What is the cause of radioactive decay?


by Katastrofa
Tags: decay, disorder, quantum, radioactive
Katastrofa
Katastrofa is offline
#1
Feb4-12, 12:14 PM
P: 13
I don't understand where radioactive decay comes from. Everything I've read discusses how the environment can influence the decay (for example, neutrons in different nuclei decay with different speeds), but I couldn't find an explanation of what is the underlying cause of the decay. If the quantum system is in its eigenstate, the Schroedinger equation tells us that it will remain in this state for ever. Then why does an elementary particle - a quantum object - decay? Is it because of some disorder affecting it "from outside"? Would a perfectly isolated elementary particle never undergo decay? What if we isolated a whole nucleus? Can the fact that we're observing the nucleus (or a sample of them) itself be triggering the decay?
Phys.Org News Partner Physics news on Phys.org
Physicists design quantum switches which can be activated by single photons
'Dressed' laser aimed at clouds may be key to inducing rain, lightning
Higher-order nonlinear optical processes observed using the SACLA X-ray free-electron laser
thedemon13666
thedemon13666 is offline
#2
Feb4-12, 12:50 PM
P: 18
In regards to nucleus stability, why isotopes and heavy nuclei decay isnt actually well known.

The actually physics of it comes from the weak nuclear force (I would recommend reading up on that)

In regards to an elementary particle decaying, yes they do, very often.

For instance a muon readily decays into an electron, an electron neutrino and a muon neutrino.

What the particles decay into varies (many factors) a large factor of that is the mass of the parent particle though. There are also laws that forbid certain decays.
AdrianTheRock
AdrianTheRock is offline
#3
Feb4-12, 04:11 PM
P: 136
A particle is unstable when it is able to participate in one or more reactions that yield particles whose combined rest masses add up to less than that of the original one. Energy doesn't like to be condensed into the rest mass of a heavy particle because multi-particle states with lower combined rest masses have higher entropy. This principle applies just as much to atomic nuclei as to fundamental particles produced in accelerators, astrophysical events, etc.

The reactions in question can also be mediated by the strong and electromagnetic forces, or sometimes a combination of these - not just the weak. Though beta decays are weak phenomena, gamma emissions are of course electromagnetic reactions. Alpha decays are also ultimately the effect of the electromagnetic force (though they also entail quantum tunnelling through the strong (nuclear) force field) because ultimately it is the electrostatic repulsion between the protons, in heavy isotopes, that prevails over the attraction of the strong force.

Its true that our understanding of nuclear physics is still less than complete because it falls into the non-perturbative domain of QCD. But a lot is, nevertheless, known about nuclei and there are some good approximations such as the semi-empirical nucear mass formula. Its five terms neatly summarise the five main factors that determine the rest energies of individual isotopes.

phyzguy
phyzguy is offline
#4
Feb4-12, 05:01 PM
P: 2,070

What is the cause of radioactive decay?


Quantum mechanics predicts the probability of an event happening, but is silent on why events happen. Most interpretations of QM say that there is no explanation possible for the randomness, and it is an inherent part of the quantum world. Thus, I believe most physicists would answer that when a Uranium nucleus sits quietly for a billion years and then one day decays, it is not possible to give a reason for why it decayed at that time and not some other time. I confess to having personally struggled with this concept, but I believe it is the conventional wisdom.
questionpost
questionpost is offline
#5
Feb4-12, 06:21 PM
P: 198
Why is there even a need for the weak nuclear force if a simple explanation is that the strong force doesn't have a big range and so can't hold large nuclei together?
The_Duck
The_Duck is offline
#6
Feb4-12, 10:20 PM
P: 790
The weak nuclear force is the mechanism by which neutrons convert into protons, or vice versa, in beta decay.
questionpost
questionpost is offline
#7
Feb4-12, 10:58 PM
P: 198
Quote Quote by The_Duck View Post
The weak nuclear force is the mechanism by which neutrons convert into protons, or vice versa, in beta decay.
Can't that still be explained by for some reason by the strong not having enough energy or range to continue to effect all of the quarks in a neutron equally? Perhaps because the nuclei is so big or contains too much energy per nucleon or too little energy per nucleon or etc?
The_Duck
The_Duck is offline
#8
Feb4-12, 11:11 PM
P: 790
I'm not sure I understand what you are suggesting here but the strong force definitely cannot turn protons and neutrons into each other. It just produces the attraction between the protons and neutrons. Meanwhile we observe many radioactive decays in which a proton turns into a neutron, or vice versa. The weak force is responsible for these processes.
questionpost
questionpost is offline
#9
Feb4-12, 11:25 PM
P: 198
Quote Quote by The_Duck View Post
I'm not sure I understand what you are suggesting here but the strong force definitely cannot turn protons and neutrons into each other. It just produces the attraction between the protons and neutrons. Meanwhile we observe many radioactive decays in which a proton turns into a neutron, or vice versa. The weak force is responsible for these processes.
I'm not saying the strong force turns protons into neutrons, I"m saying a *lack* of the strong force allows that process to happen. I thought that was already the current explanation anyway, that nuclei are so big the strong force can't contain all of it, but I guess that might just be for alpha decay.
AdrianTheRock
AdrianTheRock is offline
#10
Feb5-12, 05:31 AM
P: 136
That is correct for alpha decay. But a number of other factors come into play with beta decays.

First, because neutrons and protons have different quantum numbers, they each have their own 'shells' into which they can stack up. Second, the strong force binds protons and neutrons equally, whereas the electromagnetic replusion is between protons only. Third, the neutron is itself unstable because it is about 1MeV heavier than the proton, due to d quarks being slightly heavier than u quarks. Free neutrons beta-decay to protons with a half-life of about 10 minutes.

One way of understanding the behaviour of the weak force within nuclei is to consider the possible states of a given isobar, ie a 'bag' of N nucleons. Assuming for the moment that N is below the threshold at which alpha decays start to occur, then we can consider our isobar as a bound system in which each nucleon can be either a proton or a neutron. Within that range of possible states, there will be a ground state with the lowest energy. The other states have higher energy so can potentially decay into the ground state subject to some simple criteria.

Consider for example an isobar of 3 nucleons. The ground state for this is the 2He3 isotope, with two protons and one neutron. A commonly encountered higher energy state is tritium, which instead has one proton and two neutrons. In this case, the difference in energy between tritium and 2He3 exceeds the combined rest energies of an electron and an antineutrino, so there is sufficient delta-energy available for tritium to beta-decay into 2He3, which it does with a half-life of the order of ten years.

The other two theoretically possible states for this isobar are three neutrons and three protons. These are only ever observed as fleeting resonances, if at all. The reason for this is that the third proton or neutron has to occupy a higher energy level than the first two, because the lowest-energy shell for that particle type is already filled with one spin-up and one spin-down particle.

If an isobar has a non-ground state whose energy level differs from that of the ground state by less than the rest energy of an electron, this state is also stable because there's insufficient energy to create the particles needed to decay to the ground state. Such a state would, however, be a candidate for beta decay via neutrino capture, if the incoming neutrino supplies the extra energy needed to produce the required electron/positron in addition to the ground state of the isobar.

Note also that in isobar ground states the neutrons are perfectly stable, despite the fact that free ones are not. This is because it is the total energy of the system that is crucial here, not the rest masses of the free component particles.
thedemon13666
thedemon13666 is offline
#11
Feb5-12, 05:39 AM
P: 18
It is not possible to explain almost all fundamental particle decays without the weak force. For instance Muons decay to electrons, and two neutrinos (muon and electron). Muons are leptons and so dont interact via the stong force.
Katastrofa
Katastrofa is offline
#12
Mar17-12, 05:27 PM
P: 13
Many thanks for all your responses!

I connected the randomness with quantum fluctuations, which stimulate e.g. spontaneous emission. Even if QM Hamiltonian does not account for them, it's believable that the system does not remain in the stationary excited state forever, but under some external stimuli tends to lower its energy and goes back to its ground state. Still, the above couldn't explain the radioactive decay, but AdrianTheRock's response seems to make everything clear now. Thanks!

AdrianTheRock, can I have a few simple/crazy questions:

1) Does it mean that by directing a stream of neutrinos at a bunch of decaying material, one could speed up the decay process? It might have "interesting" consequences for the stability of Earth's nuclear weapons, if by a cosmic chance we are some day showered by neutrinos. (OK, maybe I'm jumping the shark here.) Don't you think this could also mean that our carbon dating techniques might be burdened with a HUGE error (an order larger that what we already know about their accuracy)?

2) Should I associate the random quantum fluctuations with the occurrence of neutrinos? This might suggest that they are not so random (?) and make all things very mysterious...

Maybe it's a wrong forum to post this, but what if neutrino radiation reaching the Earth is not so random, but it is a transmission from extra-terrestrial civilisation, like in Lem's His Master's Voice? :-) http://arxiv.org/abs/1203.2847
AdrianTheRock
AdrianTheRock is offline
#13
Mar17-12, 06:40 PM
P: 136
Katastrofa

1) Possibly, though only very slightly. The cross-sections for these reactions are extremely small, because at energies below the W boson mass threshold weak interaction cross-sections are proportional to the square s of the available centre-of-mass energy. (In the lab frame, this means neutrino interactions with static targets are linearly proportional to the incident neutrino energy [itex]E_\nu,\ \ [/itex]as[itex]\ \ s\ \ ≈\ \ 2\ m_{target}\ E_\nu[/itex].)

One decay route I didn't mention in my last post, and probably should have, is electron capture - this is a type of beta decay where the isotope "decays" by capturing an orbital electron and emitting a neutrino. The same reactions can also proceed by capturing an antinuetrino and emitting a positron (subject to sufficient COM energy). Either way, a proton becomes a neutron and the isotope changes. This method allows an isotope to decay to an adjacent one in the same isobar even if the mass difference between the two is less than that of an electron. Of course electron capture is fine for isotopes that want to decay into ones with one more neutron but one less electron (Z → Z-1), but isn't much use for those that would like to go in the direction Z → Z+1, because that would entail capturing a positron. There just ain't many of those about in our universe to begin with, and even if one did succeed in dodging the orbital electrons it would be kicked away electrically by the nucleus anyway! But capturing a neutrino is perfectly feasible, if infrequent.

In fact reactions like these are among those used in neutrino detectors. For solar neutrinos in particular, the advantage of using an isotope that decays by capture is that the reaction will occur for neutrinos with lower energies. This is important when trying to detect neutrinos such as those from the solar pp reaction as these are below 420 keV ie < me. The GALLEX experiment therefore used Ga71, which has a 233 keV energy threshold to trigger decays to Ge71 + e-.

2) Quantum fluctuations are quite separate phenomena from passing background particles such as neutrinos.


Register to reply

Related Discussions
Radioactive decay Introductory Physics Homework 9
Radioactive Decay Biology, Chemistry & Other Homework 1
Nuclear Chemistry: Kinetics of Radioactive Decay and Radioactive Dating Biology, Chemistry & Other Homework 1
What is the proper use and meaning of "Radioactive decay"? How much decay HAS occured General Physics 2
radioactive decay equilibrium when decay constants are equal Biology, Chemistry & Other Homework 2