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29

You have your "prove" in the wrong place. The way to prove that ground-state electrons in hydrogen atoms don't emit radiation is the following: Construct a sample of ground-state neutral hydrogen atoms. Place this sample near a detector which is sensitive to the sort of EM radiation you expect. Die of old age waiting for a signal, because ground-state ...


15

The existence of hydrogen atoms is enough to demonstrate that the electrons don't emit radiation. If they did, that energy would have to come from somewhere. The only place it could come from would be a reduction of orbital radius until the electron finally reaches the nucleus. If you accept that electrodynamics applies, then you have to accept that atoms ...


11

This is a tricky bit of intuition to get right. In essence, having a lower angular momentum expands the radial range that the electron is allowed to span - the inner turning point moves inward and the outward turning point moves outward - but the electron is moving much slower at the outward turning point, which means that it spends more time there and ...


10

The theory behind the trick is based on the Hellmann-Feynman (HF) theorem $$ \frac{dE_{\lambda}}{d\lambda}~=~\langle \psi_{\lambda} | \frac{d\hat{H}_{\lambda}}{d\lambda}| \psi_{\lambda} \rangle,\tag{A}$$ which works with a single derivative, but not with a square of a derivative, cf. OP's failed calculation (5) for the expectation value $\langle\frac{1}{r^2}...


9

Because of its wave nature, the electron in its ground state is actually smeared symmetrically about the proton (ignoring spin-spin effects), and spherically symmetric charge distributions do not radiate (there's no special direction). Accelerated charges do not always radiate em radiation. See also How to find the magnetic field due to a revolving electron ...


8

I believe some of the answer in the links are correct, others are less obvious and might even be confusing. I am not gonna repeat the arguments there, but to stress the following idea. You cannot demonstrate that using classical electrodynamics. The theory as is does not apply to quantum objects and thus it was modified. The equations are the same, they are ...


7

By "consume" we mean "convert into helium." That $6\times10^{11}\ \mathrm{kg}$ of hydrogen is part of the Sun (specifically it is found in the core of the Sun), and it is converted into $6\times10^{11}\ \mathrm{kg}$ of helium. The Sun doesn't need to suck up material from space. Note that this amount of material is miniscule compared to the $2\times10^{30}\ ...


6

In fact hydrogen is an old idea to get a high temperature superconductor, based exactly on the idea of its light mass. The problem is that one has to start from metallic hydrogen, which is a problem on itself. It has not yet been fully experimentally confirmed in the lab. You need pressures of several hundreds of GPa to achieve that (100 GPa is about 1 ...


6

The most common isotope of hydrogen has no neutrons. Other isotopes are deuterium with 1 neutron and tritium, with 2 neutrons. Since virtually all (99.98% according to wiki) naturally occurring hydrogen comes in the no neutron isotope, it seems reasonable that books show a schematic of that one when illustrating hydrogen. As a secondary motivation, the one ...


5

From the horse's mouth: The proton source is a simple bottle of hydrogen gas. An electric field is used to strip hydrogen atoms of their electrons to yield protons. Linac 2, the first accelerator in the chain, accelerates the protons to the energy of 50 MeV. The beam is then injected into the Proton Synchrotron Booster (PSB), which accelerates the ...


5

A nice overview of the problem is given in arXiv:1205.3740. I'll summarise the most important points here. Let $d$ be the number of space dimensions. Then the Laplace operator is given by $$ \Delta=\frac{\partial^2}{\partial r^2}+\frac{d-1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_S\tag{1} $$ where $\Delta_S$ is the Laplace operator on the $d-1$ ...


5

First, the idealized clean non-relativistic problem with the $1/r$ Coulomb potential is integrable: the exact wave functions and energy eigenvalues may be precisely written down using elementary functions. It is exactly true that the eigenvalues of the bound states are $-13.6\,{\rm eV}/n^2$ etc. Now, the first simple correction is the proton motion. This is ...


4

When labeling states of the hydrogen atom, one doesn't refer to the z component of the angular momentum, but rather to the total angular momentum. The total angular momentum is positive, but, as you've stated, there are two states for $J=\frac{1}{2}$ with $L=0$, and those are $J_z=\pm\frac{1}{2}$ (Or some linear combination of them) As to why this is, ...


4

The energy is defined as $$ E = \frac{p^2}{2m} + V(\vec r) $$ where the first term is the kinetic energy and the second term is the potential energy calibrated so that $V(\vec r)=0$ for $|\vec r|\to\infty$. Consequently, you may say that the energy in a given state (an analogy of an orbit in classical physics) is equal to the kinetic energy $T_\infty$ ...


4

Consider two charges $q_1$ and $q_2$ kept at some separation. Suppose we want to calculate the potential energy of the system. By definition, potential energy is the work done to assemble such distribution. We can assemble the system in two ways: Bring $q_1$ to its place; no work done during this as there is no field present. Then bring $q_2$ to its place;...


4

The deuterium-tritium fusion reaction cross-section is highly temperature dependent and peaks at temperature of about $8\times 10^{8}$ K, so I suppose these are the temperatures to aim for in a controlled nuclear fusion experiment. In fact according to this, the operating temperatures are at least $10^{8}$ K. The density of the fusion plasma is a factor - ...


4

This is really just a footnote to Rob's answer. The Sun is an absolutely terrible fusion reactor. It uses a reaction $p + p \rightarrow d$ that is hopelessly inefficient. The $d + t \rightarrow He + n$ reaction that we use in fusion reactors is (up to) 26 orders of magnitude faster. As Rob says in his answer, the power produced per cubic metre in the Sun is ...


4

$SO(4,2)$ is called the full dynamical group of the Kepler (or Hydrogen atom problem). The $SO(4)$ , $SO(3,2)$ and $SO(4,1)$ subgroups of $SO(4,2)$ are called partial dynamical groups. Unlike symmetry groups which commute with the Hamiltonian, dynamical groups do not. They have the following properties: The system's phase space is a coadjoint orbit of ...


4

In addition to the answers already given, which answer the question pretty-well, I'll say that, historically, this exact question was the one which puzzled Niels Bohr enough to inspire him to advance his famous theoretical-explanation for the several observed frequencies of the radiations emitted from hydrogen-atoms ... in general, the fact that electrons in ...


4

The short story is that the Hyperphysics article you link to is using classical and semiclassical heuristics to justify the numbers it present. As I'm sure you're aware, the hydrogen atom cannot be described in any rigorous detail using classical mechanics, and instead requires quantum mechanics for any appropriate treatment, particularly where the ...


3

The hydrogen atom has an infinite number or quantum mechanically allowed energy levels, as explained on this web page. Using that same link, scroll up the page a bit to better understand how transitions between these energy levels give rise to absorption or emission of photons of very specific frequencies. Then scroll further down to see how the hydrogen ...


3

The term to look for is Coulomb wave. These wavefunctions are well explained in the corresponding Wikipedia article. Depending on your mathematical background, you should be ready for a bit of a formula jolt, as these wavefunctions rely very intimately on the confluent hypergeometric function. If you want the short of it, then I can tell you that the ...


3

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 ...


3

Perfunctory quantum comment #1: "orbital", not "orbit". There are no little balls moving in a circles in there. These things are quantum objects. But the sort answer is "yes", the proton has a non-zero momentum distribution that mirrors the electron's. Now, because the proton is nearly 2000 as massive, the proton's position wave-function is nearly 2000 ...


3

Probably, yes. 13.56MHz was the frequency used and it was a 300W field. http://www.rsc.org/chemistryworld/Issues/2008/April/ColumnThecrucible.asp The effect doesn't appear to be well described. So while the salt is necessary, the exact role it plays isn't detailed. It could be that some complex with the salt is resonating.


3

All matter that we know is made of atoms. All atoms derived from H and He. Therefore, if a banana exists, then a banana CAN be made. YOU cannot make a banana because technology. BONUS: Matter and energy are equivalent, therefore atoms can be produced from energy, therefore a banana can be made from energy. You cannot make a banana because technology.


3

Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $ \psi_{a} $ and $ \psi_{b} $, both of which correspond to energy $ E_{0} $. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent. Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks ...


3

The relativistic generalization of the formula; and the replacement of the electron mass by the reduced mass are clearly two basically independent steps (at least in the leading approximation), and both of them have to be applied to agree with the experiment. The most accurate experimental value of the energy is $-13.59844\,{\rm eV}$, see NIST: http://...


3

In a comment elsewhere you write that you're interested in understanding how quantum-mechanical theory describes the radiation that a hydrogen atom does and does not emit. In your question you ask about another answer that suggests some significance to the electron having zero total momentum; I think that's a feature of the coordinate system choice rather ...


3

The time-independent Schrodinger equation $\hat{H} \psi = E \psi$ only holds when the Hamiltonian does not depend explicitly on time. If you start with a time-independent Hamiltonian and make a time-dependent gauge transformation, then the new Hamiltonian will depend explicitly on time, and there is no reason to expect that the (time-dependent) eigenvalues ...



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