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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 ...

9

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, ...

7

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 ...

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 ...

7

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 ...

6

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 : $$u^-_n(x,t) \sim x e^{ + \large \frac{x}{na}} ~L_{n -1}^1(\frac{2x ... 6 There is no integration of the radial part because, as you said yourself, we want the probability of finding the electron somewhere in the spherical shell between r and r+dr from the nucleus. (in a differential shell between r and r+dr, and no need to integrate over r.) 5 (I tried to post this response with chart and links, but I'm new here, so the system won't let me include images or more than two links. Please cut and paste the other links to view them in your browser.) This diagram from the NH3 Fuel Association (taken from www.nh3fuelassociation.org/about-us--why-nh3) (scroll down the page) may answer part of your ... 5 Making ammonia is an absolute bugger because nitrogen is so unreactive. The normal route is the Haber process, http://en.wikipedia.org/wiki/Haber-Bosch, but this requires high pressures and temperatures. Hydrogen is easy to produce from electricity, and I guess that's why it's the first choice for storage of energy generated by unreliable sources like wind ... 5 The physical observable is not the wavefunction, but its integral over a finite area. In spherical coordinates, this is: P({\vec x})=\int dr\, d\theta\, d\phi r^{2}\sin\theta \psi^{*}\psi This integrand is manifestly finite at r=0, even if R(r) has a \frac{1}{r} divergance. 5 The ordinary Bessel functions are perfectly well defined for complex arguments. For example, here is a plot of \Re[J_2(x + i y)]: The difference between the ordinary and modified Bessel functions is that they satisfy different equations:$$ z^2 y'' + z y' + (z^2 - n^2) y = 0, $$for the ordinary Bessel functions and$$ z^2 y'' + z y' - (z^2 + n^2) y ...

4

googling "introduction to experimental atomic spectroscopy" gives some pretty nice results. And yes, the spectrum of atomic and molecular hydrogen is radically different. This question at physicsforums correctly points the user to the NIST spectra database. One should keep in mind that not all of the possible emission/absorption lines will show up in any ...

4

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 ...

4

There are two reasons: First, the expansion of space, which was rapid in the early universe, separated the initial density fluctuations into isolated potential wells. Dark matter and ordinary matter then accumulated into these local potential wells, which eventually become galaxy clusters. Second, the temperature of the early universe was very high, so ...

4

The approximation that we all started out learning is the linear combination of atomic orbitals (LCAO) approach. The molecular wavefunction, $\Psi$, can be expressed as a sum of some set of basis functions: $$\Psi(\vec{r}) = \sum_n f_n(\vec{r})$$ and a convenient set of basis functions is the atomic orbitals of hydrogen. As a starting point we could take ...

4

When you use the reduced mass, what you have first done is to go from the variables $(r_1,p_1)$ and $(r_2,p_2)$ to $(r=r_1-r_2, p/\mu=p_1/m_1-p_2/m_2)$ and $(M R=m_1 r_1+m_2 r_2,P=p_1+p_2)$, where $M=m_1+m_2$ is the total mass and $1/\mu=1/m_1+1/m_2$ is the (inverse of the) reduced mass. As you said, this is usually introduced in classical mechanics to ...

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, $$\langle R_{n'\ell} \vert R_{n\ell} \rangle \equiv ... 4 With regards your first question: A similar (the same?) question you might reasonably ask is: how can we assume that the proton is stationary, at the centre of the problem, since it is surely going to be attracted by the electron and jiggle about a little? This is a question that would be just as valid directed at a classical system --- say, a planet ... 3 No spin measurement of proton can give a value more or less than \hbar/2. But what do we mean when we say that spin of proton is \hbar/2 ? Spin is a 'vector' quantity (at least this is what it is classically). So one should also specify its direction. The thing is that in this case direction doesn't matter much. If you think of proton as some sphere and ... 3 Mike's answer is correct that this exchange happens constantly within a system containing water. You are right, however, to suggest that heating water can cause disassociation of water molecules. I am not sure to what temperature you would need to heat steam for this to happen in the absence of other substances, but the explosions at the Fukushima power ... 3 Whenever you have a chemical reaction like 2H_2O \rightleftharpoons 2H_2 + O_2, it goes both ways. In this reaction, the products on the right have more chemical energy, so going to the right requires energy input, and going to the left releases energy, by the same amount. Under any conditions, the reaction is running in both directions, but usually at ... 3 Why did you think you were doing something wrong? The phase factor does indeed become irrelevant when you calculate the probability density. As for the factor of r^2: the proper way to interpret |\Psi|^2 is that, when integrated over some region, it gives the probability of the electron being found in that region:$$P(\text{e in }V) = ...

3

You are right, the $\phi$-dependence disappears from the probability of this state. The probability is symmetric with respect to the reflections. The differential probability in spherical coordinates is determined as $$dw=|\Psi(\vec{r})|^2 dV=|\Psi(\vec{r})|^2\cdot r^2dr \cdot sin\theta d\theta \cdot d\phi$$ You can enjoy the 3D visualizations and even ...

3

First, it is not true that $R(r)$ has to be $L^2$. Because the integration measure is $dV = r^2\cdot dr\cdot d\Omega$, and we integrate $|R|^2\cdot |Y|^2$ with this measure, it is $R(r)r=u(r)$ and not $R(r)$ itself that must be $L^2$. Now, this is not just a correction of an unrelated minor mistake in your comments; it actually answers your main question. ...

3

To get it in the momentum representation, one has to do the Fourier transform of this function. This reference can be useful: http://forum.sci.ccny.cuny.edu/Members/lombardi/publications/MOMREP-H-atom.pdf/view At the end, separation of variables after transformation to the momentum space is not trivial, and the mixing of quantum number is presented.

3

What happens, essentially, is that the S and P wavefunctions get mixed to produce eigenstates that have shifted centres. This means the atom gets an induced electric dipole moment, whose interaction with the external field either lowers or raises the eigenenergy. More specifically, consider the wavefunctions of the states $|200\rangle$ and $|210\rangle$: ...

3

$P(r)dr$ gives you only the probability in an infinitesimal spherical shell around the center. The integration you're expecting is made, when you want to know the probability in a non-infinitesimal shell around the center. For example, you'd like to know what is the probability of finding an electron between $r=1$ and $r=2$ (in whatever coordinates), you'd ...

2

Your last sentence is exactly right: the energy cost for fusion is almost all in those last few femtometers, at which electronic effects are negligible. Although there is in principle a difference between colliding neutral atoms and nuclei, at the energies required for fusion the effect is tiny. The energy differences associated with the presence, absence, ...

2

I haven't done the calculations, but I doubt that this scheme would generate net energy. As was pointed out, electrolysis uses a lot of energy. However, after the H2 and O2 rises up the water column, you could get some of the energy back with a fuel cell that would convert the hydrogen and oxygen back to water and supply additional electric power, but ...

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