Bohr said that only certain orbits of definite energy are allowed inside the atom. He said that the electrons in their ground state do not emit radiation and that they will emit radiation when they fall from higher energy levels to lower energy levels. My question is what does orbits with definite energy mean? and why do the electron in their ground state not emit radiation?
6880
One of the problems with Bohr's theory to describe the hydrogen atom, was that the electron orbiting around the nucleus has an acceleration. Therefore it radiates and loses energy, until it would collapse with the nucleus.  Now Schrödinger describes the electron as a wave function. His theory is able to describe all atoms (in contrast to Bohr's model), but how is the radiation problem solved? I understand that the wave has not an exact position in the time anymore. But the electron still "moves", so it has an acceleration anyway (because of vibrations or so).   Why is in this theory the electron not radiating anymore? And if it is, why doesn't the atom collapse?
2311
I am learning the basics of quantum mechanics and am familiar with the Schrödinger equation and its solution, but I was confused about what the familiar atomic orbital shapes represent? Do they represent nothing physical and are just plots of the wavefunction in 3D polar co-ordinates? Or do they represent the region where probability of finding an electron is $90\%$? Or something else? Levine 7th ed. states that  An atomic orbital is just the wavefunction of the electron  Wikipedia instead states that  In atomic theory and quantum mechanics, an atomic orbital is a mathematical function describing the location and wave-like behavior of an electron in an atom. This function can be used to calculate the probability of finding any electron of an atom in any specific region around the atom's nucleus. The term atomic orbital may also refer to the physical region or space where the electron can be calculated to be present, as predicted by the particular mathematical form of the orbital
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With the confirmation that elements 113, 115, 117, and 118 are indeed fundamental elements that are now to be named on the periodic table, the next question is: what is the highest atomic number possible for an element? Feynman had a go at this years ago, and he derived (according to my limited understanding) that above element-137 (informally denoted 'Feynmanium') the electrons in the nearest orbit about the nucleus would be traveling at a velocity greater than the speed of light. (Note I am considering the Bohr model of the atom in this question.) I wished to demonstrate to my friends why this is and I came up with the following explanation.  According to Bohr's quantum condition, the angular momentum of an electron about the nucleus is given by $$ L=m_evr_n=n\hslash \implies r_n=\frac{n\hslash}{m_ev},$$ where $m_e$ is the mass of the electron, $v$ its velocity, $n$ an integer, and $r_n$ the radius of the $n$th possible orbit. Because we are concerned with the nearest orbit to the nucleus, $n=1$; and thus $$r=\frac{\hslash}{m_ev}.$$   Now, according to Coulomb's law, if the electron orbits about the nucleus the centripetal motion can be described by $$\frac{Ze^2}{4\pi\epsilon_or}=m_ev^2, $$ where $Z$ denotes the number of protons in the nucleus (the atomic number), and $e$ the elementary charge. Solving for $Z$ and substituting in $r$ from above yields $$Z=\frac{4\pi\epsilon_o \hslash v}{e^2}. $$ But what is $v$? Well, the maximum velocity an electron could ever have is the speed of light, and we wish to find the atomic number associated with an orbiting electron traveling at this speed, so we set $v=c$ and obtain our final result of $$Z=\frac{4\pi\epsilon_o\hslash c}{e^2}\approx 137.521,$$ which implies that for $Z&gt;137$, the electrons at a position of $n=1$ in the Bohr model would have a velocity $&gt;c$; and thus the highest atomic number achievable on the periodic table is 137.  Again, I just want to make sure this is a correct method for deriving element-137 before presenting. Perhaps one could explain how relativity plays a role here. I know Feynman used the Dirac equation to get this result...so could anyone (subsequently of course) expatiate on this in a simplistic manner? Thanks!
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It's a common understanding that atoms emit or absorb light when the energy of the photons is equal to the the difference in the energy levels in the atom. What I don't understand is how does an electron absorb light inside an atom? Is it that the atom as a whole absorbs light or the individual electrons absorb the light packets? Also I would like to know, what does one mean when you say that an electron absorbs photon. How does an electron absorb a photon??
114937
Lets say we have an atom of hydrogen. It has one electron on $E_1 = -13.6~\mathrm{ eV} ~~(E_2 = -3.4~\mathrm{eV})$ energy level. I know that if we fire a photon with 10.2 eV energy the hydrogen atom will absorb it and the electron will jump on the next energy level E2. And bellow are my questions.   Q1: If a photon with 10.1 eV energy (insufficient to excite electron) would hit the atom of the hydrogen what would happen? Will the photon be absorbed by the atom and immediately emitted and the emitted photon (or photons?) will have the same 10.1 eV energy? Or the photon will pass through the atom or what would happen? Q2: Same question as the above one in this case our photon has a slightly more energy lets say it has 10.3 eV. Again what would happen? Will the atom absorb the photon and excite the electron but since the energy of the photon exceeds the required energy to excite the electron will the atom emit a photon with 0.1 eV energy or what will happen in this case?   I have done some research about it and got really confused. Some say that it needs the exact amount of energy ($\Delta E= E_2 - E_1$ in our case $\Delta E$ equals to 10.2 eV) to jump onto the higher energy level some say that it can jump if the energy exceeds the $\Delta E.$ What I really could not find is what happens with the extra amount of energy or maybe electron can be on $E_2$ energy level with slightly more/less energy.  Eventually I want to understand the concept of the reflection. How we see the objects, why they are transparent or glossy or red or whatever else. But this is out of scope of my question.  I'm not an expert though; so mark done the mistakes above if there are any.
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Suppose I have an atom in its ground state $|g⟩$, and it has an excited state $|e⟩$ sitting at an energy $E_a=\hbar\omega_0$ above it. To excite the atom, one generally uses a photon of frequency $\omega$ equal (or sufficiently close to) the transition frequency $\omega_0$, and this will stimulate a transition.  One thing that is often left by the roadside,* though, is the fact that the incoming photon has momentum as well as energy, and that if the atom wants to swallow the energy it also needs to swallow the momentum. So, in the nuts and bolts of states and operators,     how does one describe the transfer of momentum during an atomic transition?   In addition to this, the fact that this recoil momentum is rarely mentioned is a good indication that it is also rarely an issue. Why is it that in most circumstances we can safely ignore the photon's momentum when describing electronic transitions?  *Apart from treatments of Doppler cooling, which simply take the momentum transfers for granted and do not explain how and why they happen.
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If photon energies are continuous and atomic energy levels are discrete, how can atoms absorb photons? The probability of a photon having just the right amount of energy for an atomic transition is $0$.
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Sometimes we see energy that is negative, for example, the energy of an electron in orbit. We know energy is something that can do something. In this view does negative energy mean something opposite someway?
94291
The well known Larmor formula tells us that an accelerated classical charge radiates a power proportional to the square of the acceleration. It has been  that a charge in free fall in a gravitational field does not radiate. Since gravity is locally equivalent to acceleration, one may wonder whether a charge at rest in a gravitational field radiates. The  is that it does. However this radiation is not  to an observer in the same reference frame! Apparently, the radiation does not carry energy or momentum, or it would be observable, in this reference frame and this would mean that charged matter behaves differently from neutral matter in gravity.    How does unobservable EM radiation without energy and momentum make sense (Q1)?  Does the physical origin of charge acceleration determine whether radiation is observable or not (Q2)?  Is the answer to the question in my title "It depends?"?   Note added:  I did not find the discussion at  and  sufficiently enlightening.
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I'm working off the article, . I think it's well-written, but I'm not convinced about this part:     So what is causing the change in the particle’s behaviour?      In the example given above, when electrons are fired at a fluorescent screen, when we bombard the electrons with photons the interaction changes the state of the system catastrophically as, despite lacking mass, photons do carry momentum.   Here the author strongly implicates the momentum of photons as being responsible for the collapse. But is that correct to say?  I have read that the  verified that obscuring which-way information restores wave-like behavior. So it feels to me it's more about some "outer" state (perhaps well-characterized as "information") that we don't yet understand that collapses the system, not per se the photon-electron interaction.  Or am I wrong, and in fact for example, the process of obscuring in the quantum eraser experiments somehow "undoes" the photon-electron interaction or restores the quantum state through another interaction?
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Using electrical potential energy $V=\frac{1}{4\pi \varepsilon_0} \frac{Q_1 Q_2}{r}$ , a particle further away from nucleus has lower magnitude of energy.  Using Coulomb's law, a particle further away from nucleus experiences weaker attraction, hence less energy is needed to maintain orbit$^\star$ around that e-shell compared to a electron shell closer to nucleus, hence the one closer to nucleus supposedly should have higher energy.  $^\star$I know in reality $e^-$ does not orbit around a atom, but its position exists as a probability density or radial probability function.
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According to electrodynamics, accelerating charged particles emit electromagnetic radiation.  I'm asking myself if the electron in an hydrogen atom emits such radiation. In , Murod Abdukhakimov says the total momentum of the electron is zero, hence it does not emit radiation. Could someone prove this statement ?  It may be an obvious question, but I can't figure out why the total momentum of the electron should always be zero, in any energy state.
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We all learn in grade school that electrons are negatively-charged particles that inhabit the space around the nucleus of an atom, that protons are positively-charged and are embedded within the nucleus along with neutrons, which have no charge. I have read a little about electron orbitals and some of the quantum mechanics behind why electrons only occupy certain energy levels. However...  How does the electromagnetic force work in maintaining the positions of the electrons? Since positive and negative charges attract each other, why is it that the electrons don't collide with the protons in the nucleus? Are there ever instances where electrons and protons do collide, and, if so, what occurs?
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My somewhat basic understanding of the concept comes from lectures I've attended about the Bohr-model, which explains the phenomenon as arising from the fact that certain configurations of an atom can only absorb certain wavelengths of light and other configurations can emit the same wavelength and change into the first configuration.  Now what I cannot understand is why these effects do not cancel out, and why in some instances absorption wins out and we observe absorption lines, and in other instances it is the other way around. Also, if I haven't misunderstood there are also some instances where we observe both absorption and emission at once.  So my question boils down to this: Why?
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I am confused, hydrogen just has one electron in the 1s orbit. but why can we plot all kind of orbitals (higher energy eigenstates for that atom)? My assumption: So physically spoken these orbits around the hydrogen nucleus exist and they go to infinity, but they are just empty?   Additional question: When we want to plot orbitals of other atoms e.g. silicon, we need to use numerical methods. So instead of doing complex numerical methods, do we use simply the hydrogen solutions as approximation then?  Is there a big difference between let's say the shape of a numerically solved silicon p-orbital and an analytically solved hydrogen p-orbital?
116644
First, a bit of background. I'm a chemist who works with synchrotron-based techniques, but I'm focussed on materials analysis rather than synchrotron physics.  I'm struggling to work out where synchrotron radiation (SR) actually comes from. Why does subjecting an electron to radial acceleration cause it to emit photons? I understand, with Bremsstrahlung radiation that the energy from the decelerating particle has to go somewhere. My first thought was that SR may arise from a similar effect, as my understanding is that you do work when you radially accelerate a particle. I reasoned that, since the magnitude of its velocity remains constant, and the magnitude of the kinetic energy (KE) remains constant, SR must result from conservation of energy (i.e. the work done in the radial acceleration is converted into SR). However, I've since read that emission of SR causes the electrons in the storage ring to lose energy. And, this energy must be resupplied by the inclusion of RF cavities in the ring. This would suggest that SR comes from the KE of the electron, not from the work of the magnet on the electron. (On reflection, I can't see how a permanent magnet could "do work".)  So why is SR emitted?  (Please bear in mind, that as a chemist, I have next to no knowledge of relativity, electrodynamics, Maxwell's equations, etc. I'm really just looking for a conceptual understanding. I don't mind a mathematical answer if you can give one without assuming knowledge of such physics. But, I imagine that's unlikely.)  Edit: Any good textbook recommendations would be helpful too. All the books I've read either gloss over the issue or use physics that goes over my head.
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Position probability densities of eigenstates of hydrogen-like systems have axial symmetry, so that the wavefunction too much resembles the circular orbits in Bohr's model. I'd like to have a demonstration of correspondence principle, where an electron would be localized to look somewhat like a classical particle, and move in an elliptic (non-circular) orbit around a nucleus.  It seems though that if we try to make a wave packet too localized, then it'll disintegrate (get scattered by the nucleus) too fast to see anything resembling an elliptic orbit. OTOH, if we take it too spread out, it'll have to be quite far from the nucleus so as to avoid hitting it, and thus it'll have high total energy, which may appear to be above ionization threshold (at least partially), after which it'd be quite hard to analytically calculate evolution of the wave packet.  Thus my question is: is it possible to form a more or less localized wave packet, which would (on average) move in an obviously-elliptic orbit (major/minor axes ratio of 4:3 or higher), and only require bound states to fully represent it? If yes, then what properties (FWHM, apocenter, angular momentum etc.) should it have for this to be possible?