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In some cases it does eject a photon with a lower wavelength, if it did not do this then the laws of conservation of momentum would not be supported thus disproving many aspects of modern physics. The problem with this is without the experimental evidence or data, it is hard for someone to calculate or predict the new photons wavelength, let alone detect it. ...


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Does the specific heat of a metal should be then related to the electrons and not the atom of the metal? Specific heat (just like conductivity) depends on the molecular structure of the solid, and that depends on the atomic number, i.e. the shell of electrons. Whether the electrons are the ones conducting the heat or exchanges of vibrational ...


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As the metal heats up at one location, eg, a laser pulse, the motions of the atoms in the crystal lattice increase. If these motions are coherent you have sound, but if they are random it is heat. Most motion starts as coherent, resulting in quantized sound (phonons) but they soon exchange momentum with the free electrons, and become randomized in a few ...


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I don't quite understand what answer you exactly want, since there are too many perspectives to deal with the problem. In classical physics, you might use the degree of freedom to deal with this question, and you will find out the result fail to explain the experiment. And this problem comes from the "incorrect" Maxwell-Boltzman distribution when dealing ...


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TL; DR Short answer is the energy from the photon causes the electron to jump. Conservation of energy dictates that the photon would lose some energy, and it would be from the particle matter that it would act this way. Long Answer This is called the photoelectric effect. Basically this is caused by an energy transfer from a photon, acting as a particle ...


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It depends a little on what you mean by "extreme" electric current, but the answer is probably no. The energy scales are wrong. Electric current in a metal is a sub-electron-volt process: a potential difference of much less than a volt can displace electrons all the way through a piece of metal. The weak interaction is a keV- or MeV-scale process. And ...


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Absolutely it can - and it happens all the time. If you excite an atom, it can go through various "stages" of decay back to the ground state - with each drop in energy resulting in an emission of radiation. This happens during photosynthesis: see this page from which I copy this image: As you can see, there are multiple paths for the energy to be lost by ...


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As a general rule adding thermal energy doesn't cause electronic transitions. That's because typical electronic transition energies are a few electron volts or around 100kT at room temperature. In a metal the electrons aren't in discrete energy levels but instead reside in a continuous band of energy levels called the conduction band. While thermal energy ...


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An electron in a higher energy orbital does not necessarily decay directly into the ground state. If an electron in a hydrogen atom is excited into, say, the $n=4$ energy level, all of the following are possible decay paths: $4 \rightarrow 3 \rightarrow 2 \rightarrow 1$ $4 \rightarrow 3 \rightarrow 1$ $4 \rightarrow 2 \rightarrow 1$ $4 \rightarrow 1$ ...


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I don't know if I'm understanding your question right, but I think you are trying to pose a deeper question than it might seem at first sight... In ordinary quantum mechanics, when you study the hydrogen atom, you derive a set of solutions for the electron wavefunction using the Schrödinger equation (with different values of the energy). These are the ...


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The statement in the question is not fully correct. When an electron in a system is excited from a lower to a higher energy level the system becomes excited. That is to say, system reaches to a state which is not favourable. This is an unfavourable state for the system because the excited electron leaves a hole behind. There are several different mechanisms ...


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This is the hydrogen atom energy level solutions, as an easy example. The electron sits at the ground state, and can be kicked up to an excited state by the appropriate photon i.e. given that the photon has the quantized energy needed. For each energy level one can calculate using the solutions of the Schrodinger equation, the probability for the ...


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Actually it is interesting to calculate the radiation frequency for an electron in low orbit about a neutron star. The orbital frequency depends only on the density of the star or planet (radius doesn't matter!). Wikipedia tells me a neutron star has a density of 10^17 times as great as earth. The frequency goes as the square root of the density. The orbital ...


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Is a neutron star's residual light released similar to an exited atom the difference is gravity hold in the electrons instead of protons? No. Atomic energies are of order of keV at most, the electrons are bound in energy levels about the atom. There will only be photons produced if an electron is kicked to a higher energy level and then decays back ...


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The atom absorbs the photon that kicks up an electron to an excited state, and it is the atom that will emit a photon when it de-excites. Not the electron. Is the invariant mass of an atom higher when the electron is in an excited state? Take the hydrogen atom. The ground state energy is at -13.6eV. This means that the mass of hydrogen is less than the ...


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here's an answer from Dr.Richard Feynman http://www.feynmanlectures.caltech.edu/II_01.html#Ch1-S1 You know, of course, that atoms are made with positive protons in the nucleus and with electrons outside. You may ask: “If this electrical force is so terrific, why don’t the protons and electrons just get on top of each other? If they want to be in an ...


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1.) When you first close a circuit, there is a very brief period of time in which the electrons push each other forward "one by one", so to speak. This very brief period of time is probably on the order of nanoseconds, so we usually ignore it. After that, a steady state condition is established, and the electrons move all at once ... more or less. Don't ...


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You've asked some really good questions here. Before starting, I want to first mention that the traditional picture of particles moving through a wire in electostatics is missing some physics; for instance, it ignores the quantum mechanical nature of electrons. The reason we still teach this model is because it captures the main effects (the phenomenon of ...


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Sign conventions aren't "wrong", but they can be misleading. For example, we could re-define work done in a gravitational field so that escaping earth's gravity well would require negative work. That's an equally valid convention, but we associate positive work with effort, so reversing the convention would hinder our physical intuition for no discernible ...


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The electrons in circuits moving as a incomprehensible fluid. At least for sufficiently small electric field. So, the information of boundary on the circuit are transmitted through the electrons via equilibrium. You can imagine a lot of boxes distributed over the conductor that the circuit is made up. Each box has your own chemical potential. The information ...


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Keep in mind that for whatever magical reason electrons repel each other (like charges), and are very attracted to protons (opposite charges). Due to the omni-directional bonding present with metals (electron sea model) electrons move freely around but the metal maintains a net charge of zero. Try not to think of the electrons as "testing the water." I find ...


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It is a thermal effect, the corresponding "force" is a so called stochastic force and not a fundamental force, but rather an effective description of entropic effects. Having a temperature causes the charge carriers to move about randomly, and therefore they tend to move from regions with high concentration to regions with low concentrations. At first it is ...


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Your question has several flaws. First, you say the electron is at rest at the origin. As John Rennie noted, this implies that the position and momentum are both sharp, which contradicts the uncertainty principle. There is no such thing as an electron at rest at a particular point. An electron is described by a wave function spread over an extended region ...


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If you say: According to me the electron is at rest. that means you have measured the electron momentum to be zero, in which case the electron position is completely uncertain. So you can't be sitting on the electron. If you say: Let us say I sit on an electron. that means you have measured its position precisely so you have no idea what its ...


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The effects of gravity are really only observable to us on a macroscopic (large) scale. When a large enough number of (perfectly neutral) Hydrogen atoms come together they will gravitate towards each other. That sets things in motion for the Hydrogen to heat up. Once they reach a high enough temperature and density, they will ionize and the protons can ...


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It relies on conservation of energy and momentum and the equation for energy in special relativity: $E^2 = (pc)^2 + (mc^2)^2$. Here you go. Energy of photon: $E_\gamma = \hbar\omega = p_\gamma c$, where $p_\gamma$ is the momentum of the photon. Assume the electron is initially at rest, so it's energy is simply $m_ec^2$. By conservation of energy, the ...


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When you have two levels separated by energy $\Delta E$, you can induce a transition between them by introducing a time-dependent term to the hamiltonian (for example exciting it with a periodic electromagnetic field). This term must have a time dependence of the form $\cos(\omega t)$ with $\omega=\frac{\Delta E}{\hbar}$. The angular frequency being referred ...


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Physics is about making models of the world, if you can make them as accurate as possible why wouldn't you? Incidentally, sometimes you really need acuracy as the smallest difference in your initial conditions can make a great difference in your outcome (see chaotic systems, the best example is weather or the double pendulum) Imagine instead of taking ...


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This is too long for a comment. From Wikipedia In the most recent CODATA adjustments, the elementary charge is not an independently defined quantity. Instead, a value is derived from the relation $$e^2 = \frac{2h \alpha}{\mu_0 c} = 2h \alpha \epsilon_0 c$$ where $h$ is the Planck constant, $α$ is the fine structure constant, $μ_0$ is the ...


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Yes, this happens and it's called London Dispersion force.


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If the two neutral objects are conducting spheres, for example, the charges spread out over the surface because like-charges repel. The negative/positive charges don't want to be clumped together in one place if they can help it. So a neutral sphere will not polarize unless a net charge pushes/pulls the negatives to one side and the positives to another. But ...


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Yes, the modulus of the coefficients of the state in the energy basis are constant. But their phase changes, and that gives you changing observables.


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Some problems. First, in practice in this situation you're not going to want to use a discrete model, but rather use the continuum approximation. Second, your coefficients don't depend on $n$. Third, you totally ignore the momentum distribution of the electron, which is what relates the position and the time. So what would the continuum approximation look ...


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The wikipedia article for Particle in a Box neatly explains how it obeys the uncertainty principle. Basically, a smaller box gives the particle a wider distribution of momentum, or more uncertainty in momentum.


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If you're looking for a strict derivation of the effective mass equation, check out S. Datta, Quantum phenomena. Reading, Mass.: Addison-Wesley, 1989. What he does is take the full Schrödinger equation with the periodic potential, and write it in the Bloch state basis. He then writes the effective mass equation in the plane wave basis. By comparing the ...


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If a electron in hydrogen jumps from n1 to n2 then number of spectral lines is given by formula: {(n1-n2)(n1-n2+1}/2


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The effective masses are different because they are on different bands with different curvatures.


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Yes, everything is a detector, but you need to quantify which things your system interacts with (and how strongly). Gravity is in some sense a poor example, because the quantum details of gravity are still an unsettled question (and gravity is a weak force regardless), so let's bypass that red-herring by replacing gravity with the electromagnetic field: As ...


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"everything is a detector" This can't be true, or else there would be no such thing as persistent entanglement. As @Conifold points out, the electron's charge should be a far more potent source of environmental disturbance, anyway. Why doesn't the charge of the electron leave a trace as it passes through the slits - some persistent disturbance of the ...


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An infinitely long and wide plate will, indeed, produce a constant potential. The electrostatic force acting on a charge in a constant potential is zero, so in the ideal case the charges won't move. The solution to your paradox is that plates of infinite extensions don't exist. What we mean by "infinite size plates" are plates that are so large that the ...


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No the electrons are not leaving the atoms, they are just lowering states(when emitting light). Due to complicated Quantum Configurations(States) and the Kinetics of recombination/relaxation, the energy is released very slowly.


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An atom that gains one or more electrons will have a NEGATIVE charge. An atom that loses one or more electrons will have a POSTIVE charge. An atom that gains or loses one or more electrons is called an ION. A positive ion is called a CATION and a negative ion is called an ANION. Atoms will transfer one or more electrons to another to form the ionic ...


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If the atom is capable of $\beta$ + decay, then an electron might be lost after the event because the number of protons in the nucleus would have decreased by one. The tunneling to freedom idea in the comments is impossible because tunneling only happens when there is a finite potential barrier.


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You mean like a lone Hydrogen atom? Can the single electron in Hydrogen pack its bags and leave? Yes but only if it acquires the energy to leave, right? In this case the binding energy (due to the electromagnetic forces) in the ground state is $-13.6~\mathrm{eV}_,$ therefore if it somehow acquires this energy it will leave, forever! There are different ...



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