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19

Because the electrical force on an electron is around 10^39 times that of gravity. Given the equivalence between gravitational and acceleration forces, you would have to shake it quite hard. Before you got to the point where an electron would drop out the entire material would disintegrate and all kinds of other phenomena would take precedence over you ...


14

Because there is an energy barrier between the metal and vacuum. Consider the ions in the metal as a uniformly distributed positive charge. Near the metal surface, the free electron wave function spread out a little into the vacuum, thus near the surface of metal the electric dipole forms with the electric field points to vacuum. Thus a gradual potential ...


10

Like so many of his colleagues, Michio Kaku has written yet another "layman" book that you can safely dispose of without feeling bad. Modern physics doesn't deal in "electron clouds" any longer. That's an 80 year old paradigm that has outlived its usefulness. Today we are talking about quantum fields. A quantum field (more precisely THE quantum field) is an ...


8

The electrons are still inside the stars. A stellar plasma is electrically (almost) neutral, the electrons in a plasma are simply not bound to individual nuclei. If we could take some plasma out of a star and we would let it cool down to room temperature, most of the matter in that gas (at least from main sequence stars) would be ordinary neutral hydrogen ...


7

The thing is that not the electron is at all places at one time, but the probability to measure the electron at a certain time isn't localized at a certain point as it is with classical mechanics. The wave function of which the modulus squared gives the probability density of the electron being at a certain place is at first not much more than a mathematical ...


4

The electron itself doesn't "know". The entire system "knows" what its own energy levels are. That is, the system can make transitions only to states that exist. Don't think of the electron, think of the entire system that contains the electron. The system will stay at the excited level for some period of time depending on the strength of the coupling ...


4

Yes, the probability for the electron to be found inside the nucleus is, for some atomic orbitals, non-zero. However, you must recall that these orbitals usually assume a point charge for the nucleus, and so they may not be a valid when you "zoom in" to the nucleus. Nevertheless, there's nothing inherently wrong with the electron being where the proton is - ...


3

From Maxwells equations we know that accelerated charged particles emit em waves. This can be seen from the electromagnetic wave equation, where a second derivative of the electric field is involved. If the second time derivative of the charge density is nonzero, also the second time derivative of the electric field is non vanishing and electromagnetic ...


3

Yes, they can be. The production of neutrons by deep inelastic scattering of electrons on protons was studied in HERA experiments in the 90's. No neutrino needed, at least not as one of the colliding particles. Here are some links: Deep-Inelastic Electroproduction of Neutrons in the Proton Fragmentation Region Measurement of Leading Proton and Neutron ...


3

There is no notion of quantization of charge in classical electrodynamics. Charge is a continuous, infinitely divisible quantity there, and there's nothing at all that would indicate what carries the charge. The electron (or any other particle, for that matter) is not predicted by classical electrodynamics, and thus none of the classical notions of ...


3

This is a hypothetical question, since electrons are elementary particles and protons are composite. The solutions of the potential problem would give stable orbitals with smaller average radii. Here is a Bohr model solution for the muonic hydrogen, where the muon is 200 times heavier than the elecron. The energies become KeV instead of eV. To go to the ...


2

If you take two protons, they have a mass of about $3.34\times10^{-27}$ kg. The "Schwarzschild radius" of a two-proton black hole is given by $2GM/c^2 = 5 \times 10^{-54}$ m ! i.e. you have to get the protons closer together than this. Even ignoring that protons have a radius of $10^{-15}$ m and that there would be an enormous repulsive strong nuclear force ...


2

Gamma decay is the emission of photons. You are thinking of $\beta$ decay. When the particle is decaying, if it emits a $W^{-}$boson, it will subsequently decay and create an electron ($e^-$), and an electron antineutrino ($\overline{v}_e$), the antimatter particle to an electron neutrino. It will also flip a neutron into a proton. This is known as ...


2

Though I agree with the logic, I find I cannot reproduce this quantitatively. I get an electron number density of $1.2 \times 10^{41}$ m$^{-3}$ (is it just a unit thing?) for a Fermi energy of 30MeV. In a carbon white dwarf with 2 mass units per electron, the Fermi energy of the electrons reaches 30MeV at densities of $4\times 10^{14}$ kg/m$^{3}$ - i.e. at ...


2

Each electron has a fixed charge of $-1.602\times10^{-19}\,\mbox{C}$ (where (C stands for coulombs). If you gather $6.24\times10^{18}$ electrons, the total charge will be \begin{align*} \mbox{Total charge} &= \mbox{Charge per electron}\times\mbox{Number of electrons}\\ &= ...


2

Because of uncertainty, the electron does not exist at any single point, but exists in all possible points around the nucleus. It is all a matter of interpretation. HUP: \begin{align*} \sigma_x\sigma_p &\geq \frac{\hbar}{2} \end{align*} states that you cannot know both position $x$ and momentum $p$ exactly, at an instant. OK, say you want to ...


2

The electron in the $n$ semiconductor and the hole in the $p$ type semiconductor are delocalised and not bound to any particular atom, so arguments based on completing octets aren't useful. This diagram shows roughly how the depletion layer forms: In the $n$ type semiconductor the doping creates donor states in the band gap, and electrons from these ...


2

The drift velocity is the average velocity due to an applied electric field. In a conductor, electrons scatter around at the Fermi velocity but have a net zero average (i.e., equal scattering in all directions). When the electric field is applied, the electrons are given a small velocity in one direction. Thus, we can say, $$ v_{drift}=\eta E $$ where $\eta$ ...


2

The safety of a low voltage DC power supply is not established by the voltage on its output, but by the isolation between its input and output terminals. For example, a defective 12V power supply may have a short between the 120V AC input terminal and its negative output terminal. A user who would be connected to ground would then experience a 120V AC ...


2

The "volume in which it is contained" is an ill-defined notion for quantum objects. Consider the simplest atom, the hydrogen atom, and look at the wavefunctions for the electron states. You find that, roughly, $$ \psi(r) \propto \mathrm{e}^{-r}$$ so the probability to find an electron at a certain distance $r$ from the nucleus decreases with increasing ...


2

1) The mass of the electron is 0.5 MeV, so, for most pratical purposes, one can ignore the mass when the energy is at 1.2 GeV, the difference, most of the time, would be about 0.05%, which is usually considered small. If there is mass, not all energy is kinectic energy, you are right about that. 2) Depends on the accelerator design, but it's possible to use ...


2

"If I am not wrong, the passage says electron (not the parts of electron) can be found at many places at the same time." An electron exists at many points at once, but we may only find it at one. That distinction is the paradox at the heart of quantum mechanics. An experiment that explains this is the double slit experiment. When we fire electrons through ...


2

The following passage has been extracted from the book "Parallel Worlds-Michio Kaku": Because of uncertainty, the electron does not exist at any single point, but exists in all possible points around the nucleus. This electron “cloud” surrounding the nucleus represents the electron being many places at the same time. You say :" If I am not wrong, the ...


1

The electron is not a particle. It's not a wave either. The fact that it seems to behave like one of those under certain conditions is pretty much irrelevant. However, to understand this on a gut level, with the case of an electron in an orbital of an atom, it is very helpful to imagine the electron as a wave. The electron still has just the properties of ...


1

Low voltage sources don't have enough potential to conduct through your skin or body so touching either the positive or negative doesn't make a difference. For you to feel it or get tissue damage, current must flow through you. This won't happen with very low voltage sources unless you're covered with something more conductive like wet salt water. But 12V ...


1

Since the book is named "Parallel Worlds", it would not be inappropriate to give the Many World's Interpretion's view on this. A superposition of an electron being at different places should be interpreted as the electron being at all these places in different Worlds. So, there exists a World where the electron is in one position (but note that technically ...


1

In the quantum mechanical description of a conductor all energy levels of the conductor are filled up to some specific energy level, called the Fermi level. This is because of the Pauli exclusion principle, which says that electrons with the same spin cannot occupy the same energy level and thus causes higher energy levels to be populated. Therefore, even at ...


1

"Free" electrons and protons can combine to form neutrons (and electron neutrinos) through weak interactions. The process would usually be energetically unfavoured in low density situations because the rest mass of the neutron is about 1.29MeV higher than the combined rest mass energies of the proton and electron. This reaction, between free protons and ...


1

Is it possible for electrons to carry more than one charge? If by, one charge, you mean more electric charge than (the negative of) the elementary charge, the answer is no. More specifically, an 'electron' would not be an electron if its charge were not $-e$. However, electric charge is not the only type of charge electrons 'carry'. But this is a ...


1

Theoretically, yes. Practically, no. In order to create a singularity you would have to compress mass so that it would have zero volume, and thus an indeterminate density. We simply do not have the capabilities to produce such results on our own. Edit: Density would approach infinity as the volume approaches zero.



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