## [Physics] Curie and Becquerel

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Arminius
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### [Physics] Curie and Becquerel

Hello xkcd users!

This question actually came to me when reading this xkcd topic: Electricity from beta decay? (can't put in the link, it considers it spam)
A 10 mCi beta emitter will directly produce 3.7x108 e-/s = 59 pA. If the beta emitter is Ni-63, then each electron is carrying 67 keV. If you extracted all the energy from the electrons, then this would be 4 microwatts. A 10 mCi source of Ni-63 should mass 0.16 mg, if I've done my math correctly.

I wanted to do this calculation with other pure beta emitters (for example palladium isotopes if I'm not wrong) and went to search for the a list/table with isotopes of the most common elements and their radioactivity density (how many Ci or Bq per g). Furthermore, I was looking for the same list/table but with the energy of each electron.

I could find neither. Does any of you have an idea where to look? (wikipedia page does not provide this information).

PsiSquared
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### Re: [Physics] Curie and Becquerel

You won't find such a table, because it is very easy to calculate specific activity from the half-life of the said isotope:

Specific Activity (in Bq/g) = 4.173x1023/At

Where A is the atomic weight of the element, and t is the half-life in seconds.

(In case you wander where the number 4.173x1023 came from, it's Avogadro's number times the natural logarithm of two).

So, for Ni-63, you'll have:

A = 63
t = 96 years = 3x109 seconds

So the specific activity would be 4.173x1023/(63x3x109)=2.2x1012 Bq/g (and if you multiply this by 0.16 mg, you get an activity of 9.5 mCi, which is close to what your source stated).

As for the electron energies, they do appear in Wikipedia. Whenever Wiki gives a list of radioisotopes, the decay energy is usually given for each isotope. Just remember two things:

1. Decay energy is usually given in a unit called "electron-volts" (eV). To convert it to ordinary energy units, use the conversion rate 1 eV = 1.602x10-19 Joules. It is also common to use larger units such as keV (1000 eV) and MeV (1 million eV).

2. The energy given is the total energy output of the decay. The actual energy of the emitted electron would be somewhat lower, because of the accompanying neutrino which also has some energy. Since its a random quantum process, you cannot predict in advance how the energy will split between the electron and the neutrino in any given decay. But as a rough estimate for power generation purposes, you can assume that about half of the total energy output is available for you to do useful work.

Arminius
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### Re: [Physics] Curie and Becquerel

Thank you so much for the answer! I had believed that those results were all purely experimental.

I had a look at the isotopes of palladium on wikipedia. Let's take Nickel. http://en.wikipedia.org/wiki/Isotopes_of_nickel The numbers indicated are the isotopic mass, not the energy!

I knew about energy of neutrinos but I'm lucky you gave me an estimate. I thought that the energy of the neutrino was negligible for all intends and purposes!

P.S.: Got another source
and

Ok I don't get this. For Palladium 103 with electron capture they have an energy of 2679 KeV? Or are there Auger effect electrons that have that energy? I have been out of the field for way too long.

PsiSquared
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### Re: [Physics] Curie and Becquerel

Well, the energies used to be listed on Wikipedia. Very strange that they've removed this, because it is really basic stuff. Any decent table of radioisotopes would include this information.

So basically, you have two choices. You can look for a different source (I'm sure they exist, both offline and online. But I don't have the time to start searching for them). Or you can calculate it yourself, using E=mc2. It's not too difficult. You simply subtract the mass of the decay products from the mass of the parent nucleus, and multiply by c2. Just remember that atomic masses and energies are given in weird units (atomic units for mass, eV for energy) so the value you'll need for c2 will also be a strange number: about 931,500,000.

Take Ni-63 for example. The atomic mass, as given in wikipedia, is 62.929669. It decays to Cu-63, whose atomic mass (again, in wikipedia) is 62.929597

Subtracting the two, you get: 62.929669-62.929597=0.000072 atomic mass units. And the energy is 0.000072x9135000000 = 67000 eV = 67 keV

As for the neutrino energies, I'm afraid I've misled you a bit. The neutrinos actually take more than 50% of the energy. I've erronously assumed that the decay energy is high relative to the electron rest-mass, but it usually isn't. For a decay energy of 67 keV (which is quite low), about 95% of the energy would go to the neutrino.

Here is a table of the energy split for the typical case (i.e. the case where the electron and the neutrino have equal and opposite momentums):

10 keV - electron 1%, neutrino 99%
67 keV - electron 5%, neutrino 95%
100 keV - electron 8%, neutrino 92%
250 keV - electron 16%, neutrino 84%
500 keV - electron 25%, neutrino 75%
1 MeV - electron 33%, neutrino 67%
2 MeV - electron 40%, neutrino 60%
High Energy Limit - electron 50%, neutrino 50%

The actual average electron energy would be a bit higher, because the distribution isn't symmetric. But actually calculating the exact average is pain (it requires doing an integral over all possible outcomes), so the above numbers will have to do. You can certainly use them as a rough estimate (and as a definite lower bound).

Arminius
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### Re: [Physics] Curie and Becquerel

I actually wonder why I didn't think of it myself?

I get for Pd-103 543,0645 keV
and Pd-107 33,53400001 keV.

Just for the fun, lets do the calculation for the energy resulting from the decay of those two elements.

First line: Pd-103
Second line: Pd-107

Half life (s):
146880
2,04984E+14

Bq/g:
2,75834E+16
19025875,19

Energy:

2,75834E+16*543,0645*1000*1,602*10^(-19) = 2400J/gr
19025875,19*33,534*1000*1,602*10^(-19) = 10^(-7)J/gr

For the fun of noticing mistakes, let's analyze this "joke-response". [url]http://www.huffingtonpost.com/quora/what-is-the-theoryconcept_b_3456241.html
[/url]
Imagine we could magically extract all the energy (which is not what that concept states... I don't want to imagine with the neutrino losses you stated) and we were using pure Palladium 103 (which again is not what the response states) poor Mr Stark has a meager 2400J/gr, or for a 1kg device (it looks actually lighter): 2,4 MJ. In terms of power for 1 day: 2400*1000/(60*60*24) = 28 Watts. Enough to power a low power light bulb. Any mistake on my side?

P.S.: How did you calculate your table splitting up neutrino energy and electron energy?

PsiSquared
Posts: 126
Joined: Wed May 09, 2012 6:02 pm UTC

### Re: [Physics] Curie and Becquerel

Arminius wrote:I actually wonder why I didn't think of it myself?

I get for Pd-103 543,0645 keV
and Pd-107 33,53400001 keV.

Just for the fun, lets do the calculation for the energy resulting from the decay of those two elements.

First line: Pd-103
Second line: Pd-107

Half life (s):
146880
2,04984E+14

Bq/g:
2,75834E+16
19025875,19

Energy:

2,75834E+16*543,0645*1000*1,602*10^(-19) = 2400J/gr
19025875,19*33,534*1000*1,602*10^(-19) = 10^(-7)J/gr

For the fun of noticing mistakes, let's analyze this "joke-response". [url]http://www.huffingtonpost.com/quora/what-is-the-theoryconcept_b_3456241.html
[/url]
Imagine we could magically extract all the energy (which is not what that concept states... I don't want to imagine with the neutrino losses you stated) and we were using pure Palladium 103 (which again is not what the response states) poor Mr Stark has a meager 2400J/gr, or for a 1kg device (it looks actually lighter): 2,4 MJ. In terms of power for 1 day: 2400*1000/(60*60*24) = 28 Watts. Enough to power a low power light bulb. Any mistake on my side?

P.S.: How did you calculate your table splitting up neutrino energy and electron energy?

PsiSquared
Posts: 126
Joined: Wed May 09, 2012 6:02 pm UTC

### Re: [Physics] Curie and Becquerel

Arminius wrote:Any mistake on my side?

The calculations themselves seem fine to me.

But you've interpreted the results wrong. The energy figures you've calculated represent the amount of energy emitted every second.

So, for your first example, the actual energy output is 2.4 MJ per second, which is 2.4 megawatts!

Now, before you get all excited about the existence of such a power source, you should check how much a kilogram of pure Pd-103 would actually costs. I don't know the exact prices, but tiny medical pellets of Pd-103 typically cost tens of thousands of dollars. So a full kilogram of the stuff would probably cost billions.

You should also check how much radiation such a thing would emit. Your poor Mr. Stark would have been fried by the radiation, the moment he got anywhere near that thing.

Let me tell you something: I once toyed with the idea of using a beta-emitter to power a interplanetary spaceship. I liked the idea that unlike a conventional nuclear reactor, a decay always proceeds at a constant rate. No need to start it or control it. No fear of a "meltdown" due to an uncontrolled reaction.

Seemed simple enough. But when I did the actual calculations, I found that the whole concept is completely and insanely impractical. The constant radiation flow would be enormous, and no amount of shielding would protect you from it. I've quickly realized that an ordinary nuclear reactor would be a far more practical solution.

Arminius wrote:P.S.: How did you calculate your table splitting up neutrino energy and electron energy?

By the relativistic equation of momentum:

E2=m2c4+p2c2

And it's not a direct calculation.

What I did was work backwards: I picked a "guessed" value for the momentum, and plugged it into the above equation for both the electron and the neutrino. Summing up the two energies, I got the total decay energy. And the ratio of the two results give you the split.

I repeated this a few times with different momentum values, trying to aim at a total energy which is close to around number. Whenever I hit close to a round value (say 100 keV or 500 keV) I've posted the data in the table.

Yes, this process is somewhat unwieldy. But it works. And I suppose you can see know why I preferred to just list the numbers in a table instead of actually guiding you through the entire process.

PM 2Ring
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Location: Sydney, Australia

### Re: [Physics] Curie and Becquerel

It is possible to use a beta emitter as an electrical energy source, but there are, admittedly, some practical difficulties. High energy beta emitters tend to cause structural damage to the energy collection mechanism (and of course any particles that aren't captured constitute a radiation hazard), so practical betavoltaic devices uses friendlier isotopes, with reasonably long half-lives. So don't expect a betavoltaic device to compete with a fission reactor, or even an RTG device powered by 238Pu. But they're ok for applications that don't pull a lot of watts but need a long-term energy source.

Note that even relatively low energy beta particles have a rather high voltage, so the energy collection mechanism has to transform the beta particle energy into a more friendly form - consumer electronics generally does not cope well with a multi-kilovolt power source.

and http://en.wikipedia.org/wiki/Atomic_battery .

Arminius
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### Re: [Physics] Curie and Becquerel

Big mistake on my part was simple: becquerels are desintegrations PER SECOND... embarassing.

I have been looking all over the internet for prices of Pd103 but could not find any... a shame. Would have loved to do some number crunching on that.

Globally speaking I've seen two methods for getting energy out of high energy electrons or photons (beta and gamma): fluorescence and similar phonomena (to convert to some lower energy particles) and photovoltaïc (also it looks like gamma ray PV is in its earliest steps: http://iopscience.iop.org/0268-1242/23/8/085001).

Last question on my side: would you say the in the quora answer I added (http://www.huffingtonpost.com/quora/what-is-the-theoryconcept_b_3456241.html) the energy conversion was the biggest mistake or are there other? (I had thought it was power density but you proved me wrong)

Thanks!

PsiSquared
Posts: 126
Joined: Wed May 09, 2012 6:02 pm UTC

### Re: [Physics] Curie and Becquerel

Arminius wrote:Last question on my side: would you say the in the quora answer I added (http://www.huffingtonpost.com/quora/what-is-the-theoryconcept_b_3456241.html) the energy conversion was the biggest mistake or are there other? (I had thought it was power density but you proved me wrong)

Thanks!

I think the biggest mistake was trying to explain a comic-book device with real physics

Among other things:

1. A beam of electrons from a beta emmiter isn't going to induce electron capture in Pd-103 ions at a rate remotely similar to the listed Pd-103 half-life. A fast incoming electron (a beta particle) will have a very different effect than a nearby electron leisurely coasting in an atomic orbit.

2. Whenever a nuclear decay is involved, the energy is transformed, first and foremost, into heat. You can use this heat to drive a turbine or create electricity, but it is an unescapable intermediate step. So the whole concept of "nuclear-to-electricity without heat" is untenable.

3. What's even worse, this heat will be accompanied with tons of radiation. DEFINITELY not something you would want to activate while carrying it on your person.

The large reactor, though, definitely looks like a futuristic Tokamak. You got that part right. As for the portable gadget: I'm afraid there is no way to explain most comic-book gadgets with actual physics. And movies like "Iron Man" weren't meant to be analyzed in this way, in the first place.

Arminius
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### Re: [Physics] Curie and Becquerel

(Disclaimer: This quora post isn't at all mine)

a nearby electron leisurely coasting in an atomic orbit

I just love that expression!

Whenever a nuclear decay is involved, the energy is transformed, first and foremost, into heat. You can use this heat to drive a turbine or create electricity, but it is an unescapable intermediate step. So the whole concept of "nuclear-to-electricity without heat" is untenable.

Really? I though heat production was only the case in gamma decay and that pure alpha or beta decay did not have heat losses.

I think it's actually fun to compare and see where truth ends and imagination starts. Especially when it pretends to be so "down to earth".

PM 2Ring
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Location: Sydney, Australia

### Re: [Physics] Curie and Becquerel

Arminius wrote:
PsiSquared wrote:Whenever a nuclear decay is involved, the energy is transformed, first and foremost, into heat. You can use this heat to drive a turbine or create electricity, but it is an unescapable intermediate step. So the whole concept of "nuclear-to-electricity without heat" is untenable.

Really? I though heat production was only the case in gamma decay and that pure alpha or beta decay did not have heat losses.

Wikipedia has an article on decay heat, but it doesn't really go into details of the mechanism.

Alpha and beta particles, being electrically charged, tend to interact with matter. So particles emitted inside a lump of radioactive material transfer a lot of their kinetic energy to the surrounding matter, heating it up.

FWIW, 238Pu, which is an alpha emitter with a half-life of 87.7 years, produces around 0.54 kilowatts of heat per kilogram. Plutonium has a density of just over 19.8 g/cm3, so a 1 kg sphere has a radius of just under 2.3 cm.

I suppose you could minimize the heat produced if your radioactive material is in the form of a thin film, or a gas / plasma (so the particles have a longer mean free path). But even then, you'll still get some thermalization when you try to harvest the kinetic energy of the emitted particles.

Arminius
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### Re: [Physics] Curie and Becquerel

Minerva
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### Re: [Physics] Curie and Becquerel

Arminius wrote:list/table with isotopes of the most common elements and their radioactivity density (how many Ci or Bq per g).

The quantity you're looking for is what's called the specific activity, and a convenient way to look it up is to look up the Wolfram Alpha page for whichever radionuclide you like.

Furthermore, I was looking for the same list/table but with the energy of each electron.

It might be time to meet the table of the nuclides, for your general nuclear physics education and reference enjoyment. Personally I like the KAERI online one, but there are other options online.

http://atom.kaeri.re.kr/

Drill down to find the nuclide you like on the table, and it will tell you the decay energy, which is 224 keV for Pm-147 for example.
(Note that with a beta decay because you've got that continuum energy distribution the decay energy is the maximum beta energy, not the mean beta energy which is lower.)
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