plante1999
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''Random'' question about Nuclear physic
I have a question about Nuclear physic , If I add thorium dioxide on a peltier chip does the Beta radiation will make electricity?
[Edited on 18-10-2011 by plante1999]
I never asked for this.
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Ozone
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No. Only heat of decay would cause a current to flow through a thermocouple (what you have if you run a Peltier in reverse). The half-life of 232Th is
waaaay too long to impart a specific activity anywhere near high enough to be useful for this purpose. 238Pu and 90Sr-90Y, however, have found use in
this respect. See RTGs (radioisotope thermoelectric generators).
O3
-Anyone who never made a mistake never tried anything new.
--Albert Einstein
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plante1999
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Wath common isotope could do the job and are easily find in common material wich is not too dangerous.
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ScienceSquirrel
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They are simply no isotopes that are readily available that could be used. Anything radioactive enough to heat up would be far to radioactive to be
close to.
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plante1999
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Thanks , It is simply for a model of a nuclear pile that look like this: http://en.wikipedia.org/wiki/Betavoltaics , so if it make 0.01V it should be okay. How exacly the radiation make electricity in this process?
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ScienceSquirrel
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That uses tritium which emits beta particles to drive the cell. Tritium is found on luminous watch dials, etc in minute quantities.
I suspect that having enough to make a cell that would produce 0.01V would require some heavy licensing.
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plante1999
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I suspect that there is other isotope that cant emits beta particles to drive the cell wich does not require licensing.
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stygian
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Any amount of radioisotope that does anything beside exists (for a while) probably requires licensing.
Though could it really be that hard to get .01v?
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fledarmus
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0.01 V is a lot at atomic levels!
Do some back of the envelope calculations - how many electrons would need to be released to generate this much potential? Back-calculate using rates
of decay - how much material would have to be packed into what volume to generate that many electrons per second?
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unionised
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OK, here's the "back of an envelope" calculation.
Imagine I have some tritium. Each time an atom of it decays it fires out an electron with about 15000 electron volts of energy. Most beta decays are
about a hundred times more energetic
For some odd reason you think 0.01 is a lot.
How much tritium would I need to get 1 electron per second, well, it's certainly not more than 1 Bq because that is 1 electron per second.
What does that mean in practice?
According to this
http://www.physics.isu.edu/radinf/natural.htm
the natural tritium level in water is about 0.6 mBq/litre so I'd need about 6000 litres of water. Or, if you prefer, from the same site, I have about
about 23 Bq of trituium in my body. so I'm at least 23 times more active than your suggestion.
Actually, because that single electron has a lot of energy it will probably ionise a lot of molecules as it slows down so I could probably use 10
thousand times less tritium.
That's the radioactivity of a glass of water.
Not very impressive.
Of course it has nothing to do with the original question. There's simply no way that you could legally get hold of enough radioactive material to
generate enough heat to notice.
[Edited on 20-10-11 by unionised]
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Mr. Wizard
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There are many types of "nuclear battery". The most common one uses the heat from short lived isotopes that get warm from the relatively fast decay.
This heat is used to warm one junction of a thermoelectric generator, with the other junction kept as cool as possible. These generators can be
designed to deliver a fraction of a watt to thousands of watts. Their life-span is determined by the isotope, and design considerations. A very short
lived isotope would generally get warmer and make more power, but not last as long. Each lead telluride junction would deliver millivolts, but at a
higher amperage.
There is also a 'plasma thermocouple' or thermionic conversion which uses the heat generated to boil electrons off a conductor, through a plasma, to
another cooler conductor.
There is another type of battery that depends on the electrons being emitted from the isotope through an insulator and caught by another conductor.
These 'Beta Cells' are high voltage low current (1000 VDC at microamp current).
More information can be found where I found this: "The Sourcebook of Atomic Energy" Chapter 17:189- 17:202 published 1967.
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plante1999
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Beta cells are exactly wath I want to make. Where can I find the ''The Sourcebook of Atomic Energy''?
[Edited on 20-10-2011 by plante1999]
I never asked for this.
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ScienceSquirrel
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I suspect that any beta source strong enough to drive a cell will not be available to anyone outside a licensed research lab or similar.
It is going to have to kick out a fair few electrons to do anything at all so it will be pretty 'hot' in radiation terms.
For instance potassium 40 is a natural beta emitter but it is present in only very small quantities and it has a very long half life http://en.wikipedia.org/wiki/Isotopes_of_potassium so you would need to use a very large bag of it and wrap it in your capture material to attempt
to make a practical device.
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aaparatuss
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dr. Paul brown
Always remember dr. Paul brown.... he was on to something
before is untimely death
http://www.rexresearch.com/nucell/nucell.htm
i always pick up up my smoke detectors when i see them cheap, but i need like 500,000 thousand of them....
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plante1999
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Because of the majority of comment conserning the quantity of radioactive material , I desided to change somewath my question , I know that alfa
radiation make ligth when they ''hit'' glowing material like Ag doped ZnS (there is probably some compound that have a much longuer service life) , it
is possible to simply make '' glow'' a compound with alfa radiation and then use the glow to make electricity?
I never asked for this.
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plante1999
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I find a plastic that if hited with alfa/beta radiation it is suposed to emit glow that could be used with a solar panel to make electricity... The
name of this polymer is Polyethylene naphthalate.
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IrC
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If only you could power something using the output of a photomultiplier tube. However you cannot so the idea will not work.
"Science is the belief in the ignorance of the experts" Richard Feynman
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