Tag Info

New answers tagged

2

They do not need to necessarily collide like balls. I guess the picture in your book is illustrative. Conservation laws apply to any kind of interaction between them. Note that details of the collision are not even provided in the question but you still can calculate the answer. The detailed theory of photon-electron interactions is called Quantum ...


4

Here are real events relating to the last page of the pdf link you gave: Fig.1 This bubble chamber picture shows some electromagnetic events such as pair creation or materialization of high energy photon into an electron-positron pair (green tracks), the Compton effect (red tracks), the emission of electromagnetic radiation by accelerating charges ...


0

Current is movement of charge. In conductors that's usually electrons doing the movement (flowing). One of the things keeping them from moving quickly is that they bump into each other and all the metal atom nucleus's constantly. Secondly, because your battery isn't sufficient, it creates a very limited potential difference. Even if the electrons never ...


1

David Zaslavsky has given a solid, relatively model-independent explanation of the empirical bounds on the size of an electron based on particle-physics experiments that probe short distance scales by using collisions at short wavelengths. There is also another way of getting at this question, which has been studied by people who have tried to model quarks ...


3

A current is nothing than charged particles moving. Since those charged particles also have a mass, it follows that they cannot possibly reach the speed of light. In a real material that conducts electricity, the average net velocity of charges is actually very, very low, because they bump into atoms all the time, which basically sends them flying off in ...


0

Let's try just considering what could happen based on conservation laws. The two electrons have a charge of -2e, so the end product must as well. Lepton number conservation is required also, and we have $L_e=2$ here. At this level, it looks difficult to produce additional particles which satisfy just these two conservation laws. If you work in QED the only ...


4

As far as I know, nobody has ever done this, at least not at what we currently consider high energy. (Electron-electron collisions happen at low energy all the time, of course.) I doubt that anything interesting would happen, primarily because electrons are mutually repulsive, and they have a low mass. That means two colliding electrons would just bounce ...


2

Energy and momentum has to be conserved. That the electron / photon has to have enough energy for the excitation is obvious. What is interesting is what happens when they have too much energy. For radiative transitions between bound states the orbital anuglar momentum has to change by 1. This means that the photon has to be absorbed which in turns means ...


0

The answer is YES, it is possible for current to flow in an open circuit. The only requirement is that the current be "alternating" current. A capacitor is essentially an open circuit, and alternating current will "flow" trough it.


0

This isn't what you're asking, but it is possible for currents to flow in an open circuit. Even a plain old block of copper is an "open circuit", for instance, but you can induce eddy currents to flow in it by applying a changing magnetic field. In a regular circuit this effect would be very small, though, as the copper wires have small width.


2

Think about a 2D Fermi surface (or a 2D section of a 3D Fermi surface). Now look in the extended Brillouin zone - thats the one where you take many copies of the first Brillouin zone and use it to tile the plane. Now we are in 2D so the Fermi surface is a bunch of curves in the extended BZ. There are three possibilities about the shape of the shape of this ...


0

Essentially what you are asking about is the "photoelectric effect". The intensity of the light does not contribute to energy needed for the material to expel a photon. The Minimum energy needed to release a photon is given by the work function $\Phi=hf$ where h is Planck's constant and f is frequency. Here is a flash simulation of what is going on. What ...


1

There is a difference between an electron gaining energy and the whole system (atom) gaining energy. Electron orbital energy levels are quantized and make discreet jumps. For this reason "centripetal force" for electrons doesn't make much sense. There is an electron orbital binding energy though which is equivalent to the "work function" in your question. ...


0

You are quite correct that the enthalpy change to form an isolated Na$^+$ and an isolated Cl$^-$ ion from gaseous Na and Cl atoms is positive. However you need to include the very large and negative enthalpy change when the ions come together to form solid NaCl. A quick Google found this energy diagram: To calculate the enthalpy of formation of NaCl from ...


0

$$Na+Cl\longrightarrow [Na^+Cl^-]\ \ \ \ ...\Delta H$$ $$\text{Includes:}$$ $$Na\longrightarrow Na^+ + e^- \ \ \ \ \ \ ....\Delta H_1$$ $$Cl+e^-\longrightarrow Cl^-\ \ \ \ \ \ \ \ ....\Delta H_2$$ $$\text{Where:}\Delta H=\Delta H_1+ \Delta H_2$$ After some addition I can't see a problem.


2

The the easiest way to see that time reversal transforms electrons into positrons relies on the fact that PCT (parity, charge conjugation and time reversal) combined are a symmetry of every Lorentz-invant QFT. Using $P^{-1} = P$, $C^{-1} = C$, $T^{-1} = T$, i.e. a parity transformation is undone by a second parity transformation etc. you can see that $$1 = ...


1

I think every fundamental definition is kind of going in circle. I would say an electric charge is something that obeys Maxwell's laws. But to write those laws, you have to know $\vec{E}$ and $\vec{B}$ which need a definition of an electric charge. At the end you just group things that look/react alike and named them. The problem arise when you have to ...


1

Practically, for a macroscopic body such as a chunk of metal, the charge on that body is the difference between the number of electrons and protons in the body. It is hard to knock a proton out (can be done, though), but for conductors we can push and pull electrons out by supplying a bit of energy (back to that in a moment). However, instead of saying the ...


1

I would guess the question is asking you what the maximum charge on the sphere is. Suppose the photon energy ($hc/\lambda$) is $E$, then the kinetic energy of the electron leaving the surface (in electron volts) is $E$ - 4.47. As you increase the positive charge on the sphere you increase the work needed to remove an electron to infinity, and for some ...


0

This formula is derived using conservation of charge principle and so it's valid for the superconductors as well. There's a critical magnetic field that above which a superconductor becomes normal conductor and it's a function of temperature. If a large current is to pass through a superconductor, a magnetic field will be produced that disrupts ...


2

A hydrogen atom ion $H^{+}$, with an atomic mass number of A=1, charge number Z=1, is the same as a proton. A hydrogen ion thus usually just refers to a proton. Depending on context, however, you may also have a hydrogen ion which is (a) an ion of a deuterium atom, in which case it is a bound state of a neutron and a proton, with atomic mass number A=2, ...


0

Look at it this way. An electron is a charged particle. A moving, electrically charged particle creates a magnetic field, and the particle itself already has an electric field. If the particle is accelerating, then you're going to have a ripple effect from the electric and magnetic field, or an electromagnetic ripple, ie an electromagnetic wave. So an ...


2

On a uniform circular orbit, even if the speed does not change in norm, it does change in direction so that the speed vector change over time and $\frac{d\vec{v}}{dt}\neq\vec{0}$. In fact, in polar coordinates, you have $$\vec{a} = \frac{d\vec{v}}{dt} = -\frac{v^2}{R}\,\vec{e_r}$$ Imagine a car taking a turn at constant speed: if the turn is left, you feel ...


5

In uniform circular motion: $a=\frac{v^2}{r}$


3

Let us clear some misunderstandings: if an electron absorbs a photon to get exited to a higher energy level, It is not the electron that absorbs the photon to go to a higher energy level. It is the whole atom, which is represented by a potential well with energy levels filled by electrons up to a point. A photon with the correct energy, i.e. an ...


0

I think your questions concerns somehow another question: what is the relation between the macroscopic observables (like electrical current, temperature) of system consisting of many particles and parameters of single particles forming this system. The answer on this question is given by statistical physics. First, it will be useful to think about an ...


0

The absolute velocity of the electrons actually doesn't matter for joule heating. Think about it this way, if there is no current flowing there wouldn't be any joule heating. So, even if electrons are moving quickly and randomly when no current is flowing, we know no joule heating would occur and that joule heating is really about the net change in effect ...



Top 50 recent answers are included