# Tag Info

5

As usual, you can re-express the wave-function in momentum space (it's just a Fourier transform away from the spacial wave-function for bound state which is really nice). But that does not tell you how the electron moves anymore than the spacial wave function tells you where it is. Instead, it tells you the probability distribution function for results of ...

4

The options 1,2 are actually physically identical because the electrons are identical particles. Once we have two electrons, we can't say which of them is "Paul" and which of them is "Peter". When the addition is slow etc., the option 1=2 violates the conservation law for the angular momentum. So it is indeed 3 that has to happen: the ion will refuse to ...

4

Although not a complete answer, one place to start is with the coldest naturally occurring place in the universe, which is the Boomerang Nebula, a planetary nebula that is around 1 K. As best as I can tell, this cooled below the CMB temperature simply by adiabatic expansion, and is insulated in its interior from CMB heating. Is this a feasible way to get to ...

4

Think of an atom as a family with cats. The husband (neutron), wife (proton) and cat (electron) live in the house (atom). It's a bit more complicated because each house can have several husbands, wives and cats, so its more like a 1960s communal house, but anyway... The electron, like a cat, is somewhere in the house but you're never quite sure where. ...

3

I will try to address the points you have made in your question, one by one. First, what are electrons? Thanks to @CuriousOne, electrons are best treated as perturbations in a quantum field, which is explained by quantum-electrodynamics. In the Standard Model, they are considered elementary particles of the 1st generation lepton family with a mass 1/1836 ...

3

A system can absorb a photon if the energy of the photon matches an excitation in the system. So the hydrogen atom can absorb a photon if its energy matches one of the frequencies in the hydrogen spectral series. A proton is a composite object and it does have a spectral series. However the excited states of the proton involve rearrangements of the energy ...

3

Does a proton have a "bandgap"? If yes, what happens when a photon is absorbed by a proton? For single protons, as in a plasma , there exists Compton scattering . The photon transfers part of its energy to the proton and scatters off at a lower energy/frequency, the proton taking up the energy-momentum balance. This is a continuous spectrum, from very ...

2

The question is within a Bohr model of an atom, and the Bohr model worked well for the hydrogen atom by postulating fixed orbits, but it is not the real description of what happens at the atomic dimensions. The real description for the hydrogen atom is given by the solutions for the hydrogen potential of the Schrödinger equation with the postulate of the ...

2

If you treat the 1s ground state's probability distribution as a classical charge density distribution (not really accurate, but I think the simplest way to interpret the problem), then there isn't one. This state is spherically symmetric, so the electric field is always radial and depends only on the radial coordinate and not on the angular coordinates. So ...

2

Sure, it's been done for a long time. Google "atomic force microscopy". And click on images to see lots of pictures of individual atoms/molecules/etc, e.g., http://iopscience.iop.org/0953-8984/labtalk-article/48480

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Bohr's atom became famous for reproducing the Rydberg formula for spectral lines of hydrogen, which Rydberg presented in 1888 and published the next year. Bohr remarked that it was Rydberg's switching from wavelengths to wavenumbers that allowed him to make the discovery. His inspiration came from Balmer's 1885 formula, which was a particular case, and had ...

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A good starting point would be the series of papers by Bohr, starting with "On the constitution of atoms and molecules" Philos. Mag. 26, 1 (1913) and checking the references therein.

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If your photon has not enough energy to excite the electron then it will just not be absorbed and will pass by, and if you have an electron with an excess energy, it will be absorbed and a photon with the excess energy will be automaticaly emitted and the electron will jump to an excited state. So yeah in your case, you might have a photon with $0.1 \space ... 1 In quantum mechanics and particle physics, spin is an intrinsic form of angular momentum carried by elementary particles, composite particles (hadrons), and atomic nuclei. Spin is one of two types of angular momentum in quantum mechanics, the other being orbital angular momentum. The orbital angular momentum operator is the quantum-mechanical counterpart to ... 1 Spin and orbital angular momentum are two different things, as already pointed out in Aniket's answer, but there is a good reason why we still call spin a "spin". This is because the Einstein-de Haas-Richardson experiment shows that electron spin is indeed of the nature of an angular momentum, although not exactly due to a "spinning electron". In fact, ... 1 It depends on what you want how you choose the polarization of the light. The polarization of your light determines the recoil of your electron and your ion. In photoelectron vmi you would like to see the angular distribution of how the electron detaches from the molecule, so you should select the polarization of your light such that the velocity vector of ... 1 Two isolated hydrogen atoms cannot form an$H_2$molecule for the simple reason that they have too much energy. Any system formed from the two atoms will have an energy greater than the dissociation energy of$H_2$so no bound state will be formed. Observation tells us that the process must happen because there is a lot of$H_2$around. It happens when ... 1 It has been shown experimentally that the formation of H2 can happen in the presence of free electrons: so a photon is emitted in this two step process, taking energy away . Note that this is for low densities. The three body process in the answer by John dominates with increasing density, as discussed in the link. 1 In this link you will see the radial hydrogen wavefunctions. It is only the l=0 states, S states, that have a value different than zero at r=0. The other angular momentum states get a very small contribution to the probabilities from r>0 to r=1 fermi ( the charged radius of the proton) as 1 fermi is of order 10^-15meters, and the probability is the ... 1 One calculates the probability that the electron is inside the nucleus by integrating$\psi^*\psi$over the volume of the nucleus. For example, the radial part of the hydrogen ground state wavefunction is$\psi=\frac{e^{-\frac{r}{a_0}}}{\sqrt{\pi a_0^3}}$, so the integral is$\frac{1}{\pi a_0^3}\int_0^b e^{-2r/a_0} 4\pi r^2 dr$. In the above,$a_0\$ is ...

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