# Tag Info

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 ...

25

One cannot tell by the light spectra. Hydrogen and antihydrogen would give the same lines in the spectrum. The prevalence of matter over antimatter from other evidence indicates matter is predominant in the observable universe, and here is a nice review. How do we really know that the universe is not matter-antimatter symmetric? The Moon: Neil ...

22

There is a rigorous formal analysis which lets you do this. The true problem, of course allows both the proton and the electron to move. The corresponding Schrödinger equation thus has the coordinates of both as variables. To simplify things, one usually transforms those variables to the relative separation and the centre-of-mass position. It turns out that ...

19

Heavy water is easy to separate from regular water because the difference in mass is quite large. The molar mass of heavy water is 11% heavier that regular water. However if we take uranium separation, then the percentage weight difference between $^{235}$UF$_6$ and $^{238}$UF$_6$ is only 0.9%, so the relative difference is far smaller. So it's a lot ...

16

The degeneracy of energy-levels can be traced to the fact that the hydrogen atom possesses an enhanced $SO(4)$ symmetry caused by (among other things) the conservation of the Laplace-Runge-Lenz vector operator, see e.g. this Phys.SE post and Ref. 1. References: G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapter 9. The pdf file is ...

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 ...

14

I assume you're talking of the hydrogen atom; the hamiltonian of the nucleus + electron system is $$H = \frac{p_e^2}{2 m _e} + \frac{p_n^2}{2 m _n} - \frac{e^2}{|r_e - r_n|}.$$ You can do a change of coordinates (center of mass coordinates) $$\vec{R} = \frac{m_e \vec{r}_e + m_n \vec{r}_n}{m_e+m_n} \\ \vec{r} = r_e -r_n$$ and find the conjugate momenta to ...

12

Bohr postulated that electrons orbit the nucleus in discrete energy levels, and electrons can gain and lose energy by jumping between energy levels, giving off radiation of frequency $\nu$ according to the formula $\Delta E = E_2 - E_1 = h\nu$ where $\nu = \frac{1}{T}$, where T is the period of orbit, as in classical mechanics. Now during the transition, ...

12

As dmckee says in his comment - Population III stars have no metals (a tiny bit of lithium and beryllium), but they are not "pure hydrogen stars", they still have the big bang fraction of Helium. Taking the second part of your question first. These "stars" will last for ever. Their final fate is to become a completely degenerate ball of helium, supported by ...

11

The Hamiltonian for the hydrogen atom $$H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r}$$ describes an electron in a central $1/r$ potential. This has the same form as the Kepler problem, and the symmetries are similar. There is an obvious $SO(3)$ generated by the angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. In other words, the components of $\... 11 This is a tricky bit of intuition to get right. In essence, having a lower angular momentum expands the radial range that the electron is allowed to span - the inner turning point moves inward and the outward turning point moves outward - but the electron is moving much slower at the outward turning point, which means that it spends more time there and ... 10 The problem with attempting to fuse two protons is that there is no bound state$^2$He, for the rather obvious reason that there are no neutrons present to hold the two protons together. The fusion of two protons requires one of them to undergo beta plus decay while the two protons are close, and the probability of this is vanishingly small. It happens in ... 10 In a neutral hydrogen atom the ground state has the electron and proton spins anti-parallel i.e. lined up with each other but pointing in opposite directions. The state with the spins parallel and pointing in the same direction has a slightly higher energy, and transitions between these two states produce the notorious 21cm hydrogen line. Since the ... 9 I am highly skeptical of this result, primarily because the theories promoted by Black Light Power are improbable to the point of being gibberish. The energy states of hydrogen can be calculated exactly, and have been both calculated and measured spectroscopically to extremely high precision, and experiment and theory are in perfect agreement. If the modern ... 9 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 ... 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 The infinitesimal probability for the electron to be in the volume$dV$around a point$(r,\theta,\phi)\leftrightarrow (x,y,z)$is given by $$dP = dV\cdot |\psi(x,y,z)|^2 = dV\cdot |R(r)|^2\cdot |Y_{lm}(\theta,\phi)|^2 =\dots$$ as you can see if you substitute your Ansatz for the wave function. However, the infinitesimal volume$dV=dx\cdot dy\cdot dz$may ... 8 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 (x,... 8$\newcommand{\d}[1]{\,\mathrm{d}#1}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}\newcommand{\p}{\psi_{100}}\newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$The hydrogen ground state is the following: $$\psi_{100}=Y_{00}\frac{2}{a_0^{3/2}}e^{-r/a_0}$$ The expectation value of the position operator on this state is the following: \begin{... 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 (I tried to post this response with chart and links, but I'm new here, so the system won't let me include images or more than two links. Please cut and paste the other links to view them in your browser.) This diagram from the NH3 Fuel Association (taken from www.nh3fuelassociation.org/about-us--why-nh3) (scroll down the page) may answer part of your ... 7 The physical observable is not the wavefunction, but its integral over a finite area. In spherical coordinates, this is:$P({\vec x})=\int dr\, d\theta\, d\phi r^{2}\sin\theta \psi^{*}\psi$This integrand is manifestly finite at$r=0$, even if$R(r)$has a$\frac{1}{r}$divergance. 7 The scattering states must be included in the perturbative calculations if the result is to be highly accurate. In particular, it is not justified to ignore the continuous spectrum at energies close to the dissociation threshold. The Hilbert space in the position representation is the space of square integrable functions on$R^3\setminus\{0\}$with respect ... 7 With a potential$V(x) = - \frac{\alpha}{|x|}$, with the notation$a = \large \frac{\hbar^2}{m \alpha}$, solutions are : $$u^+_n(x,t) \sim x e^{ - \large \frac{x}{na}} ~L_{n -1}^1(\frac{2x }{na}) e^{ -\frac{1}{\hbar} \large E_nt}~~for~~ x>0$$ $$u^+_n(x,t) = 0~for~~ x\le0$$ and : $$u^-_n(x,t) \sim x e^{ + \large \frac{x}{na}} ~L_{n -1}^1(\frac{2x }{na}... 7 This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? After all, there's no particular reason for an electron to be in an eigenstate. Good question! The function \psi does not need to be Hamiltonian eigenfunction. ... 7 The graph shows the probability of finding the electron between the distances r and r + dr. This probability is given by:$$ P = \psi^* \psi dV $$where dV is the volume element:$$ dV = 4\pi r^2 dr $$So we get the probability:$$ P(r,r+dr) = \psi^* \psi 4\pi r^2 dr$$and therefore when$r = 0$the probability$P = 0$. It isn't that the ... 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

There is no integration of the radial part because, as you said yourself, we want the probability of finding the electron somewhere in the spherical shell between $r$ and $r+dr$ from the nucleus. (in a differential shell between $r$ and $r+dr$, and no need to integrate over $r$.)

6

I'm not sure what you mean by "collapse", but if I interpret that as "no hydrogen is formed" or "the electron is not captured", then 2 things can happen: 1) Elastic electron-proton scattering: the electron and proton just "bounce" off each other under some angle theta. By observing the cross section of the scattering versus the theta angle it was shown that ...

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 ...

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