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In the early phases of the big bang only the constituent parts of nucleons (quarks and anti-quarks) plus leptons (e.g. electrons, positrons) and light existed. As the universe expanded and cooled, quarks were able to combine and form the basic building blocks of nuclei - the neutrons and protons. A proton is of course a hydrogen nucleus; any heavier elements ...


31

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 ...


26

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 ...


26

Of course the reaction is possible. It doesn't even require special environmental conditions. Having no charge the neutrons don't need to overcome a strong Coulomb barrier to interact with atomic nuclei and will happily find any nuclei that can capture them at thermal energies. KamLAND (for instance) relies on this reaction as the delayed part of the delay-...


24

Warning: long post ahead. To simply watch an animation, scroll to the bottom ;) Requirements on the wavepacket First let's do some estimations, using the exact analytic expression for evolution of a Gaussian wave packet in free space. Hopefully, it'll not be too wrong in the case when Coulomb potential is on. The expression looks like this: $$\...


23

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 ...


23

It is known that orbital motion is accelerated motion under the influence of a central force; and that in a hydrogen atom the electron orbits the proton and is therefore accelerated Your initial assumption is wrong. Electrons do not orbit atoms like a tiny planetary system. The angular momentum of the electron in the ground state of hydrogen is zero. So the ...


21

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 ...


21

This is an interesting and non-trivial problem. Basically the Coulomb potential assumes a point particle but, if the proton is modelled as a solid sphere of finite radius, part of the electron wave function would be "inside" the proton, where the assumption of point charge no longer holds. To account for this one must modify the Coulomb potential from $1/...


20

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 ...


19

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 ...


18

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 ...


18

In general - yes, if an electron is in superposition of eigenstates it can radiate its energy. In order to describe this, of course, we need also to introduce into our model the electromagnetic field, so the electron will be able to radiate its energy to something. We can do this and calculate transition probabilities and rates etc. And this is of course ...


17

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 ...


17

As John Rennie says the hydrogen atom is a case where the electron is in the s shell, which means it has no angular momentum. We can think of the electron if it were measured as being at a point above the proton, where the electron as a wave would then spread into a spherical shape around the proton. One might imagine a sort of Zeno machine that keeps the ...


17

Especially the hydrogen atom, with a proton in the nucleus and an electron revolving acting as a dipole This is a problematic way of understanding the hydrogen atom ─ it basically tries to insist on treating it within classical mechanics, and this is doomed to fail. Instead, the hydrogen atom must be treated within quantum mechanics. This introduces a bunch ...


16

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 ...


16

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 $\...


16

$H_2$ contains 2 electrons in the same ground-state orbital; by Pauli exclusion, one must be spin-up and the other must be spin-down. The 21cm line is generated in a normal hydrogen atom when an electron's spin flips from being aligned with the proton to being anti-aligned with the proton. In $H_2$, an electron's spin cannot flip because it would then be ...


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

Basically, it's the Schrödinger equation for a free particle, but it's important to note that that particle isn't the proton - it's the entire atom's center of mass. This is covered in reasonable detail in suitably rigorous textbooks in quantum mechanics (though I can't think of a specific example at the moment), and the basic idea goes like this: You ...


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 ...


12

Between about t=$10^{-12}$ and t=$10^{-6}$ seconds, the universe was filled with a quark-gluon plasma. Temperatures were too high for mesons and baryons (like protons and neutrons) to remain bound, should they briefly form. Then, between about t=$10^{-6}$ and 1 second, during the "Hadron epoch," the temperature was low enough that quarks that stuck together ...


11

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 ...


11

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 $d-1$ ...


11

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}...


11

In the Bohr model of the hydrogen atom we take the potential energy of the electron to be zero at infinity, so the potential energy becomes negative as the electron approaches the hydrogen atom. However kinetic energy is always positive. In the Bohr ground state the potential energy is -27.2 eV. Note that as described above this energy is negative. The ...


11

By conservation of momentum, the center of mass of the atom is what actually stays fixed. This implies that there is a perfect correlation between the wavefunctions $\Psi$ of the electron and $\Phi$ of the proton: $$\Phi(x)=\Psi(-(M/m)x),$$ where $M$ is the mass of the proton and $m$ is the mass of the electron. The effect on the energy levels is to ...


10

From what I have read in "American Prometheus: The Triumph and Tragedy of J. Robert Oppenheimer" Teller was the first one to express this concern before the Trinity test. Also quoting from: http://www.sciencemusings.com/2005/10/what-didnt-happen.html Physicist Edward Teller considered another possibility. The huge temperature of a fission explosion -- ...


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