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19

Note well: What we perceive as color is bit of a tricky subject. This is a different question, one that has been asked and answered multiple times at this site. Per the typical human eye response, sunlight at the top of the atmosphere is about as "white" as "white" can be. Some of that incoming sunlight is reflected back into space, some is absorbed by the ...


18

The sky does not skip over the green range of frequencies. The sky is green. Remove the scattered light from the Sun and the Moon and even the starlight, if you so wish, and you'll be left with something called airglow (check out the link, it's awesome, great pics, and nice explanation). Because the link does such a good job explaining airglow, I'll skip ...


16

Orbitals are solutions to time-independent quantum wave equations. That is, there is no time-dependence. There is no little ball in there moving around, the electron has a quantum characteristic and exists with neither a well defined position nor a well defined momentum.


11

You are missing nothing. The Bohr model of the atom is false, and nowadays we replace the idea of the semi-classical "orbit" of Bohr with the fully quantum mechanical notion of orbitals or electron clouds, which give a probability distribution for the position of the electron around the nucleus, but do emphatically not imply that the electron is moving in ...


11

It is a very interesting question that allows to point out the differences between a Neutron Star and Nuclei. Although the dedicated article in Wikipedia Neutron Star fully covers the information, it is relevant to summarize here the elements. Nuclei are essentially different to Neutron Stars and some reasons are: Different bounding force: while Nuclei ...


8

The hand waving explanation in your question is called Rayleigh scattering Rayleigh scattering results from the electric polarizability of the particles. The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency. The particle therefore becomes a small radiating dipole whose radiation ...


7

The following passage has been extracted from Bohr's Nobel lecture: While in contradiction to the classical electromagnetic theory no radiation takes place from the atom in the stationary states themselves, a process of transition between stationary states can be accompanied by the emission of electromagnetic radiation, which will have the same ...


7

The nuclei of heavy elements (lead, gold, ...) approach the asymptotic density of extended nuclear matter (and therefore the density of neutron stars). The lighter elements do not. That said, it would be an error to refer to nuclei as "miniature neutron stars" because the binding force and dynamics are different. Nor are nuclei protected, shielded or held ...


5

The sun is technically green because the peak of its black body spectrum is near green wavelengths. When light scatters parallel to the plane of incidence (i.e., during the day time), it is blue-shifted. When light scatters perpendicular to the plane of incidence (i.e., sunset or sunrise), it is red-shifted. The light that is not scattered but makes it ...


5

dmckee is right. However I would like to add some notes to provide an intuitive connection between the asker's question and the answer. When we conceive of the orbital as a 2-dimensional surface in 3-dimensional space, as in the video above, we are not looking at the orbital. We might be tempted to say we are looking at the outline of the orbital, but the ...


5

The orbitals, which recently have been observed for the hydrogen atom, are probability distributions. These probability orbital distributions have been calculated using quantum mechanical solutions of the Schrodinger equation which give the wave function, and the square of the wave function is the probability distribution for finding the electron at that ...


4

The probability density of the ground state is time independent, so there is no motion in this sense. Yet the expectation value of the kinetic energy is non-zero, so there is movement in this sense. How are these notions of movement reconciled? First off, classically, if we had a particle in a $1/r$ potential and released it from rest, it would indeed bob ...


3

Almost but not quite. Qualitatively the spectrum is the same with the $1/n^2$ spacing, but the scale of the spectrum is set by the reduced mass $\mu$, $$\mu = \frac{1}{\frac{1}{m_l}+ \frac{1}{m_p}}$$ where $m_p$ is the proton mass and $m_l$ is the lepton (muon or electron) mass. Since $m_p \approx 2000 m_e$, it is not a large error to take $\mu = m_e$ for an ...


3

I think your confusion is cleared up quite simply: you're confusing the terms "orbital" and "electron shell". An orbital is characterized by the three quantum numbers $n,\ell,m_\ell$. This terminology makes sense, because these three numbers together completely determine the spatial component of the wave function. However, this leaves a freedom in the spin ...


3

The time-independent Schrödinger equation for the hydrogen atom is $$-\frac{\hbar^2}{2m}\vec \nabla^2\psi-\frac{e^2}{4\pi \epsilon_0r}\psi=E\psi $$ If your aim is just to verify that the $1s$-wave function $$\psi_{100}=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}\hspace{2cm} a\equiv \frac{4\pi\epsilon_0\hbar^2}{me^2} $$ is indeed an eigenfunction, then your task ...


3

In quantum mechanics, one can form a solid understanding of these matters. The conceptually simplest example which demonstrates the essential physics is, of course, the hydrogen atom. In solving it, we find that the (spatial part of the) allowed eigenstates of the Hamiltonian are parametrized by three quantum numbers, $n$, $\ell$ and $m$. So, we have ...


3

It's not really a product of functions, it's a tensor product. In a sense, you have two systems: The spatial system and the spin system. The combined system is thus the tensor product of the two spaces by the fundamental postulates and an eigenfunction is a (tensor!) product of eigenfunctions. Not that an addition of two wave functions wouldn't make ...


3

The orbital wavefunctions of the hydrogen atom, which obey the eigenvalue equation $$ \left[-\frac{1}{2\mu}\nabla^2-\frac{e^2}{r^2}\right]\psi_{nlm}=E_{nl}\psi_{nlm}, $$ are functions of the separation vector $\mathbf r$ which points from the proton towards the electron. This is a standard trick in the two-body problem and it is done in both the classical ...


2

Tyson claims that an electron disappears from one orbital and appears in another and claims that this is like going from the second floor of a building to the fourth floor without existing in between. This doesn't actually happen. What happens instead is that each possible state of the system has a continuous amplitude associated with it. In a transition ...


2

I read a book on pop sci book on quantum mechanics and the author said that electrons do not fall into the nucleus due to quantum mechanics- which principles suggest this (I think it was Heisenberg's Uncertainty and Pauli's Exclusion Principle) and why? The basic argument can be based on two things from non-relativistic Schroedinger theory: 1) for ...


2

Disclaimer: I will try my best to make this rather long story short. First of all, we must say that we are talking about the so-called equilibrium geometry - molecular geometry that corresponds to the true minimum on the potential energy surface, a surface which describes the energy of a molecule as a function of nuclear coordinates. Secondly, potential ...


2

Rather than write something unintelligible, I'll quote from a page on cesium clocks. According to quantum theory, atoms can only exist in certain discrete ("quantized") energy states depending on what orbits about their nuclei are occupied by their electrons. Different transitions are possible; those in question refer to a change in the electron and ...


2

The potential energy function is the same for both. The energy level solutions are the same for both. The key difference is that in (most modern interpretations of) the Schrodinger model the electron of a one-electron atom, rather than traveling in fixed orbits about the nucleus, has a probablity distribution permitting the electron to be at almost all ...


2

If you had a single hydrogen atom, and you watched for a single transition, then yes, you would only see emission at a single frequency. There would be one line in your spectrograph so to speak. But rarely do you have just one atom. And quite often an atom undergoes multiple transitions while you are watching it. When looking at a large ensemble, whether it ...


2

Like garyp says, the electrons are not discrete particles, but rather exist as a smear (a cloud) with the most intensity of their existing in the spaces so described by the wavefunction. Now, all of the electron needs to interact at once, so when an interaction (measurement, chemical reaction, etc.) happens, the wavefunction of the electron changes as well ...


2

Various physical properties of materials including shear strength are connected to the strength of attractive forces between molecules -inter-molecular forces. The type of inter-molecular force determines the force and energy by which the molecules stick to one another. Inter-molecular force types include ionic bonds, hydrogen bonds, dipole forces and ...


1

The problem here is how to quantize systems whose classical hamiltonian involves factors of the form (for example) $p^nx^m$, because these cannot be unambigously represented in a formalism where $p$ and $x$ do not commute. As such there are many alternatives (all of which are classically equivalent) but only one is quantum-mechanicaly relevant. In most ...


1

"The process of imparting energy from the nucleus to an orbital electron is a quantum process and may be seen as taking place by means of a virtual photon. In that sense the photon involved can be considered as a "virtual gamma ray"..." (Wikipedia) The electromagnetic force is indeed a long range force - virtual photons bind the electrons to the nucleus. ...


1

In H2O the center atom has four valence electrons, in CO2 the center atom has six valence electrons. H2O forms two simple bonds, while CO2 forms two double bonds. Why should it behave the same? Simple qualitative answer: Think about the ball and stick model (Not sure, how these are really called. I mean the ones with plastic spheres and soft plastic bonds). ...


1

The modern mind picture of electrons bound to a nucleus is that of an orbital rather than of an orbit as you seem to be thinking. However, your ideas are still somewhat meaningful in this modern picture in that region of electron delocalisation is more tightly confined around the nucleus for lower energy orbitals than it is for higher energy ones. You might ...



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