# Tag Info

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It is a very interesting question that allows to point out the differences between a Neutron Star and Nuclei. Although the dedicated article in Wikipedia Neutron Star fully covers the information, it is relevant to summarize here the elements. Nuclei are essentially different to Neutron Stars and some reasons are: Different bounding force: while Nuclei ...

8

The photon couples to all particles with electric charge or magnetic moment. This includes all of the quarks, the charged leptons $e,\mu,\tau$, and their antiparticles. It also includes particles composed of quarks and charged leptons: the proton and neutron (though the neutron only magnetically), the charged mesons, etc. Many electrically neutral mesons, ...

7

The nuclei of heavy elements (lead, gold, ...) approach the asymptotic density of extended nuclear matter (and therefore the density of neutron stars). The lighter elements do not. That said, it would be an error to refer to nuclei as "miniature neutron stars" because the binding force and dynamics are different. Nor are nuclei protected, shielded or held ...

6

The new particles are baryons, and baryons are composite particles made up from three quarks. So the new particles aren't fundamental in the way that the Higgs is. To make an analogy, suppose physicists discover a new element. That's interesting, but like all atoms it's made of electrons, neutrons and protons. So the new element is just another way of ...

5

Does the fact that protons and neutrons have larger mass than electrons mean they're bigger in size? No. The electron and muon are both believed to be "point-like" (which really means smaller than we can measure" despite having $\frac{m_\mu}{m_e} \approx 200$. That is not to say the proton isn't bigger---it is---but that mass does not imply size in any ...

4

To begin with electrons are not composite. It is baryons and hadronic resonances that are composites of quarks. Hadrons are held together by the strong forces between quarks. These forces, in contrast to the electromagnetic ones which fall with distance as 1/r^2 (and thus allow us to detect free electrons, whose potential falls like 1/r), they behave ...

4

You apply a (net) force (i.e. push it). Recall that the generalized version of Newton's 1st law is that force is proportional to the rate of change in momentum: $$\vec{F} = \frac{\mathrm{d} \vec{p}}{\mathrm{d}t} \,,$$ or in the language of impulse ($J$) $$\vec{J} = \Delta\vec{p} = \langle \vec{F} \rangle \Delta t \,,$$ with $\langle \rangle$ meaning ...

4

The pion does indeed annihilate into photon pairs. But it is an EW process, so the lifetime is large and the pion is long lived. Actually, setting EW couplings to zero the pion would be stable since there would be no lighter hadron it could decay into.

4

I can't give a complete answer because it seems there is still some research ongoing. Unlike what most people have been taught, water is not colorless. At least, large masses of water will be seen blue, such as the sea or a swimming pool. (Left: tube if filled with (light) water. Right: empty tube.) The fact is that water absorbs mostly the red ...

3

Perhaps I do not understand the question. When, for instance, a photon is observed in a state of circular polarization it is simultaneously in a superposition of linear polarization states. Every pure quantum state $\psi$ is always a coherent superposition of other quantum states eigenstates of observables which are not defined in the state $\psi$. A ...

3

The proton and electron exchange information via a gauge boson, in this case, a virtual photon. This is how the electromagnetic interaction is mediated. As for your other question, the electron will get decelerated and deflected and emit a photon, releasing some of its energy in a process called Bremsstrahlung

3

I think you are correct in being confused. The earth's magnetic field, or any external to the atom magnetic field , can distort orbitals but is not the creators of them. Orbitals are the locus in space where the probability of finding an electron is large enough to be measurable. In the quantum mechanical framework orbitals play the role orbits have in ...

3

The good thing about this discovery is that those particles were predicted by the standard model but never measured before. So it is, yet again, another good news for the standard model, it seems to work perfectly. In science anyway you are more exited when you discover something that you didn't expect. In this sense the Higgs boson was unusual: we were ...

2

But im struggling to understand where exactly the potential energy is stored in mass. It is largely in the binding energy of the protons and neutrons in the matter. This represents the vast majority of the rest mass of the nucleons. However, the fundamental particles also have a non-negligible rest mass themselves. This is true for the quarks that ...

2

The term "element" is reserved for atoms that have a nucleus that is a combinations of at least one proton and optionally one or more neutrons. Also, only a difference in the number of protons makes a nucleus considered that of a different element. Changing just the number of neutrons only makes a different isotope. Changing the number of electrons is ...

2

To start answering this question let us talk about a diatomic molecule. In that case we can define the internuclear axis as the z axis in the molecular frame, and talk about the orientation of the various atomic orbitals with respect to that. There is a natural choice of orientation along the bond. For an isolated atom there is no preferred z axis to ...

2

Orbitals (s, p, ...) are special wave functions $\psi$ that are used to approximately describe one electron in spherically symmetric potential due to nucleus (and for many-electron atoms, also due to other electrons). $\psi$ is a mathematical function that is useful in the process of finding some interesting numbers, like expected average electric moment of ...

2

On the electron 'cloud' pictures: what is usually shown is a surface where the probability (per unit volume) of finding an electron (the magnitude of the wave function squared) is constant. 'Inside' the volume, the probability (per unit volume) is higher in this case, so indeed these shapes show the volume where the probability of finding an electron is ...

1

C: See "Quantum Mechanics". More specifically you're wanting solutions to the Schrodinger Equations that represent your system. In this case that is an electron represented by a probability density function $$\psi (\vec{x},t )$$ under a potential $$V(\vec{x})$$ which is the potential energy function of the system. This contains information regarding the ...

1

it is not just that every transition would result in the change of spin this occurs only some times which is explained below Electron Spin The Pauli Exclusion principle states that two electrons in an atom cannot have the same four quantum numbers (n, l, ml, ms) and only two electrons can occupy each orbital where they must have opposite spin states. These ...

1

You're misinterpreting the separation energy $S_\mathrm n$. If you wanted to start with $^{236}$U and end up with $^{235}$U and a neutron at rest, you'd have to add 6 MeV. When the neutron is captured, it's into some nucleon orbital 6 MeV above the $^{236}$U ground state; as the excited nucleus cools that energy gets distributed among all the nucleons. ...

1

Specific charge is indeed the ratio of charge and mass, but since an atom is made up of neutrals and charged particles, you need to account for them. Thus, you'd use $$\eta=\frac{q\left(n_p-n_e\right)}{n_pm_p + n_nm_n + n_em_e}$$ where $\eta$ is the specific charge (my own variable, don't believe it's standard), $m_i$ is the mass of $i$ (neutrons, ...

1

For example, how many quarks are in my brain(easy to find out once you know how many atoms there are)? Actually it's easier to count how many atoms are in your brain than how many quarks are in your brain. As you may know there are three quarks per nucleon in your brain... but this is not the whole truth. The force the binds quarks together creates a ...

1

If one uses the sphere volume as $\pi^{n/2}/\frac n2 !$ and then apply stirling's approximation to $n! = (n/e)^n$, then one gets $(2.\pi.e/n)^{n/2}$ as the approximation for the volume of an n-sphere. This then equates to a cube of edge $\sqrt{2\pi e/n}$. Of course, one can use by way of limits, that even a cube of side $1/2$ must 'go to zero', even ...

1

Quantum physics is a fascinating place, sometimes likened to a zoo, full strange noises and unexpected surprises! If you take a tour of the subatomic zoo, you will see that there are many, many subatomic particles, not all of which are considered 'fundamental' or 'elementary' within the current theory of particle physics (known as the Standard Model). In ...

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