# Tag Info

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 ...

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

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 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 ... 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

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 $|210\rangle$: ...

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 ...

2

The problem is that you're thinking of the electron as a particle. Questions like "what orbit does it follow" only make sense if the electron is a particle that we can follow. But the electron isn't a particle, and it isn't a wave either. Our current best description is that it's an excitation in a quantum field (philosophers may argue about what this ...

2

Just to make sure we're all on the same page, I'll take a step back before taking two steps forward. The normalized wavefunctions for the $n = 2$ energy level are usually written as \begin{gather} \psi_{2,1,0}(r, \theta, \phi) = \left(\frac{1}{32\pi a_0^3}\right)^{1/2} \frac{r}{a_0} \mathrm{e}^{-r/2a_0} \cos(\theta), \\ \psi_{2,1,\pm1}(r, \theta, \phi) = \mp ...

2

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 ...

2

http://en.wikipedia.org/wiki/Recombination_(cosmology) The first phase change of hydrogen in the universe was recombination, which occurred at a redshift z = 1100 (400,000 years after the Big Bang), due to the cooling of the universe to the point where the rate of combination of an electron and proton to form neutral hydrogen was higher than the ...

2

If you pour water on a 3000 degree celsius fire it goes boom. Because water dissociates into H2 and O2 before H2 detonates.... That is why firefighters use specially made foams to estinguish these kind of fires. http://en.wikipedia.org/wiki/Water_splitting

2

It is only a question of definition. There is the operator of interaction of particle with an externally-produced magnetic field: $\hat{H}_{int}=-\hat{\boldsymbol{\mu}}\cdot\mathbf{H}$, where $\mathbf{H}$ is a magnetic field and $\hat{\boldsymbol{\mu}}$ is an operator: $\hat{\boldsymbol{\mu}}=\displaystyle \frac{g e}{2 m} \hat{\mathbf{s}}$ By the ...

2

As Lubos points out, you are allowed to make any substitutions you want as long as the new function is equivalent to the old one. In this case, substituting $u(\rho)=\rho^l e^{-\rho}v(\rho)$ is allowed because you can recover $u$ if you know $v$. The question is, though, why would you choose exactly such a form? If you're just given it outright, it does ...

2

A hydrogen atom ion $H^{+}$, with an atomic mass number of A=1, charge number Z=1, is the same as a proton. A hydrogen ion thus usually just refers to a proton. Depending on context, however, you may also have a hydrogen ion which is (a) an ion of a deuterium atom, in which case it is a bound state of a neutron and a proton, with atomic mass number A=2, ...

2

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.

1

First, before we go any further, we need to understand pressure. Imagine piling sand on top of you on the beach--just a little layer, you don't feel much weight, but as you get buried deeper and deeper, you feel more and more force pushing down on you. This makes sense, since your body has to hold up the weight of all of the sand on top of you. It's ...

1

First, Hydrogen bond is not the bond in a Hydrogen molecule. A hydrogen bond is another kind of bond. Second, chemical bonding cannot be described by the Schrödinger equation alone because this equation only describes isolated systems and an atom in a molecule is anything except isolated! The Hydrogen molecule is trivial, there are only two atoms and are ...

1

The Hamiltonian for the spin-spin interaction is: $\Delta H_{SS} = \frac{\gamma_p e^2}{m m_p c^2 r^3} \Big( \frac{1}{r^3} \big(3(\vec{s}_p \cdot \hat{r})(\vec{s}_e \cdot \hat{r})-(\vec{s}_p \cdot \vec{s}_p) \big)+\frac{8 \pi}{3} (\vec{s}_p \cdot \vec{s}_p) \delta^{(3)}( \vec{r} ) \Big)$ For the cases where $l \neq 0$ the term with the delta function ...

1

According to Wolfram Alpha the gas density in the exosphere is $10^{-13}$ to $10^{-15}$ kg per cubic metre. Suppose your spaceship has a collector with a 10 square metre area, then it would have to travel at least $10^{12}$ metres to collect 1kg of hydrogen. For comparison, $10^{12}$ metres is about 2,500 times the distance to the moon. I suspect that makes ...

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