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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

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, ...


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

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 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 ...


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

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

(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

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 is a lot of confusion on this issue, and indeed plenty of textbooks have got their terminology mixed up. The brief story of the opaqueness of the universe is as follows: In the beginning, everything was a plasma. The photons coupled to the free protons and electrons, unable to travel far before scattering. As the universe expanded and cooled, neutral ...


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

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

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

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

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

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

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

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

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

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

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 ...



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