# Tag Info

1

The obvious Google search finds various articles on the subject, including this one that has a graph of excitation lifetime against temperature: The lifetimes vary from about 600$\mu$s to about 3ms, so a 5 kHz signal (200$\mu$s) would indeed appear steady.

1

The accepted airglow answer might be technically true, but it does not answer the question! The existence of an additional and very faint source of green light in the atmosphere does not explain the absence of the green light in the sunset sky gradient. I wasn't satisfied with other answers either. The only satisfactory answer I could find is this one. ...

1

Soft x-ray optics typically uses grazing reflections on suitable mirror materials and gratings (e.g. Pt coated optics) or crystals and the detection can be done with scintillators using CCDs or PMTs. Commercially available instruments have approx. 0.2eV resolution (2000 lines) while research grade instruments can achieve much higher resolutions e.g. 30000 ...

1

It is possible to measure wavelengths of light to many decimal places. When you see accurate determinations of atomic energy levels, they were done spectroscopically, looking at absorption or, more commonly, atomic fluorescence. Since $E=hc/\lambda$, one can accurately convert between wavelength and energy. When excited, atoms emit many wavelengths of ...

1

Disclaimer: I do particle physics / cosmology, so this is definitely outside my field, apply grains of salt to this answer appropriately. I think Reference [29] (Lin et al, arxiv reference: 1008.4864) honestly does a better job of explaining what is going on (which makes sense, the impression I get is that 1008.4864 is a foundational paper in this ...

-2

Hmm, interesting! Some of the rock must be superheated enough to vaporize and ionize as the shockwave, plasma, and EMP travel outwards. If it's like a meteorite crater, then the "spherical" radius that the shockwave travels from the source before it dissipates enough to "stop" is the diameter of the resultant hole.

1

Why is not really important, how is. If you ask yourself why then the answers can be many, for example Why does gravity make two masses attract each other? The answer is because it does, what is really important is how and for that you have a first theory, Newton's Law of Gravitation, this theory is only true for relatively small masses (or masses with ...

9

Well, we could say, yes, that is simply how quantum mechanics works. But these are not the axioms of quantum mechanics, and the exclusion principle in particular is really only understood in the context of quantum field theory. The electron does not "spiral in" because it doesn't move in the classical sense at all. At the scale of the size of an atom, ...

30

There is an interesting diagram in the wiki article on underground nuclear testing - the picture file is here This shows that the crater you get from a nuclear explosion depends on the depth of burial: I think the most interesting diagrams are the ones labeled (e) and (f) - where the explosion happens at great depth. In that case, you get a "tight ...

6

I would think that there is a cavity after the explosion, as well as - Rock is compressible, and the surrounding rock is squeezed together - There is a bulge on the surface, it's just too flat to be noticeable I don't think the rock would be porous since that would allow radiation to leak out of the site. Added: From Wikipedia: When testing went ...

-3

If we calculate the energy required to assemble an electron from a thinly spread out cloud of charge as e, then e is almost equal to the rest mass of the electron. (This is taking into account the experimental size of the electron from collision experiments and uniform charge density in the electron). Though this information does not explain the stability ...

1

There is no such thing as "proving a postulate" theoretically, because your proof will always need to start from somewhere. What you can hope for is to show that a set A of postulates, which seems clunky and ad hoc, actually follows from a simpler set B of postulates. The clearest example of this is when Einstein showed that the Lorentz transformations, ...

2

As another side note, if you know the ground state, it IS possible to use the variational method to get the next-lowest energy. You simply restrict your trial wavefunctions to be orthogonal to the ground state, and then minimize as usual. If you know the two lowest ground states, you can use the variational method to find the third-lowest energy, etc. Also, ...

1

This is just an additional remark – the answer by NowIGetToLearnWhatAHeadIs is fine. But you can get all states from the variational method, just not with the trial function method (unless your trial function space contains the exact excited states). While the excited states are usually not extrema of the variation functional, they are stationary points! ...

2

1. Why can you only get the ground state? According to the wikipedia article linked to, the variational approach works by taking a wave function with some parameters (for example, a gaussian with mean $\mu$ and standard devation $\sigma$) and minimizing the energy with respect to these parameters. The minimum energy parameters give you a wave function that ...

2

Laser modes are the eigen-modes of a laser resonator: only specific distributions of electro-magnetic field can "resonate" in each particular resonator. Due to the 3D nature of our space, each mode is described by 3 numbers, or indices, $m$, $n$, $q$. The latter is the longitudinal mode number, and is easy to understand: to form a standing wave, the ...

3

Yes it's possible, and it's called (unsurprisingly) two photon emission. However the probability of two or multi photon emission is generally lower than for single photon emission by several orders of magnitude, so it's hard to observe.

1

This is going to take a bit, so be prepared. Also, some of what I say will not be exactly correct, in order to get the larger point across, but bear with me. First, the Rutherford model really did not speak to the question of emission lines. It simply noted that, on the basis of Rutherford's scattering experiments, all of the protons in an atom had to be ...

0

Hydrogen is a stable atom and its spectrum is discontinuous and was fitted mathematically by the Rydberg series. Classical electromagnetic theory predicts that the electron, due to its continuous acceleration on the orbit, would emit photons continuously and would drop to a lower energy until it fell on the proton and neutralized it. To explain the ...

1

Distinct spectral lines (rather than an infinite range of energies and hence white light) are a result of only specific transitions being allowed, which implies a set of distinct and unvarying energy levels. Classical models of the atom before quantum mechanics couldn't explain why there should only be certain electron energy levels.

0

The numbers you quote are the X-ray fluorescence lines, i.e. the K, L, M and N lines, not the energies of the electrons in an iridium atom. For example the first energy, 76.1 keV, is the $K_\alpha$ line and corresponds to the energy change when an electron falls from the $2s$ level into the $1s$ level. That is, the X-rays eject an electron from the $1s$ ...

1

In the real world transitions are rarely (never?) forbidden because the assumptions we make rarely hold exactly. You are quite correct that the 21 cm transition has $\Delta\ell = 0$ and is therefore forbidden. However it can occur (very slowly) as a magnetic dipole transition. The lifetime of the excited state is around $10^{15}$ seconds, and we can observe ...

0

The key in your mercury diagram is the phrase "a few energy levels". In the hydrogen diagram, the lines were labeled with "n-1, n=2" etc; the mercury lines were labeled only with letters. You could write down the energy level of all possible states of all electrons in all orbits - you would find that there are many of them. However, just because an energy ...

0

The energy levels can be found in the solutions of the time-independent Schrodinger equation for a system with a harmonic oscillator. For atoms with more than one electron, the situation becomes complicated: the Hamiltonian operator depends on a potential that has to take into account the relative position of not only the electron to the nucleus, but also to ...

1

Considering the way matter waves are associated with all moving particles, it seems inconceivable to me that electrons cannot move in other than elliptical orbits. Close examination of the harmonics & resonance effects of phase waves & matter waves, it becomes apparent that they need to move in elliptical orbits in order to harmonize the way they do. ...

0

Yes, you can be sure because both $S = 1/2$ and $S = 3/2$ are fermions that must respect the Pauli exclusion principle.

1

If we write: $$\psi_+ \propto \frac{r}{a_o}e^{-r/2a_0}\sin \theta e^{i\phi}$$ and: $$\psi_- \propto \frac{r}{a_o}e^{-r/2a_0}\sin \theta e^{-i\phi}$$ then because any superposition of solutions is also a solution we can write the solutions: $$\psi_{px} = \psi_+ + \psi_-$$ and likewise: $$\psi_{py} = \psi_+ - \psi_-$$ which gives us: $$... 1 With such a two-electron spin wave function both electrons cannot be in the 1s state due to the electron parity (the wave function should be antisymmetric relative to exchange of two particles). When both electrons are in the orbital 1s, the only possible spin function reads$$\frac{1}{\sqrt{2}} \left( |\uparrow \downarrow >-| \downarrow \uparrow> ...

1

Even in the classical model, an infinite amount of levels doesn't necessarily mean that it occupies an infinite amount of space. You can divide any finite distance into infinitely many bits (for instance, $1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots$). EDIT: I'd forgotten about the $r\sim N^2$ relation that the OP mentions below, so yes, although ...

3

The mass of an atom is always less than the sum of the masses of the particles that compose it. The lack of mass (or energy, from E = mc^2) is called binding energy and it is the energy expended by the particles to remain confined inside of the atom. When fission occurs, not more spending of energy to hold together the individual particles. So the energy ...

1

Apparently, the CANDU reactor can accept a variety of fuels, including what would be considered "waste" from other reactor types, although some amount of reprocessing is involved.

18

A lot of different forms, but mostly kinetic energy. A good table is given at Hyperphysics. The energy released from fission of uranium-235 is about 215 MeV. This is divided into: Kinetic energy of fragments (heat): ~168 MeV Assorted gamma rays: ~15-24 MeV Beta particles (electrons/positrons) and their kinetic energy: ~8 MeV Assorted neutrons and their ...

3

The energy that is released when a the absorption of a neutron causes a heavy atom nucleus to fission into two daughter nuclei comes from the tighter binding energy of the two daughter nuclei compared to the weaker (smaller) binding energy of the original nucleus. This extra energy is mostly released in the form of the kinetic energy of the two daughter ...

0

In an esoteric sense, the helium has enough components to be helium but there isn't quite enough room, due to the surrounding pressures. The variable density of electrons is 'simply pressuring the helium' so after the helium is sufficiently annoyed ( and starts producing the cyan output) increasing the density doesn't increase the helium's output, it's only ...

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