# Tag Info

10

That's a really good question! There are three cases, the third of which is the most fundamental and most interesting. The first case is incomplete absorption, such as a gamma ray knocking loose a few electrons as it passes. In that case the differences are taken care of locally and fairly trivially by allocating energy, momentum and spin appropriately ...

6

That's a great question! Unfortunately, the only honest answer is "that's what we see in nature, with great precision and complete reproducibility." There is no deep theoretical understanding. The more exotic form of your question is phrased in terms the self-energy of an electron, and it's a question that plagued Nobel Laureate Richard Feynman his entire ...

6

Arriving at the same answer as quantum mechanics for one particular scenario by making a bunch of ad hoc assumptions (for example - the calculation didn't work, so we'll make the orbital planes perpendicular) isn't useful. QM allows you to calculate much more than the ground states of atoms. Any competing theory - and that paper doesn't contain anything ...

5

Ne is used, Because it caused the red glow inside the tube, infact you can get a whole array of colors using different noble gases. eg. Ne => Red, Xe => Whitish Blue, Ar => Blue etc. Check Wikipedia for more. Because even when it exist as plasma, it doesn't react with the filament inside the tube or the glass walls, this helps in the longer life of lamp, ...

5

Relativistic effects are those that disappear in the non-relativistic approximation $1/c\to 0$, usually small corrections to the non-relativistic approximate results that are proportional to $1/c^2$ or higher powers of the inverse speed of light. Let me correct a typo: "cannot account for GR" should have read "cannot account for the special theory of ...

5

Gold foil is quite easy to hold you just hang it from a paperclip. The only difficulty is if there is a lot of static electricity in the air which makes it stick to things. (This is the main reason for the cold damp Cambridge's supremacy in early particle physics) Photographic film at the time wasn't sensitive and so in Marsden and Geiger's experiments ...

5

It's not often that dmckee and I differ (mainly because he's usually right :-) but we differ on this on. Or at least we differ if I've correctly understood what you're asking. In a hydrogen atom the 1s, 2s, etc wavefunctions are (subject to various approximations) good descriptions of the single electron and have well defined angular momentums. In ...

5

Here is an experimentalist's answer: You state: The probability of a photon having just the right amount of energy for an atomic transition is 0. You must be aware that the statement falls just by the existence of lasers, so your question should have a how is it possible to have lasers. 1) An individual photon cannot be labeled as continuous. It has ...

5

Yes, the photon energy has to be equal to the total energy of the atom before the transition (including the kinetic energy) minus the total energy of the atom after the transition (including the kinetic energy). In practice, the kinetic energy of the atom is negligible. The mass of a nucleus is at least 1 GeV/$c^2$ or so (the mass of the proton) while the ...

4

In order to find the possible ways of an how an electron acts in the presence of a proton we solve the Schrodinger equation with a coulomb potential, $\frac{k q}{r}$. At the outset of solving an equation, from a strictly mathematical viewpoint, the equation you are solving might have no solution, 1 solution or infinitely many solutions, and also the ...

4

What do we actually mean when we say that matter is a wave? We mean that particles like electrons and photons exhibit wave-like phenomena, such as superposition and diffraction. This is more properly known as wave-particle duality. The thought experiment that best illustrates this is the double-slit experiment, in which electrons behave as waves while ...

4

There's an infinite number of orbitals, and thus an infinite number of possible state transitions. However, the energy of the orbitals asymptotes to a finite value - for example, a hydrogen atom has energy levels given by the formula $E_n = -\frac{13.6\text{ eV}}{n^2}$ (ignoring some very tiny quantum corrections). If you let $n$ go up to infinity, the ...

4

The problem with his claims is that they don't include entanglement, which was the major prediction of new quantum theory, as opposed to the Bohr model. At least he correctly is attacking the source of the quantum weirdness-- entanglement was experimentally demonstrated from the He atom ground state originally. The main point of this attack on QM is to ...

4

I will address the premise that the electron is in an orbit around the hydrogen atom. This is a classical picture overlayed on the basic quantum mechanical one. The electron around the hydrogen atom is in a "spherical" probability cloud about the proton. The above is the n=1,l=0,m=0 probability distribution, which is the lowest energy state. A single ...

3

The spin-orbit interaction is responsible for splitting the $5p$ level in two, hence the two identified $5s \rightarrow 5p$ wavelengths. To calculate the splitting magnitude, you want to be looking at the difference $\Delta E$ between the two transition energies. By the way, I'm seeing the coupling constant typically characterized in inverse centimeters: ...

3

The scale of the atom has Planck's constant in it, so it's quantum. The force that keeps the electrons near the nucleus is the electrostatic attraction between the electron and the nucleus. To understand why the electron doesn't fall into the nucleus formally, you can solve the Schrodinger equation, but there are seat-of-the-pants arguments that are correct ...

3

Because the black area is half the box below. To explain: move the dipole from an area of no field to an area of field strength E. As you do, there's a force proportional to the dipole moment and to the gradient of E. For a fixed dipole, this force depends only on the gradient (horizontal dashed line). But for an induced dipole, the dipole moment depends ...

3

A fluorescent lamp or fluorescent tube is a gas-discharge lamp that uses electricity to excite mercury vapor. The excited mercury atoms produce short-wave ultraviolet light that then causes a phosphor to fluoresce, producing visible light. Taken right from Wikipedia! http://en.wikipedia.org/wiki/Fluorescent_lamp End of life The end of life ...

3

The Darwin term can be obtained from the low energy approximation ($|p|^2/m^2<<1$) of the Dirac equation of the electron in a central field. An elegant way to perform this task is by means of the Foldy-Wouthuysen Transformation. The same approximation leads also to the other terms detected in the Hydrogen atom fine structure including the spin-orbit ...

3

Though @Christoph and @poorsod cover the mathematical concepts, the basic meaning of attributing a wave nature to matter is not emphasized enough. It is not a matter wave in space time, it is a probability wave that is described by quantum mechanics. A probability tells me what are my chances to find the particle at a particular (x,y,z,t) and nothing more ...

3

There is always a possibility, in any experiment, that some unknown effect was the main cause for the obtained measurements. This can never be ruled out. The reasonable way to approach this is to first ask what is our null hypothesis. In this case we would like to test general relativity, so our null hypothesis is that there is no relativity and that ...

3

The Berry connection/Curvature can be formulated as the connection/curvature of a principal bundle over the parameter space, in this sense it can be thought of as a "gauge theory". It can also contain topological information, such as the first Chern number (measuring "magnetic charge"), second Chern number (measuring "instanton charge") etc., so ...

3

Your teacher is referring to the LCAO approximation as a way of calculating molecular orbitals. Suppose you bring two hydrogen atoms together i.e. create a hydrogen molecule. To calculate the electronic structure you need to solve the Schrodinger equation, but even for something as simple as the hydrogen molecule the Schrodinger equation is too complex to ...

3

Your seemingly unrealistic gedanken experiment is in fact a quite realistic. First, one can kick out the proton with help of a fast neutron. Next, to increase your "delay" time, you can consider a Rydberg atom with a high enough $n$, so the electron velocity is rather small with respect to the light (and the maximum proton) velocity. What happens to the ...

3

In physics we distinguish between the physics of "atoms and molecules" and nuclei. Atoms and molecules are described by the same theory, thus I will ignore those molecules here completely and only consider the difference between nuclei and atoms. I suppose you recognize that an atom is a bound system, so is a nucleus a bound system. Maybe you have seen how ...

3

The Legendre polynomials occur whenever you solve a differential equation containing the Laplace operator in spherical coordinates with a separation ansatz (there is extensive literature on all of those keywords on the internet). Since the Laplace operator appears in many important equations (wave equation, Schrödinger equation, electrostatics, heat ...

3

Though some atomic levels can be theoretically discrete (for example, the ground states), the frequencies of atomic transitions are defined by the difference of two atomic levels and are not discrete, the relevant bands have finite width (so called natural line width). Furthermore, there is Doppler broadening, collisional broadening of the band, and so on. ...

3

Yes. Lubos answered quite well, so I will simply expand on his answer with the calculation of the energy transfer. The answer, once you know that the energy is conserved and you are in the non-relativistic limit, is obvious. The advantage of this derivation is that you can see the steps and see where the recoil energy will be non-negligible (when the ...

3

Since you used the tag wave-particle-duality, I imagine you mean the frequency $f$ that corresponds to an electron's energy $E$ via Planck's relation, $$E=hf,$$ where $h$ is Planck's constant. That is a valuable question and nothing to get picked on for. After all, if the electron is a wave with wavelength and so on, it surely has a frequency, right? It ...

3

These are all good questions! Based on your description I assume you haven't had an introduction to solid state physics yet? Let's take your image of an electron that "jumps" from atom to atom. In my understanding I wouln't describe it that way, to me it's a wavefunction of the electron that is almost independent from the valence electrons and you can use ...

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