Picture a simple hydrogen atom with one electron which is bound to a proton (nucleus). When trying to impart momentum to the atom, we may specify the photon wavelength to be $\lambda = 121.57$ nm to access the 1s-2s electronic transition. But in this picture, where exactly does the photon's momentum go and how does it get there?   We know that the electron can only occupy quantized states and the transition is instantaneous. The photon energy has decreased the electric potential energy between the proton-electron pair by $hc/\lambda$. But do not the quantized orbits occupied by electrons have well-defined momentum, too? Since the electron's momentum now that it has gone from 1s to 2s may not necessarily be the photon's total original momentum, $h/\lambda$, where does the rest of this momentum go? And is this momentum transfer an instantaneous process as well?   I believe a partial answer is that the entire atom has momentum imparted to it, but what is the mechanism to make the nucleus start moving if the photon was wholly absorbed by just the electron and the electron cannot simply drift off its quantized orbit (which would then pull on the nucleus via Coulomb interaction)? Perhaps answering this question may help elucidate the above: is imparting momentum by photons with wavelengths corresponding to particular transitions more efficient than photons with arbitrary wavelength?
50508
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.
87166
I have been recently wondering, if I take a powerful enough energy source (photon) and I have an perfect mirror exactly in front of it and assume an "emitter" shot the light towards the mirror.   As perfect mirrors absorb no energy of ANY kind from photons, should this mean that the perfect mirrors would never move due to transfer of momentum of the light?
7223
My question isn't how they receive the energy to jump, but why.  When someone views an element's emission spectrum, we see a line spectrum which proves that they don't exist outside of their orbitals (else we would see a continuous spectrum).  Electrons can be released in the form of beta decay, thus proving that they are capable of traveling outside of orbitals contrary to the statement my teacher said that they stay within orbitals. Then, to add to the confusion, the older model of rings floating around a nucleus has, from what I can tell, been outdated, which would support this model. My teacher's explanation was that the electrons made a quantum jump of some kind. How do electrons move between orbitals or do we know how they jump, excluding the reason that energy causes them to jump, and why are positrons formed sometimes instead of electrons in Beta decay?  When I'm asking "how do electrons jump" I would like to know how an electron can jump between each orbital such as how it moves and how it knows where to jump since it appears to be a jump where the electron doesn't slow into a orbital position. Specifically how they jump what is this Atomic electron transition, I understand that they jump and that they do this through absorbing and releasing energy but what is this Atomic electron transition other than what is already on the wikipedia article .
90933
So I was taught that:  Kinetic Energy (of electron) = Energy (of photon) - Ionization Energy  If E(photon) = IE, then KE=0 of the electron.   What does this physically/theoretically mean?   My current thoughts/interpretation is that enough energy/force is applied to ionize the electron so it is 'sufficiently far' from the atom, and then I guess it just moves with whatever speed it is moving at with natural kinetic laws, since no more energy/force is being applied....?  Any clarifications would be greatly appreciated, thanks in advance.
98227
Although I've read answers to similar questions, they did not answer my doubt:  If a photon has an energy less than that required to move an electron from one energy level to another, then what happens when the electron and photon collide? Is the photon absorbed, or does it by some mysterious method, just pass through the electron?
90047
This is an extract from the astrophysics chapter in my book:     Hydrogen atoms fuse to form helium.  At the same time, lots of of gamma photons and neutrinos are produced. The photons take thousands of years to "fight" their way to the surface of the Sun, but then escape into space as visible or near visible photons at the speed of light.   I am really confused about how the sun produces light. I understand that through nuclear fusion, energy is formed, but isn't this energy in the form of thermal and gamma photons? I do not understand how visible light is emitted which is the reason to why we see the sun. Could someone please elaborate the paragraph from my book.
7852
This might be a stupid question, but nonetheless, it has been bothering me.  If you take a photon, make it go through some atoms in a solid, liquid or whatever, then you have the chance of this photon being absorbed by an electron, and thereby exciting the electron. This requires the photon to have enough energy to actually excite the electron to another energy level.  My question is then: How does the photon know if it has enough energy or not? Do they interact very quickly to determine if it's okay or not, or is it just something it "knows?"
88235
As I navigate through physics stackexchange, I noticed .  Electrons radiate light when revolving around nucleus? Why is it so obvious?  Note that I do not know anything about quantum mechanics - and I just want classical stories.
69880
As a student in a highschool physics class, my teacher has repeatedly told me that photons are massless. Yet, I have also heard from other sources that photons have momentum. If photons were to have momementum, that would mean that they have mass as according to p = mv.  Do photons really have mass?  Also, how would this mass be calculated?
53693
I've just answered , and while I'm confident my answer is correct I'm less certain about the exact consequences. To simplify the situation consider a diatomic molecule falling into a Schwarzschild black hole with its long axis in a radial direction.    The inner atom cannot interact with the outer atom because no radially outwards motion is possible, not even for electromagnetic fields travelling at the speed of light. However I find I'm uncertain exactly what the outer atom would experience.  We know that the outer atom feels the gravitational force of the black hole even though gravitational waves cannot propagate outwards from the singularity. That's because it's experiencing the curvature left behind as the black hole collapsed. Would the same be true for the interaction of the outer atom with the inner atom? Would it still feel an electromagnetic interaction with the inner atom because it's interacting with the field (or virtual photons if that's a better description) left behind by the inner atom? Or would the inner atom effectively have disappeared?  If the latter, presumably the fanciful accounts of observers falling into the black hole (large enough to avoid tidal destruction) are indeed fancy since it's hard to see how any large scale organisation could persist inside the event horizon.  Later:  I realise I didn't ask what I originally intended to. In the above question my molecule is freely falling and the question arose from a situation where the object within the event horizon is attempting to resist the inwards motion. I'll have to go away and re-think my question, but thanks to Dmitry for answering what I asked even if it wasn't what I meant :-)
23316
So when learning about the  of hydrogen and de Broglie waves, it was shown that treating the electron of hydrogen as a de Broglie wave results in the relationship $$L=n\hbar, \qquad n\in\mathbb{N}.$$ However, when learning about the azimuthal quantum number, it was stated that $$L=\sqrt{\ell(\ell+1)}\hbar.$$ So how come in the ground state ($n=1, \ell=0$), these two equations give different values for angular momentum? I feel like I'm missing something really important here. If it's the case that the Bohr model doesn't accurately describe the angular momentum of the electron in the ground state, why is the angular momentum zero?
3247
From what I understand, the  states that there is a fundamental natural limit to how accurately we can measure velocity and momentum at the same time. It's not a limit on equipment but just a natural phenomenon.  However, isn't this just an observational limit? There is a definite velocity and momentum, we just don't know it. As in, we can only know so much about the universe, but the universe still has definite characteristics.  Considering this, how do a wide range of quantum mechanical phenomena work? For example, quantum tunneling - its based on the fact that the position of the object is indefinite. But the position is definite, we just don't know it definitely. Or the famous light slot experiment? The creation of more light slots due to uncertainty of the photon's positions?  What I am basically asking is why is a limit on the observer, affecting the phenomenon he is observing? Isn't that equivalent to saying  because we haven't seen Star X, it doesn't exist? It's limiting the definition of the universe to the limits of our observation!
80391
I understand that mass-energy equivalence is often misinterpreted as saying that mass can be converted into energy and vice versa.  The reality is that energy is always manifested as mass in some form, but I struggle with some cases:  Understood Nuclear Decay Example  In the case of a simple nuclear reaction, for instance, the total system mass remains the same since the mass deficit (in rest masses) is accounted for in the greater relativistic masses of the products per $E=\Delta m c^2$.  When a neutron decays and you are left with a fast proton and a relativistic electron.  If you could weigh those two without slowing them down, you would find it weighed as much as the original neutron.  Light in a Box  This becomes more difficult for me when moving to massless particles like photons.  Photons can transmit energy from one heavy particle to another.  When a photon is absorbed the relativistic mass (not the rest mass) of the (previously stationary) particle that absorbs it increases.  But if my understanding is correct, the energy must still be manifested as mass somehow while the photon is in-flight, in spite of the fact that the photon does not have mass.  So let's consider a box with the interior entirely lined with perfect mirrors.  I have the tare weight of the box with no photons in it.  When photons are present the box has an additional quantifiable amount of energy (quantified below) due to the in-flight photons.  Say there are $N$ photons... obviously assume $N$ is large.  $$\Delta m = \frac{ E }{ c^2 } = \frac{ N h }{ \lambda c}$$  Interactions are limited to reflections with the wall, which manifest as a constant pressure on the walls.  If I hold this box in a constant gravitational field (like the surface of Earth) then there will be a gradient in the pressure that pushes down slightly.  Is this correct?  Wouldn't there still technically be mass as the photons are in-flight, which would cause its own gravitational field just as all matter does?  How is this all consistent with the assertion that photons are massless?  Is it really correct to say that photons don't have mass?  It seems to be a big stretch.  Please offer a more complete and physically accurate account of this mirror-box.
6782
Lamb 1969 states,     A misconception which most physicists acquire in their formative years is that the photoelectric effect requires the quantization of the electromagnetic field for its explanation. [...] In fact we shall see that the photoelectric effect may be completely explained without invoking the concept of "light quanta."   The paper gives a description in which an atom is ionized by light, with the atom being treated quantum-mechanically but the light being treated as a classical wave.  Is it true that all the standard treatments in textbooks are getting this wrong?  Lamb and Scully "The photoelectric effect without photons," in "Polarization, Matière et Rayonnement," Volume in Honour of A. Kastler (Presses Universitaires de France, Paris, 1969) -- can be found online by googling
16399
In his  Feynman calculates the field momentum of a moving charged sphere with charge $q$, radius $a$ and velocity $\mathbf{v}$. He finds that the total momentum in the electromagnetic field around the charged sphere is given by:  $$\mathbf{p} = \frac{2}{3} \frac{q^2}{4\pi \epsilon_0} \frac{\mathbf{v}}{ac^2}.$$  He calls the coefficient between the field momentum, $\mathbf{p}$, and the velocity, $\mathbf{v}$, the electromagnetic mass:  $$m_\textrm{elec}=\frac{2}{3} \frac{q^2}{4\pi \epsilon_0 a c^2}.$$  He claims that this electromagnetic mass $m_\textrm{elec}$ has to be added to the standard "mechanical mass" of the sphere to give the total observed mass of the object.  Would this view be accepted by most physicists today?  Have any experiments been performed that show the effect of the additional electromagnetic mass on the dynamics of a macroscopic charged object?  I guess the problem is that such an effect would only be large enough to be observable for charged particles like electrons. In that case it would be difficult to distinguish mechanical mass, presumably due to the Higgs field, from electromagnetic mass. Maybe one could perform a high energy/short length scale experiment on an electron that excluded the effect of the electromagnetic mass?
47641
I'm just starting to learn physics and I have a question (that is probably stupid.)  I learned that energy levels that the bound electron can have are discrete. I also learned that when an electron transitions from one level to another, a photon with a specific wavelength (energy) is created and released into the wild.  My question is this: are there holes in the electromagnetic spectrum, values of frequency that the created photon can't possibly have?
30946
I understand that if a particle approaches a finite potential barrier of height $V_0$ with energy $E &lt; V_0$, there is still a finite probability of finding the particle on the other side of the barrier due to quantum tunneling.  My question is, since the wavefunction is nonzero inside the barrier region, is it possible to actually make a position measurement and locate a particle inside the barrier?  I mean, if we can say that "there is a nonzero probability that the particle is inside the barrier", surely this would suggest we can do so?  If not, why not? Am I understanding the whole wavefunction/probability distribution thing right?