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By definition, Thomson scattering is the elastic scattering of light by a free charged particle. Atoms cannot be described as such, but the electrons in an atom may approximate to free electrons if their binding energy is much lower than the photon energy. This might be true for X-ray wavelengths, although if the photon energy gets too high then elastic ...


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This is one of the first examples of energy levels for electrons within the atom! If we take the Bohr model, which imagines that electrons circle the nucleus on set orbits Each of these orbits has a corresponding energy. The electrons are more stable at lower energy levels, and thus, prefer to be there. When you provide energy to the electrons (in the ...


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Is it possible to prove in 2016 that the universe is made up of more discrete units than say an atom or quark? Physics is not mathematics, it does not prove anything. It measures and observes and fits mathematical models to data. These models become validated as long as their predictions are fulfilled. A wrong prediction falsifies a model and ...


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This is closely related to the questions: Why do non-hydrogen atomic orbitals have the same degeneracy structure as hydrogen orbitals? Which electron is first ionized $n=2,\ell=1, m=?$ Strictly speaking polyelectronic atoms do not have atomic orbitals because atomic orbitals exist only for a central force. In polyelectronic atoms the repulsion between ...


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Assuming I understand you correctly the quantity you refer to is the second electron affinity i.e. the energy absorbed in the gas phase reaction: $$ X^- + e \rightarrow X^{2-} $$ (it's the energy absorbed because a negative second electron affinity means energy is released) If so, there are no elements for which the second electron affinity is negative. ...


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The pre-factor is exactly the same thing, look up what $R_H$ is equal to. The (Z-1) term, is due to the inner electrons shielding part of the nuclear charge such as the electrons involved in the X-ray transitions only see a reduced electric field. It's explained in the link you provided.


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Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $ \psi_{a} $ and $ \psi_{b} $, both of which correspond to energy $ E_{0} $. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent. Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks ...


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De-exictation is the process of transitioning from a high energy state to a lower energy state; the photon emitted has energy equal to the difference in the energy of the two states. If we're tacitly assuming that we started with a system in the ground (i.e. lowest energy) state, then something had to put the electron into the higher energy state before we ...


2

The color of the photon is related to its frequency $f$, which can be related to the energy of the photon by the expression $E = hf$, where $h$ is Planck's constant. Thus the different colors of the emitted photons describes their different energies. The next step is to determine why specific elements emit certain colors. This has to do with the different ...


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One possible application of AMO physics would be inertial guidance systems based on atomic interferometers--similar systems are currently being investigated for missile guidance and have shown much higher accuracy than other methods. Inertial navigation systems don't rely on a network of GPS satellites, which obviously wouldn't be present on Mars (yet!)


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When the electron falls to its ground state, it will emit the photon in a random direction.This means that the photon might not travel in the direction as the rest of the white light. In the absorption spectrum of the sun, there will be some photons of the characteristic frequency of helium, but significantly less than the rest of the spectrum, meaning a ...


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The diffraction pattern is due to elastic scattering from the "ion core", which is the stationary net charge of the atomic nucleus and it's bound electrons; these elastically scattered electrons don't lose any energy. The electrons which interact with the free electrons are inelastically scattered, and contribute a foggy background to the diffraction ...


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The spin orbit coupling can be derived from the nonrelativistic limit of the dirac equation and is given by $$ H_{\text{s-p}} = \frac{\varepsilon_0}{2m_e^2c^2}\mathbf{\hat{s}}\cdot\left(\mathbf{E}\times\mathbf{\hat{p}} \right) $$ $\mathbf{E}$ is the total electric field acting on an electron, which consist of a microscopic electric field ...


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Angular momentum is conserved only if there's no external forces, in this case the electron gains energy by light or by heat wich is kinetic energy. They are both external forces so the conservation of angular moment does not apply.


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In short, you must consider the total elements of the system for conservation of momentum. In this case, nearly all of the momentum is exchanged between the electron and a photon that is absorbed or radiated away (the light). Momentum is conserved, and is largely balanced by this electron-photon interaction, although smaller amounts may be exchanged with ...


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The entropy of a single atom does not make sense per se, unless you specify the preparation. The entropy of a single isolated atom, fixed at a point, is indeed not defined – the entropy is, after all, a property of an ensemble not of a system. The entropy of an ensemble of isolated atoms prepared at a specific energy, on the other hand, is well defined (this ...


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You can only have an inelastic collision between two bodies if one or both of the bodies have some internal degrees of freedom that can absorb energy. For example if you have a rigid sphere then the only type of energy it can possess is kinetic energy. If we collide two rigid spheres then conservation of energy means the sum of the kinetic energies before ...


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As @EntropicallyDriven mentions, matter-wave interferometry can be used for inertial navigation. Better clocks (in terms of both performance and size / weight / power) would help with pulsar navigation.


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As phrased, this question suggests suggests a very Bohr-like image of electrons. This answer is intended to nudge you in the direction of a quantum model, but is by no means complete. Honestly, there's no way to understand QM without delving deep into the math, and even then its difficult to make a math <-> reality correspondence. Firstly: quantum ...



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