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7

Photons mediate the electromagnetic force. Atoms are not necessary for photons to exist. You just need charged particles (electrons, protons, etc) to interact with each other from a distance. There are many ways for a photon to be created and destroyed. Depending upon its wavelength, as it propagates in free space , it could "disappear" and a pair ...


6

There are several different things that need to be explained / explored here. First - the speed of light in vacuum is independent of frequency / wavelength. The same is not necessarily true for light in any medium other than vacuum: this is why we can see rainbows! Second - not all objects emit "white" light. The emission spectrum of a star depends, among ...


5

At the Large Hadron Collider we have studied matter down to a length scale of about $10^{-19}$ metres, which is about a billion times smaller than an atom. All the results so far confirm our existing theories. So it seems very unlikely that an undiscovered class of small atoms exists. The size of an atom depends on well understood physical principles. At ...


4

The fact of the matter is that there are no stable mesons, that might conceivably form states bound by the strong force, as the nucleus is bound. Within the nucleus there exist virtual mesons, i.e. described as pions etc but not on mass shell To a large extent, the nuclear force can be understood in terms of the exchange of virtual light mesons, such as ...


4

As anna mentioned, there are non quark models which clarify exotic hadrons. In principle, they are allowed in Quantum Chromodynamics (QCD). Non-quark models predict 1.hybrid mesons: Include quark anti-quark pair and gluon. 2.Glueballs: Gluons are their own bound states. 3.Exotic hadrons as in figure below which exchange pion at low energies (couple ...


4

The quote is correct but a bit misleading. The statement "In doing so it also liberates particles known as neutrinos" includes electrons also which are the other particle that is released in neutron decay, and is the way that beta decays were discovered. The neutrino was discovered because neutron decay showed a three body momentum spectrum for the ...


4

Chadwick didn't discover the neutron on purpose, of course. After the discovery of the nucleus by Rutherford in 1911, alpha particles were used to probe its structure. These kind of experiments were pioneered by Rutherford himself (as an example, he discovered in 1917 that the nitrogen nucleus contains hydrogen nuclei, i.e. protons). In 1930, Bothe and ...


3

The kinetic energy of an atom is not a uniquely defined quantity because it depends on the inertial frame. In the rest frame of the atom the kinetic energy is zero. However if by energy of the atom you mean the energy associated with its electron configuration then yes this can have an effect of the interaction between atoms. For example the ground state of ...


3

First of all the $\beta $ particle emitting from nucleus is too energetic to be captured in atomic orbit to form atom.So it is definitely not the case. Infact thta's why we are sure that it is coming from nucleus not from atomic orbital. More over while writing for radio active decay we are looking for change in the nucleus as it is basically a nuclear ...


3

There's no universal notion of "up" or "down", just like there's no universal notion of "left" or "right" in the universe. Speaking of "spin up" and "spin down" involves two completely arbitrary choices - a choice of axis and then a choice of "up" and "down" along that axis. You can swap the meanings of "up" and "down", and you can choose another axis to ...


3

The energy shifts of transitions including the s-orbitals of various isotopes of hydrogen are dependent on the proton's charge radius and are a surprisingly sensitive tool for this kind of thing. Recently this has been checked with muonic hydrogen, with surprising results. Paper at http://dx.doi.org/10.1126/science.1230016, and references therein. Related ...


3

[The figure shown in the OP question above ...] is experimental data for the ratio $R = $ [...] as a function of the centre of mass energy $\sqrt{s}$ The so called cross section ratio $$R[~\sqrt{s}~] = \frac{\sigma^{(0)}[~e^+~e^- \rightarrow \text{hadrons}, \sqrt{s}~]}{\sigma^{(0)}[~e^+~e^- \rightarrow \mu^+~\mu^-, \sqrt{s}~]}$$ ?? Actually, ...


2

$^1_1H$ is a special case as the proton does not have any other nucleons to bind onto. I suppose that you could call the binding energy per nucleon zero which means that you require no energy to split up the nucleus of $^1_1H$ into its constituent parts? Note also that the binding energy per nucleon is not necessarily the full measure of whether a nucleus ...


2

One kind of nuclear response to a photon happens at very high photon energies (like, gamma rays up to thousand-kilovolts range, an order of magnitude beyond X-rays). That is the energy requirement to excite a nucleus above its ground state (you wouldn't want to be in the room when that is happening). The only useful exception, is if an excited nucleus ...


2

'Crossover diagrams' exist when the final products are indistinguishable. When using Feynman diagrams for computations, you need to add all distinct diagrams that depict the same process, i.e., have the same external lines. Both the crossed and uncrossed diagrams for the electron-electron process have the same final products, since you cannot tell the ...


2

The nucleus obeys quantum mechanical laws. It is not the number of combinations that controls the existence of an element (or an isotope of the element) but the number of solutions of the specific quantum mechanical equation that describes the nucleus. This is a many body problem and has been approached by quantum mechanical models as the shell model, which ...


2

Lightning is a large-scale natural spark discharge that occurs within the atmosphere or between the atmosphere and the Earth’s surface. On discharge, a highly electrically conductive plasma channel is created within the air, and when current flows within this channel, it rapidly heats the air up to about 25 000°C. The lightning channel is an example of ...


2

In 1930 two German physicists Walther Bothe and Herbert Becker dicovered that when alpha particles were directed at beryllium some form of ionising radiation was produced, which they thought was gamma rays. Later Irene Joliot-Curie (Marie Curie's daughter) and her husband discovered that this radiation could eject protons from paraffin wax. Chadwick's ...


1

The parton model was proposed by Feynman before the existence of quarks and gluons within the proton was established experimentally: In particle physics, the parton model was proposed by Richard Feynman in 1969 as a way to analyze high-energy hadron collisions. At that time the parton model was set up as a uniform distribution of partons within the ...


1

The reason the number of elements is finite is because of the proton electric charge. For larger atoms the electromagnetic repulsion between protons eventually overcomes the nuclear binding from the strong force and the nuclei become unstable.


1

Probably the best way to think of this, and the way I suspect it works in the minds of people who write exactly this sort of equation in physics textbooks, is that it's an equation describing a nuclear process, rather than a chemical process. The electrons initially bound to the nucleus are irrelevant spectators. A less ambiguous way to describe the ...


1

Maybe you should look up the definition of what an orbital is again. Your question doesn't make much sense. https://en.wikipedia.org/wiki/Atomic_orbital#Shapes_of_orbitals The "Shapes of Orbitals" section should help the understanding. I suppose by orbital you mean an atomic orbital, which is basically nothing else than a designated area in which the ...


1

Oh yes, certainly. Of course there is, by your definition, no way to determine this, even in principle. Since these "ghost atoms" are unobservable, they emit no radiation that we can sense. Unlike dark matter, whose existence we have deduced from gravitational effects, they do not affect us gravitationally. They do not collide with regular atoms, and they do ...



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