that electron motion is not explicitely modelled. In ...


9

You should not think a particle as point like. Classically, the probability of two point like particles colliding with random location and velocity is 0, that is why you said it never happens. However, at quantum mechanical level, these particle are described by wavefunction. It means that there is spreading in its spatial location, say 0.1nm (the minimum ...


8

In the rest frame of the atom there are of course no changes, so I assume you're asking what the atom will look like to the stationary observer watching the moving atom. First note that electrons don't orbit the atom like planets orbiting a star. The electrons in atoms exist as a delocalised probability distribution. This distribution can have a non-zero ...


8

The degeneracy of energy-levels can be traced to the fact that the hydrogen atom possesses an enhanced $SO(4)$ symmetry caused by (among other things) the conservation of the Laplace-Runge-Lenz vector operator, see e.g. this Phys.SE post and Ref. 1. References: G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapter 9. The pdf file is ...


8

I think the problem here is with E&M, and it is in the assumptions implicit to the question. Here is the assumption: the electron orbiting around the nucleus has an acceleration. Therefore it radiates and loses energy, until it would collapse with the nucleus. This statement can be demolished in short order, considering the topic of a ...


7

If we take a system, and let it evolve for some indefinite amount of time, it will be in an incoherent mixture of energy eigenstates. Many systems we encounter in nature have been sitting for some time, and not interacting with the environment (much). These can be considered to be in energy eigenstates. For example, suppose we consider an atom in a gas. ...


7

No, it wouldn't change due to the energy conservation law or mass conservation law – these are the same laws in relativity where $E=mc^2$. The nuclear energy extracted from the nuclei by fission would be converted to other forms of energy inside the box – ultimately the thermal energy. High temperature means that particles are moving at higher speeds and ...


7

I would say that one experiment that demonstrates the atomic nature of things is the observation of Brownian motion. But it is not the experiment itself that convinces that things are made of atoms, rather its theoretical explanation given by Einstein in one of his 1905 papers (actually Einsteins work for his PhD was on the subject of atomic theory and there ...


7

As suggested by M.Sameer I convert my comment into an answer: Dear M.Sameer: It seems that you are missing that the $n+\ell$ Madelung rule is not an exact result derived from first principles, but rather a rule of thumb, that holds for, say, approximately 95 percent of all the elements, with important exceptions, cf. this wikipedia page. Nevertheless, ...


7

Arriving at the same answer as quantum mechanics for one particular scenario by making a bunch of ad hoc assumptions (for example - the calculation didn't work, so we'll make the orbital planes perpendicular) isn't useful. QM allows you to calculate much more than the ground states of atoms. Any competing theory - and that paper doesn't contain anything ...


7

If it were possible for one object to pass through another object, then it would be possible for one part of an object to pass through a different part of the same object. Therefore the question asked here is equivalent to the question of why matter is stable. See this question on mathoverflow. That question was more about the stability of individual atoms, ...


7

When one is in the micro level of particles one has to stop thinking classically, i.e. with terms we have developed from our macroscopic observations. "Touch" at the particle level can be defined as "interaction". Our feeling of touch actually does involve the electromagnetic interactions, we touch with the field of the molecules in our hand the field of ...


6

That's a great question! Unfortunately, the only honest answer is "that's what we see in nature, with great precision and complete reproducibility." There is no deep theoretical understanding. The more exotic form of your question is phrased in terms the self-energy of an electron, and it's a question that plagued Nobel Laureate Richard Feynman his entire ...


6

The covalent bonds form when electrons attached to nearby nuclei can exist in a superposition state where they can partly be on another nucleus. This happens when the electron state they are mixing with is unfilled. For example, for H2, two hydrogen nuclei are close, there is no electrostatic energy for this in the first approximation because the electron ...


6

Gold foil is quite easy to hold you just hang it from a paperclip. The only difficulty is if there is a lot of static electricity in the air which makes it stick to things. (This is the main reason for the cold damp Cambridge's supremacy in early particle physics) Photographic film at the time wasn't sensitive and so in Marsden and Geiger's experiments ...


6

The Legendre polynomials occur whenever you solve a differential equation containing the Laplace operator in spherical coordinates with a separation ansatz (there is extensive literature on all of those keywords on the internet). Since the Laplace operator appears in many important equations (wave equation, Schrödinger equation, electrostatics, heat ...


6

We say that a free, unbound electron has zero energy (that's convention, you could just as well put another number there). This means that the level $n = \infty$ is fixed at $E_\infty = 0 \text{eV}$. Since the other levels lie lower, i.e. possess less energy, this forces all other bound states to have negative energies - which then represent that we need to ...


5

Yes, the photon energy has to be equal to the total energy of the atom before the transition (including the kinetic energy) minus the total energy of the atom after the transition (including the kinetic energy). In practice, the kinetic energy of the atom is negligible. The mass of a nucleus is at least 1 GeV/$c^2$ or so (the mass of the proton) while the ...


5

Since you used the tag wave-particle-duality, I imagine you mean the frequency $f$ that corresponds to an electron's energy $E$ via Planck's relation, $$E=hf,$$ where $h$ is Planck's constant. That is a valuable question and nothing to get picked on for. After all, if the electron is a wave with wavelength and so on, it surely has a frequency, right? It ...


5

Optical pumping will at most only achieve equal population of a two-level system. This is because the probabilities for raising an electron to the upper level and inducing the decay of an electron to the lower level (simulated emission) are exactly the same! In other words, when both levels are equally populated, the numbers of electrons "going up" and ...



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