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19

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 ...

12

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 ...

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 ...

10

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, ...

9

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 ...

8

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 ...

7

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 ...

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 ...

4

Air is lighter because there are fewer molecules per unit volume compared with a unit volume of liquid water. A mole of water is 18 grams, so a liter of water contains about 55 moles (1000 grams). A mole of air at standard temperature and pressure, however, occupies a volume of 22.4 liters, much more. Dividing a mole of 02 (32 grams) by 22.4, you have ...

4

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 ...

4

No, the radial parts of the wavefunctions are not orthogonal, at least not quite to that extent. The radial components are built out of Laguerre polynomials, whose orthogonality only holds when leaving the secondary index fixed (the $\ell$ or $2\ell+1$ or whatever depending on your convention). That is, $$\langle R_{n'\ell} \vert R_{n\ell} \rangle \equiv ... 4 With regards your first question: A similar (the same?) question you might reasonably ask is: how can we assume that the proton is stationary, at the centre of the problem, since it is surely going to be attracted by the electron and jiggle about a little? This is a question that would be just as valid directed at a classical system --- say, a planet ... 4 The idea here is increasingly complex depending on how deep into modern physics you want to delve, but also key to understanding quantum mechanics. So, I'll give a bit deeper explanation than it seems you've seen, but there's plenty more. It's understood that a photon acts both as a particle and a wave. As a particle it has an amount of energy associated ... 4 This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? After all, there's no particular reason for an electron to be in an eigenstate. Good question! The function \psi does not need to be Hamiltonian eigenfunction. ... 3 This calculation agrees with experimentally measured spectral lines, but why would we expect it to be true, even if we accept that the electron moves according to the Schrodinger equation? Your puzzlement arises because you are putting the cart in-front of the horse. The cart is the theoretical model of quantum mechanics and the horse is the data. As ... 3 These relations are based on the fact that both the position and the momentum distributions are centred around zero, which is in turn due to the symmetry of the atom. Given that, the width of the position and momentum distributions (\Delta x and \Delta p) is of the same order as a typical position or momentum within those distributions (r and p). 3 The time-independent Schrödinger equation for the hydrogen atom is$$-\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

The electron is the lightest lepton and the proton is the lightest baryon, so it's hard to see what reaction could occur without violating lepton number or baryon number. I suppose if proton decay (to a pion and positron) occurs then there could be a reaction to give a pion and two photons.

3

The symmetries that you're missing are conservation of baryon number $B$ and lepton number $L$. We strongly suspect that baryon number is not an exact symmetry, because the universe appears to contain very many baryons and very few antibaryons. Actually, a better metric for the baryon asymmetry of the universe is to compare the baryon density to the density ...

3

What you're trying to do is called radiometric calibration. The problem with doing it with an ordinary incandescent light bulb is that the bulb itself would have to be radiometrically calibrated to get a precise spectral calibration of the detector. That's because the bulb's spectral emissivity deviates from an ideal blackbody source with an emissivity ...

3

The numbers are not measured, they are part of a model that explains why the hydrogen atom emits/absorbs only at those wavelengths. The first series of Hydrogen lines to be discovered was the Balmer Series, and nobody knew why they were discrete lines instead of a continuous spectrum. Johann Balmer discovered that the lines all had wavelengths equal to, ...

2

It's because there is another vector quantity $A_i$ conserved in addition to the angular momentum $L_i$. Furthermore, the commutation relations of $A_i$'s and $L_i$'s are those of $SO(4)$. See for instance this reference : http://hep.uchicago.edu/~rosner/p342/projs/weinberg.pdf

2

The Wikipedia article answers most of your questions. What are the requirements for hydrogen atoms to go through fusion? Two atoms must overcome the coulomb barrier, which can be done by forcing two atoms very close together, or by leaving them moderately close for long periods of time, which allows them to tunnel through the barrier. Is it a ...

2

Air is mostly dinitrogen and dioxygen. At room temperature water is liquid and the molecules are close together due to strong attraction of water molecules to each other. This intermolecular interaction is called "hydrogen bonding". The oxygen atom of water has a partial negative charge and the hydrogens partial positive charges. A hydrogen atom of one ...

2

depends on the energy of the electron. For low energies, a bound state will be formed due to electromagnetic interaction between the two. In the case of higher energy, the proton can be transformed into a neutron.

2

Hydrogen bonding arises when a chemical bond is polarised to one end of it has a slight positive charge and the other has a slight negative charge. In the case of o-nitrophenol it's mainly the OH bond that is polarised - the H atom has a slight positive charge and the O atom has a slight negative charge. The charge separation means the OH bond has an ...

2

user26143 gives a correct argument about the effects of entanglement in two-atom processes, but specifically to your question, I have to reiterate the answer 1) You are not talking about entanglement 2) Yes, this effect gets effectively blocked out in the cases you consider 3) QFT does not really bring any new insight to this problem In theory, all ...

2

To add to John Rennie's answer and Hennes's Answer, an answer to your last sentence: How will the aliens know what this means? is well given in a thoroughly interesting and enjoyable tale by Carl Sagan's novel Contact. I can't recall whether the following is made clear in the film with Jodie Foster. In the novel, signals received from the Vega ...

2

No, the difference in lifetimes shouldn't be due to energy difference. For one thing, these energy differences are nearly zero. The states of a given $n$ in a hydrogen atom should be nearly degenerate regardless of $l$-value. (I say "nearly" since there is some energy splitting due to the Lamb shift, spin-orbit coupling, etc., but these are expected to be ...

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