# Can nuclear transmutation be observed in real time?

Ignoring the quantum zeno effect (if possible?), can we observe in real-time the transformation of one element to another? I'm talking about an amount visible to the naked eye where one could see obvious changes in colour, reflectivity, phase, surface finish etc. occurring in say, seconds or minutes.

• What exactly do you mean by 'transformation from one element to another'? How would you classify chemistry, for instance?
– Danu
May 21, 2014 at 10:25
• Well, as an example, say potassium-40 decaying to argon-40. This has a massive half-life but this is what i am asking about in principle. The decay of one pure element into another that can be seen happening in real time. Maybe what i'm asking is not possible. May 21, 2014 at 16:17
• I am personally more interested in what you might mean by "observe in real time". One can make precise counting experiment to detect each decay as it happens, and with a mass spectrometer one can measure the change in the chemical composition of a samples over time; but each individual decay happens over a very short time-scale (an issue distinct from when it happens). May 21, 2014 at 16:42
• if you are satisfied with an answer the rules are that you check it as the accepted answer. Jun 2, 2014 at 4:16
• If you were observing such a thing with your naked eyes, you'd be killed instantly by the radiation.
– user4552
May 13, 2018 at 22:29

I think that you are asking whether there's an example of a naturally radioactive material, or an irradiated material, whose decay is quick enough that you can prepare a sample with one set of physical and chemical properties, wait a finite amount of time, and have a sample that is visibly changed.

This would require you transform a chemically significant amount of material, which in general can't be observed by eye in a small laboratory.

For example, let's suppose we have a reaction where the decay energy is 1 MeV. If we wanted to transmute one mole of this material, the total energy released would be $$\mathrm{ 1\,MeV \cdot 6\times10^{23}\,atoms = 10^{11}\,joules }$$ If you wanted the transformation to take place over a year ($\pi\times10^7$ seconds) you'd have a constant power of about 3 kW (mostly carried by fast decay products) that you'd have to remove from your sample.

That sounds nice and everything, but there just aren't any reactions in that energy and speed range. The best-known reaction whose rate can be engineered is uranium fission, where each fission releases about 200 MeV. Typically less than 5% of the mass of uranium fuel undergoes fission in a several-week fuel cycle. I assume you have some idea of the precautions necessary to handle spent nuclear fuel — it's doable, but not a lab demo.

As another example, if each fission releases 2–3 neutrons and 200 MeV of energy, the ~60 terajoule explosion over Hiroshima in 1945 involved about half a mole of fissioning uranium and about a mole — one gram — of free neutrons.

Your other option for an observable transmutation would be the decay of tritium to helium, which has a fairly short half-life (12 years) and quite low decay energy (around 0.020 MeV). Of course, both tritium and helium are colorless gases when pure at room temperature, so you'd have to use some other property to observe the decay. (For instance helium-3 has twice the pressure of hydrogen-3 at a given mass density, since hydrogen forms H$_2$ molecules and helium is monoatomic.)

• Seconds in a year, $\pi \times 10^7$. Nice!
– Danu
May 21, 2014 at 22:07

Certainly with a cloud chamber you can. Here is a nice video of using one: http://www.youtube.com/watch?v=Efgy1bV2aQo

There are many instructions on the internet for making your own cloud chamber and observing decay of radioactive americium-241 from an ionizing smoke detector for example.

Building on @rob's example of tritium decay, you can start with tritiated water, where hydrogen is replaced with tritium. As tritium decays, the liquid tritiated water (for example, in a transparent container) will turn into gases - oxygen and helium (helium-3, actually).

EDIT (5/13/2018): Let us consider a numerical example. If initially we have 22 g of tritiated water ($T_2 O$), it is approximately 12 ml as the density of tritiated water is 1.85 g/cm^3 (https://en.wikipedia.org/wiki/Tritiated_water). As halflife of tritium is 12.32 years (https://en.wikipedia.org/wiki/Tritium), in an hour we will have 22.4 1000. (2. (1. - 0.5^(1./365./24./12.32)))=0.29 ml of He3 (as the volume of a mole of a gas at normal conditions is 22.4 liter/mole, and as if all 6 g of tritium were converted to He3, that would be 2 moles of He3). That amount of gas would be visible (and I did not take into account the resulting amount of gaseous O2). Another approach - tritium decay generates a significant amount of energy (How hot would tritium water be?), so you can start with solid tritiated water and pretty soon have some liquid.

I must point out that both dmckee and DavePhD are incorrect when they imply you observe a nuclear transformation when you count clicks in a geiger counter or observe tracks in a cloud chamber. This is not how quantum mechanics works. There is not a one-to-one correspondence between decay events and detection events. There is a sample over here, and from time to time decay events are presumed to occur; and there is a detector over there, and from time to time a detection event occurs. There is neither a theoretical nor an experimental basis for claiming that each detection event corresponds to a particular decay event.

• I strongly disagree with this answer. For instance, a track in a cloud chamber (or other ionization detector) consists of many scattering events between the condensation centers and an ionizing particle. Already in 1929, Mott was asking why an $s$-wave alpha particle should make a straight cloud-chamber track, since its wavefunction is symmetric under rotations; the answer is that its multiple interactions are correlated. Modern magneto-optical traps can load hundreds or tens of radioactive atoms at once and observe their decays singly. You're wrong.