# Tag Info

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You ask: Do atoms of a surface excite to reflect the light? but it's actually the other way round. Reflection occurs because the oscillating electric field of the light produces oscillating dipoles in the electrons in the substrate. These oscillating dipoles in turn radiate light isotropically and the reradiated light interferes constructively only in ...

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Short answer: Light is absorbed and re-emitted all the time. Colors and reflectivity all depend on the electrons in the material, and what the material is. Sometimes photons are converted to heat in a material. Other times photons pass through a material or are re-emitted (so it may look like they just "passed through"). Depending on what photons are ...

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Light of different color is understood as a wave of different frequency of oscillation. Matter in surfaces has characteristic behaviour for each of these frequencies, based on its chemical composition: electrons in the surface oscillate and absorb a lot of light when its frequency is around one of resonance frequencies of the matter; and the electrons ...

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The expression you have there looks like that of the electron relative to the proton.

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In nuclear fussion electrons and protons can fuse to form neutrons with the release of photons.

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To provide a graphic version of Punk_Physicist's answer, we have the Feynman diagram for that particular interaction: This diagram evolves in time from bottom to top, ie take a piece of paper, and run it upwards along the diagram to see how the system evolves. We have an electron and a proton coming towards each other, then we see them interact by ...

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Part 1: Conceptual/physical intuition Since there is an electrostatic attraction between the 2 particles, then when they are apart they are at a higher potential energy then when they are together. Here's an analogy: Physically, this situation is like having a ball at the top of a hill overlooking a valley or well. The ball will roll down the hill and ...

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Check this link here. This is a quote from the website: When an electron falls from infinity towards a proton it acquires 13.6 electron Volts of energy to reach the ground state “orbital” around the proton. I have always wondered why it does not go all the way. Apparently, its Debroglie wavelength has to fit” around the “orbit radius” for it to occupy a ...

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Note: ChocoPouce's answer is the same as mine but is more mathematical. You have a (spherically symmetric) probability density distribution $\rho$ in space (which we get from the square of the amplitude). The "radial probability density" is roughly the chance that the electron is at a given radius, say $r = 0.1\mathrm{nm}$? In other words, how much of this ...

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The link uses a spherical shell element which is $4\pi r^2 \mathrm{d}r$ and has the dimension of a volume ($r^2$ is a surface and $\mathrm{d}r$ is a length. The wavefunction of the ground state is spherical, if it weren't the calculation should have been made using a spherical volume element such as $r^2\mathrm{d}r\mathrm{d}\theta\mathrm{d}\varphi$. There ...

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This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? Your puzzlement arises because you are putting the cart in-front of the horse. The cart is the theoretical model of quantum mechanics and the horse is the data. As ...

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This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? After all, there's no particular reason for an electron to be in an eigenstate. Good question! The function $\psi$ does not need to be Hamiltonian eigenfunction. ...

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To have emission (or absorption) of photons you must have a Hamiltonian that includes those degrees of freedom also. If your system consists of (a) the electromagnetic field and (b) a hydrogen atom, you can specify the state with (a) for each frequency, the number of photons with that frequency and (b) the state of the hydrogen atom, in your favorite way, ...

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Conservation of energy. If we measure the energy of an atom, we will always report an eigenvalue, because we are forcing it into an eigenstate (this is something like the quantum mechanical definition of measurement). Now suppose that we measure the energy of an atom twice, before and after it emits a photon. For conservation of energy to hold, the energy ...

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The idea here is increasingly complex depending on how deep into modern physics you want to delve, but also key to understanding quantum mechanics. So, I'll give a bit deeper explanation than it seems you've seen, but there's plenty more. It's understood that a photon acts both as a particle and a wave. As a particle it has an amount of energy associated ...

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1) does "atomic density" correspond to "atoms in a volume unit that partecipate to the stimolated emission process"? "Atomic density" means number of total rubidium atoms per volume of gas, not just the ones that happened to participate in the absorption/emission process. 2) is $\sigma$ the cross section for a particular transition? (for example the ...

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It is not empty in the sense that you still have quantum fields present. You also have quantum fluctuations happening all the time, and other quantum phenomena. Quote from wikipedia: "According to modern understanding, even if all matter could be removed from a volume, it would still not be "empty" due to vacuum fluctuations, dark energy, transiting gamma- ...

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The speed only affects if there is an acceleration. If the atom is moving at a constant speed, you can do a galilean transformation, move with the atom, and it will be at rest. The two simplest reasons for an atom to accelerate are collision with other atoms and bombardment. The first one happens everytime you have a gas or a liquid. The atoms wiggle ...

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The numbers are not measured, they are part of a model that explains why the hydrogen atom emits/absorbs only at those wavelengths. The first series of Hydrogen lines to be discovered was the Balmer Series, and nobody knew why they were discrete lines instead of a continuous spectrum. Johann Balmer discovered that the lines all had wavelengths equal to, ...

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Morsley's experiment was based on Bohr's atom model, it showed that the value of the nuclear charge increased by one, as the atom's place increases across the periodic table. The method was to knock out an electron, and look for the emmission-spectrum as another electron fell from the outer shell to the inner one. This would produce a hydrogen-like ...

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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 ...

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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

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While DumpsterDoofus is right, perhaps this explanation might be helpful. A dipole is an asymmetric separation of charge, like this: $+ -$. A dipole can have many charges. The total charge must be 0. The center of charge for the $+$ charges and the center of charge for the $-$ charges must be at different places. A dipole can exert electrical forces on ...

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So, why molecules with mirror symmetry have no permanent dipole moment? That's actually not true. For example, hydrogen cyanide has an infinite number of mirror planes of symmetry, but has a nonzero permanent dipole moment. Also, formaldehyde has two mirror planes of symmetry, and a permanent electrical dipole moment. However, for any molecule with ...

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If a molecule has mirror symmetry then charge distribution will be uniform. Charge distribution on the left of the molecule will be equal to the charge distribution on the right effectively cancelling out, resulting in no permanent dipole moment. On the other hand if a molecule has no mirror symmetry then there will be a direction where charge distribution ...

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You should not try to rephrase things in terms of some fuzzy smeared out averaged classical physics, or that the electron "really" is moving in an orbit or something like that. What is angular momentum? Angular momentum is the quantity that is conserved in a rotationally symmetric system. You can work out that in quantum mechanics the (orbital) angular ...

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In an atom, the electron is not just spread out evenly and motionless around the nucleus. The electron is still moving, however it is moving in a very special way such that the wave that it forms around the nucleus keeps the shape of the orbital. In some sense the orbital is constantly rotating. To understand precisely what is happening you need to use ...

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The extra kinetic energy appears as heat, which is vibration-energy at the atomic scale. Energy forms like this is largely where much of the chemical energy comes from. The heat derived is then used to drive an engine of some form, which produces mechanical, and then prehaps, electrical, energy.

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The electron is still translating in space with a certain momentum. This can be represented with an angular momentum operator. Even though the translation of the electron is not "continuous" as in classical mechanics there is still a quantised way of translating about the nucleus, with associated momentum, and position.

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I do not understand the source of the orbital angular momentum. Is it intrinsic like the spin? No. Despite the fact that the picture of the orbiting electron is not quantum-mechanically correct, orbital angular momentum is still written in terms of position and momentum of the electron. More precisely, orbital angular momentum is represented as a ...

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Your statement $U=-kq^2/a_B$ makes the assumption that the total energy $E_\text{tot}=U+K$ is zero at the transition between bound and unbound. To find the ionization energy then, you want to find the energy input required to bring the total energy $E_\text{tot}$ up to zero. I say up to zero because for bound states the total energy is negative. To proceed, ...

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The electron cloud is not exacly centered at the proton in the hydrogen atom (or about the nucleus in other atoms). This is analogous to the Moon not exactly orbiting about the Earth, but the Earth and the Moon each orbiting about their barycenter (center of mass of the Earth - Moon System). In solving the Schrodinger equation for the hydrogen atom, it ...

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To rephrase: an electron approaches an cation. Most likely result: a photon is emitted and the electron is captured. Added: The way I see it: an ion and an electron at a large distance apart. This would probably be an electron in a very high orbital of the ion. As the electron approaches, all the natural forces affect the electron and the electron slows ...

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This is an excellent question. At the heart of it is this: you compare the clock to another copy of the same clock. Well, actually you need to compare three identical clocks to each other to make a strong statement about the clock noise, but lets not worry about that for now. If the noise of your clock is stationary (which it better be for a good clock), ...

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Let's first recall that there is a strict definition for (how to determine) whether a given clock $A$ is "accurate" (or "good"); namely: if for any three of its indications, $A_J$, $A_K$ and $A_Q$, the durations of clock $A$ between pairs of those indications, say $\Delta \tau_A [ \small{\text{from }} A_J \small{\text{ until }} A_K ]$ and \$\Delta \tau_A [ ...

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Terry is correct that to be sure of the configuration requires a complex calculation. However for the vast majority of the known elements the electron configuration is correctly predicted by the Madelung rule. This gives the order in which the atomic orbitals are occupied. As it happens copper is one of the few elements that violate this rule. The Madelung ...

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In theory, yes: If you assume a point-like positive nucleus and use Feynman's QED theory, you can predict the electron configuration of any element. Feynman even gives some rough examples in his Lectures on Physics, which now are online (yahoo!... as in yippee!, versus a certain web site). Since this is a homework question, I heartily recommend looking up ...

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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 ...

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The degeneracy pressure does not directly apply here, I think. It will be important for neutron stars. The particles that matter is made up with (electrons, protons, neutrons) all have spin 1/2. That means that they are fermions. The wavefunction, which contains all the information about the system, has so change its sign when two particles are exchanged. ...

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That's a fundamental problem with all measurement devices. What you can do is to compare your clock with other clocks. If yours differs significantly from the average then you know your clock is inaccurate. Of course, this does not protect you from overall errors (i.e. a fundamental flaw in the mechanics/physics behind the device).

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