# Tag Info

34

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 ...

13

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 ...

8

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 ...

7

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 ...

7

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 ...

7

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

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 ...

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

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 ...

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

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 ...

5

Relativistic effects are those that disappear in the non-relativistic approximation $1/c\to 0$, usually small corrections to the non-relativistic approximate results that are proportional to $1/c^2$ or higher powers of the inverse speed of light. Let me correct a typo: "cannot account for GR" should have read "cannot account for the special theory of ...

5

Ne is used, Because it caused the red glow inside the tube, infact you can get a whole array of colors using different noble gases. eg. Ne => Red, Xe => Whitish Blue, Ar => Blue etc. Check Wikipedia for more. Because even when it exist as plasma, it doesn't react with the filament inside the tube or the glass walls, this helps in the longer life of lamp, ...

5

In order to find the possible ways of an how an electron acts in the presence of a proton we solve the Schrodinger equation with a coulomb potential, $\frac{k q}{r}$. At the outset of solving an equation, from a strictly mathematical viewpoint, the equation you are solving might have no solution, 1 solution or infinitely many solutions, and also the ...

5

What do we actually mean when we say that matter is a wave? We mean that particles like electrons and photons exhibit wave-like phenomena, such as superposition and diffraction. This is more properly known as wave-particle duality. The thought experiment that best illustrates this is the double-slit experiment, in which electrons behave as waves while ...

5

It's not often that dmckee and I differ (mainly because he's usually right :-) but we differ on this on. Or at least we differ if I've correctly understood what you're asking. In a hydrogen atom the 1s, 2s, etc wavefunctions are (subject to various approximations) good descriptions of the single electron and have well defined angular momentums. In ...

5

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 ...

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

Is there a more sophisticated answer than "This is done because isolated low-temperature systems tend to give up energy more often than they receive it, and thus, gravitate toward their ground states. Thus, even if they start in a mixed energy state, the system will radiate away energy until it reaches its ground state, which is obviously an energy ...

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 ...

5

Three countries signed a treaty banning all nuclear tests but underground ones in 1963 (https://en.wikipedia.org/wiki/Partial_Nuclear_Test_Ban_Treaty ), partly for ecological reasons. EDIT (08/3/2013): The official reasons are given in the preamble of the treaty (http://www.state.gov/t/isn/4797.htm ): "The Governments of the United States of America, the ...

5

These statements show great confusion in the concepts of modern physics. I read that the reason solids emit continuous spectra is that they don't have time to let their electrons decay-they are too close together. It is confusing to be talking of time with respect to emissions and you give no link. To start with at the atomic level, in any phase of ...

4

Though @Christoph and @poorsod cover the mathematical concepts, the basic meaning of attributing a wave nature to matter is not emphasized enough. It is not a matter wave in space time, it is a probability wave that is described by quantum mechanics. A probability tells me what are my chances to find the particle at a particular (x,y,z,t) and nothing more ...

4

There's an infinite number of orbitals, and thus an infinite number of possible state transitions. However, the energy of the orbitals asymptotes to a finite value - for example, a hydrogen atom has energy levels given by the formula $E_n = -\frac{13.6\text{ eV}}{n^2}$ (ignoring some very tiny quantum corrections). If you let $n$ go up to infinity, the ...

4

A fluorescent lamp or fluorescent tube is a gas-discharge lamp that uses electricity to excite mercury vapor. The excited mercury atoms produce short-wave ultraviolet light that then causes a phosphor to fluoresce, producing visible light. Taken right from Wikipedia! http://en.wikipedia.org/wiki/Fluorescent_lamp End of life The end of life ...

4

It's a very good question. The electron is described by a wave field which resembles a charge distribution, so it is natural to wonder why it doesn't repel itself and spread out all over. However, the wave is not a classical wave but is quantized, i.e. the energy in a given vibration mode has to come in discrete bundles. One can count how many excitations ...

Only top voted, non community-wiki answers of a minimum length are eligible