Pila atomica fai da te

a me sembra un modellino , se leggi la scritta in piccolo in verticale nell'imggrossa...

E sarebbe?

Cioè fatemi capire, a cosa servirebbe?

Mi sbagliero' ma puo' funzionare solo in quelle dimensioni.

Montandolo in macchina mi auguro di non schiantarmi su un muro,se no quell'ampolla potrebbe finirmi in bocca!

Salve,

qui c'e' qualcosina: https://www.forumcommunity.net/?t=2558155

Riguardo quel sistema, non ho guardato attentamente ma credo che sfrutti la cattura delle particelle dovute al decadimento per creare energia elettrica. Il punto è che purtroppo è il metodo meno efficace. L'uranio già è caldo di per se a causa del decadimento, sfruttare quel calore tramite una termocoppia è più semplice. Forse però la risonanza che si cita rende il processo più efficiente. Quindi è su quella che ci si deve concentrare nell'analisi. Ovvero la domanda secondo me è: Che effetti ha la risonanza sulla barretta di uranio?

A voi la risposta, cosi' si movimenta la discussione

Salve,

qui c'e' qualcosina: https://www.forumcommunity.net/?t=2558155

Riguardo quel sistema, non ho guardato attentamente ma credo che sfrutti la cattura delle particelle dovute al decadimento per creare energia elettrica. Il punto è che purtroppo è il metodo meno efficace. L'uranio già è caldo di per se a causa del decadimento, sfruttare quel calore tramite una termocoppia è più semplice. Forse però la risonanza che si cita rende il processo più efficiente. Quindi è su quella che ci si deve concentrare nell'analisi. Ovvero la domanda secondo me è: Che effetti ha la risonanza sulla barretta di uranio?

A voi la risposta, cosi' si movimenta la discussione

Bravo Sierra

The following was posted March 11, 2006 by User:Erewhon.


Ok, I hate to lead one of my first comments off on Peswiki quite like this, but this is what we call "bravo sierra" in the engineering biz.

Are there ways to convert radiation to electric power, yes. He's even hit on one of the terms: betavoltaics. Betavoltaic converters (real ones) typically use doped semiconductor PN junctions similar to photocells. If an incoming beta particle produces an electron-hole pair near the junction, and the electron thus produced crosses the diode junction, it cannot return. This creates a voltage difference which produces power. Basic semiconductor and nuclear physics. However, the device in the picture is not a betavoltaic converter.

You can also make power with energetic alphas through a fairly nifty trick involving decelerating the alpha through an electric field so that they just kiss the collecting plate, you'll be hearing more about that one in conjunction with another system within the next couple of years. But it's a "well-known" approach. The device in the picture doesn't do that either.

Following the links to the description of the Nucell reveals some other problems, quite a few. Starting with Burke's cell, which seems fairly implausible too. Note that any conductor is replete with electrons which are very loosely bound to the atoms of the metal, to the point that they freely interchange places and migrate. You can almost treat it as a gas (aka Drude gas). Ionization states added to a metallic conductor would be fleeting, because the Drude electron gas is free to recombine with the produced 'hole', if you could even call it that in a conductor, and would do so in picoseconds.

Since you are creating a "hole" for every electron you free by ionization, you will "eat" a free electron from the Drude gas for every one you liberate, for a net gain of zero. The ions created by the passage of a beta particle will be shot in every direction in the wire as well, statistically, you will have as many going the wrong way as you do the right way.

Thus, the net power production of a nuclear particle striking a conductor is zero, except, of course, that any energy thus absorbed will be available as heat. Further inspection of Burke's patent reveals that he also believes that heating the radioactive source material will increase the energy in the emitted particles; bogus. So, I suspect Burke never actually built or validated the cell.

Next. The classic test: the claim is that the energy in the system comes from the radioactive decay of the U238 rod claimed to be in the picture. I say claimed, because I can't see where either Paul Brown or Peripheral Systems has a license for nuclear materials. This doesn't mean that they don't, but it's sure not in any listing that I have access to.

But let's be kind and say that the item in the photo really is 12.5g of pure U238. I hate to say it, but fully depleted uranium (U238) is not very radioactive at all. That's why it has a looooong half-life. U238 emits alphas only, it decays into Thorium 234 and Protactinium 234 which are beta emitters, and that cascades down into a bunch of lesser daughter products, each step emitting gamma radiation in addition to alphas and/or betas. The entire decay chain is too long to go into in this post, but if you take the first three decay steps of the chain into account, because they dominate the quantity of daughter products present, the decay energy produced is about 6.7MeV per mole per second. Well, U238 is obviously 238 grams per mole. He's got 12.5 grams of it. So there's about 0.053 moles of U238 in his setup, for an emitted power of about 0.352MeV per second. That sounds like a lot, doesn't it? But it's only about 0.16 MICROWATTS/sec. Now, that's if all the emitted particles make it out of the material. And they don't. Alphas in particular won't penetrate much of anything, and if they're emitted in the center of that rod, they won't make it to the surface. Their energy will be converted to heat in the rod, not that you'd be able to easily measure that small of an amount.

But to be kind, again, let's say it all somehow emerges to strike the conductors. Given the small cross-section of the conductors surrounding it, that's not too likely, but even if magic happens and they DO, you're still talking 0.16 MICROWATTS/sec here. That's all the power that will be coming out of the rod. If an emitted particle striking a conductor DID cause a net current flow, and it doesn't, that's still all the power you'll get.

There's some other hooha about extracting energy from a moving charged particle using a magnetic field. If you put a moving charged particle in a magnetic field, it does not decelerate or exchange energy with that field, it goes in circles. That's how mass spectrometers, your CRT's deflection yoke and cyclotrons work. You can extract energy from a moving charged particle with an ELECTRIC field; that's how the alpha particle trick I mentioned above works.

You should note that electric and magnetic fields are not going to have any effect on the decay rate of U238, nor on the types or energies of particles emitted from it. So no magic Tesla resonators or what-have-you are going to improve on that terrifically tiny power emission.

If you think I'm wrong, feel free to point out what you feel are the incorrect bits.

Further pickings: the header says the cell produces energy from nuclear decay, then says it does not rely on a nuclear reaction. Nuclear decay IS a nuclear reaction. The equations for the ones in U238 are easy to find, just google for "u238 decay chain".

Also, there's some comment made about deliberate mis-tuning to avoid "a runaway reaction". In order to have one, you'd have to have some sort of reaction that has positive feedback. A betavoltaic, as you say, DOES extract energy from nuclear decay. But it doesn't CAUSE nuclear decay. In order to have a "runaway reaction", the external circuit would have to have some sort of positive feedback mechanism with the nuclear decay rate. However, there isn't one. No amount of diddling around externally with electric or magnetic fields is going to change the decay rate. You can do what you want to with that circuit, and there's not going to be any "runaway reaction".

Summary: the story's pretty hole-y. There's nothing here to see.