Explain uranium to me

Factual Questions
81

Those reddit snippets should be fully ignored. The linked article answers your question incorrectly as well (while handwaving around a few unrelated true things here and there.) :man_facepalming:

The “cloud” is the answer to the question, “Where am I likely to find the electron?” It’s a probability distribution. While it’s mathematical form is a little messy, it’s no different from a Gaussian distribution or whatever in terms of how it functions. So your question becomes, “What sets the shape of the probability distribution for the location of the electron?” and/or “Why isn’t it an infinite spike at zero radius?”

Already at this stage it is required to take the leap from saying (classically), “A particle has a location,” to saying in QM, “A particle has a probability distribution for its location.” This is essential to QM. You can still measure the location of a particle, but the result will be random according to the probability distribution.

Classically, if you know the starting conditions for a particle and you can write down all the forces it feels from the world, you can predict it’s position for all time. You might start by writing Newton’s 2nd law (or an equivalent formulation of the physics) to relate forces (or potential energies) to changes in the particle’s momentum. The evolution of the momentum and, in turn, the particle’s position can then be extracted to provide a description of the position for all time.

In QM, if you know the starting conditions and the forces, you can predict the particle’s position’s probability distribution for all time. Newton’s 2nd law is not the relevant dynamical principle anymore. It needs to be something new that fundamentally works with probability distributions of positions or something related to those (in practice: the “wavefunction” – but we won’t need that detail here.) The relevant dynamical principle – the engine of the physics – is now the Schrödinger equation.

The math is a step up, but the principle is the same: you write down the forces (or potential energies) involved, take any needed starting conditions, and you can work out how the probability distributions behave.

Not any ol’ shape of probability distribution can result from the Schrödinger equation. For any specific force you impose, only certain shapes satisfy the equation. This is where the discrete nature of the physics comes to light. Analogy time:

Discrete answers to a spatial question shows up in many classical systems, for different physical reasons but equivalent mathematical reasons. A plucked guitar string is a common example. The vibrating string has a fundamental (lowest) frequency vibration that corresponds spatially to a one-humped vibration, as in the first figure on this wiki page. The next fundamental shape isn’t arbitrary; it’s the next pattern (two-humped) that satisfies the underlying dynamics while also respecting the fact that the ends of the string are pinned down and can’t move at all. This combination of having “constraints” on the system due to the forces and the boundaries and having a specific dynamical principle at play (here, Newton’s laws) is enough to yield this set of discrete behaviors for the string. These specific shapes of vibrations – plus combinations of them occurring simultaneously – fully describe the possibilities.

The shape of the probability distribution in QM for a particle subject to forces and constraints follows the same mathematical principle as the plucked string: certain shapes are fundamental. Further, different fundamental shapes for the probability distribution generally correspond to different amounts of energy in the system. These different “states” of the system are all that can exist (along with combinations of them).

Since the different states have different energies, the stable configuration is the one(s) with the lowest total energy. But even that lowest energy state has some specific shape for its probability distribution. For the electron in hydrogen, that lowest energy state has a probability distribution that is spherically symmetric and falls off exponentially with radius. That’s the dynamical “end of story” of this isolated system. There’s nothing further for it to evolve into. There is no more-squished probability distribution available because no viable shape with smaller energy exists given the specific forces involved and the dynamics embedded in the Schrödinger equation.

For a plucked string, the ends of the string being pinned down is like a “brick wall” constraint for the system. For the electron attracted to a proton, there isn’t a hard boundary like this. The proton provides not a brick wall defining a spherical enclosure but rather a spherical gradual “valley” that influences the electron. All the same, though, these forces lead to a suite of special shapes that underlie all resulting behavior. The electron system also has considerations of angular momentum that lead to much more complexity in the suite of shapes, and this is why your “hydrogen wavefunction” image has so much going on. (That image doesn’t actually include the lowest-energy state at (1,0,0), which is simplest in shape.) But the fundamental reasons for “Why doesn’t the electron fall in?” don’t require all that complexity.

82

The OP may find those articles interesting:

https://www.damninteresting.com/the-isle-of-doctor-seaborg/
I find the second one, without the preview, better, although it is older. Of course it is not at the same level as Pasta’s, Chronos’ and Dr.Strangelove’s contributions.

83

As @Pasta helpfully explains, and you can see from the pictures and graphs, there is no need to to consider multiple electrons— if you consider a single electron (i.e., hydrogen atom), even in the ground state the electron is not infinitely concentrated at the nucleus (any more than the Earth orbiting the Sun “collapses” into the centre— though an electron is not at a well-defined unique radius from the nucleus, hence the talk of “clouds”).

I suppose that is exactly what we teach the kids— you are supposed to draw diagrams representing particles and interactions, and there are rules so that, for instance, a vertex with an electron, an electron neutrino, and a W-boson is valid. Protons are not even fundamental particles at this level.

84

I would say instead that they just work differently. Chemical reactions require that the molecules collide in a particular way. If you increase the mass, you decrease the speed, since at a given temperature all molecules have the same average kinetic energy. But then, you also increase the momentum, since velocity goes down with the square root of mass, while the momentum increases linearly. So a D2 molecule will move at 0.7x the speed of an H2 molecule, but also have 1.4x the momentum. Of course, molecules that just contain H/D (water or otherwise) will be more subtly affected, but the principle is the same.

Reading around a bit, it seems that the main effect (when you get to high concentrations such as 50%) is that it suppresses cell division, and it’s mainly the function of enzymes that are disrupted. So the death is indirect, like with radiation poisoning. Your cells don’t get replaced as they die off. I’d suppose that as with radiation, the first effects are going to be with the cells that naturally get replaced most frequently, like blood and the stomach lining, but that’s just a guess.

85

This is an absolutely fundamentally quantum-mechanical question. I would go so far as to say that, once you have enough understanding of quantum mechanics to even be able to pose the question properly, that’s the point at which you’ve answered it.

86

There is a slight complication in that picture. The Earth doesn’t collapse into the sun because there’s no way for it to shed the energy it has in order to do so (except gravitational waves, but that is a slow process). But an electron is electrically charged, so classically it can (and must) shed energy by emitting electromagnetic radiation. The reason why it doesn’t is purely quantum mechanical.

87

If I understood what I read somewhere a while back the reaction time is a bit different, different enough to throw things off if enough of them are deuterium ions.

88

See Kinetic Isotope Effect. It’s very large in mechanisms involving proton tunneling.

89

I think the difference in strength between gravity and EM is responsible for the difference in how quickly waves of gravity or EM would cause orbital collapse (absent QM effects)

90

It’s partly that, and partly also that gravitational systems usually have both much lower frequencies and much larger energies. Gravitons and photons have the same relationship between frequency and energy, and a graviton emitted by the Earth orbiting the Sun would have a frequency of 1/year, giving it an absolutely pathetic energy, especially when that energy is coming from the fairly large orbital energy of the Earth.

91

Also, electromagnetism can emit dipole radiation, while gravitational waves are only quadrupole and above. The quadrupole moment tends to be significantly weaker than the dipole.

92

Thanks. That’s not completely unrelated to the strength of gravity, though, right? The frequency would be higher if gravity was stronger.

On the other hand, this is an independent difference between gravity and EM

93

Depends on what else you’re keeping constant. If you had a planet in a circular orbit at a distance of 1 AU from a Sun with the same mass but a stronger gravitational constant, then yes, the frequency would be higher. Though, no matter how strong you made gravity, there’s a limit to how high a frequency you could get from a 1 AU circular orbit. To a large degree, it’s low-frequency just because it’s big.

94

Thanks. That makes sense

95

For systems involving just a few fundamental charged particles, you are spot on. If the discussion was more about large-body systems, though, it is all dependent on the specifics of the physical system. For orbiting astronomical systems, the objects tend to be approximately electrically neutral and the masses and orbital periods span a bonkers range of orders of magnitude. So, there’s no off-the-cuff way to say whether EM or gravity would win the battle in who contributes more to orbital decay.

For the earth-sun system, synchrotron radiation beats gravitational radiation only if the earth has an excess charge beyond 10^{29} electron charge units (plus or minus).

For compact binary systems (like binary neutron stars or black holes), gravitational radiation is a clear winner. And in the end-of-life stage where the orbit has gotten very small, the system can be so gravitationally bright that it radiates solar-mass-equivalent amounts of energy in just a few seconds.

The energy per graviton doesn’t provide an answer because the rate of graviton emission is an independent consideration. The Earth-Sun system radiates 1042 gravitons per second (assuming gravitons exist), so it’s still just back to the overall power emission (which is about 200 watts, incidentally.)

96

Like others are stating, there is no such thing as non-isotopic. Isotopes refer to all kinds. The best example I can think of is hydrogen. Hydrogen has 1 proton and generally 1 electron. When we say “hydrogen”, we refer to the hydrogen atoms with no neutrons (I know there is a technical name for it, but that isn’t generally used). When we say deuterium, we refer to the ones with one neutron. They are both isotopes of hydrogen.

That being said, when you refer to uranium without the atomic mass number (for example, U instead of U235), I believe you are referring to the isotope with 146 neutrons (feel free to correct me if that is incorrect).

Question two is a misunderstanding, as explained.

As for question number three, I don’t know what I’m talking about enough to try and answer. Several other responses explain it better than I could get close to.

97

Not really that much. Only ~10 billion coloumbs. You could fly a balloon with a 1000-amp cable and a low-energy, high-current electron gun and create that charge difference in a few months. Unless there’s some neutralizing force that would outcompete it… which there probably would be.

Hmm, maybe we can get rid of the Van Allen belts by just charging up the Earth and attracting the protons.

98

Eh, the capacitance of the Earth is still less than a millifarad, so to get that 10 billion coulombs, you’d need to bring the surface to over 10 trillion volts. Doesn’t sound plausible, to me.

99

I didn’t follow your design (are you trying to shoot electrons off an ever-more-positively-charged earth?), but in any case, the interplanetary plasma from the solar wind neutralizes everything quite efficiently.

100

Yes, just shoot the electrons off. Probably the atmospheric density at balloon heights is still way too high to get electrons through. I suppose we’ll need a space elevator instead. And I haven’t worked out the energy required once the Earth gets a little charge. The first electrons are easy; one electron at even 1 eV is well above escape velocity. I’ll have to work it out sometime how it scales with increasing charge.