I'm aware my question might be considered a duplicate of this one: Radio-dating and the age of the earth

I read that one and I looked everywhere and I still can't find my peace of mind. I would really like to understand the following points:

  • does one uranium nucleus start the radioactive decaying process the instant it is born inside a supernova ? (I understand that for a single nucleus the chance of decaying which is an instantaneous event, grows from 0% to 100% in the time from the beginning to the half-life)

  • while talking about a collection of atoms, Wikipedia states here that "This predictability allows the relative abundances of related nuclides to be used as a clock to measure the time from the incorporation of the original nuclides into a material to the present."

  • @JohnRennie helped me understand that chemically speaking zircon and lead do not mix whereas zircon and uranium do. So basically macroscopic contamination via accretion with flying lead is an event too rare to take into consideration right ? Also... I understand that the uranium-lead decay series has very short half-lives for every nuclide except for the beginning and the end (uranium and lead) and so, from a cosmological perspective it's quite improbable that zircon would mix with anything other than uranium, should a chemical compatibility exist with another nuclide.

Please bare with me and excuse my "highly scientific" vocabulary.

I have an experiment I would like to imagine in order to discover what I'm doing wrong:

  1. A star goes supernova
  2. We concentrate our attention on one uranium nucleus which is fused together during the violent event
  3. The uranium nucleus travels for millenia towards the an accretion cloud surrounding a newborn star
  4. The nucleus collides with brand new molten zircon crystal and becomes part of it.

Let's say that event 4. was happening after 0.99 uranium half-lives and the uranium nucleus still hadn't decayed. Isn't it imminent that the nucleus will decay pretty soon after event 4. ?

I mean, I understand that if the nucleus would've become thorium or protactirium or lead it wouldn't've gotten stuck in the crystal for chemical reasons, but if it was still uranium what does that say about the internal clock of the nucleus which had been ticking for 99% of the half-life ?

Is that clock reset because of external electromagnetic reasons the moment the nucleus becomes part of the crystal ?

Thank you for your indulgence

  • 1
    $\begingroup$ "Isn't it imminent that the nucleus will decay pretty soon after event 4. ?" -- Note that the probability of a nucleus for having decayed within the first half-life period is per definition 50%, not 100%. $\endgroup$
    – JimmyB
    Apr 25, 2014 at 10:06
  • 3
    $\begingroup$ Your confusion is due to a misunderstanding of how nuclear decay works. See How does a half-life work? for what's essentially the question you should've asked. $\endgroup$ Apr 25, 2014 at 11:18
  • 2
    $\begingroup$ This is actually a variant of the Gambler's Fallacy. It's a very common mistake of intuition that the non-occurrence of some random event for a long time means that it is more likely for the event to occur in the near future. Flipping a coin and getting heads 100 times in a row doesn't make it more likely that the next coin flip will come up tails. $\endgroup$
    – Dan Bryant
    Apr 25, 2014 at 15:06

6 Answers 6


Here's the key point:

If [the nucleus] was still uranium what does that say about the internal clock of the nucleus which had been ticking for 99% of the half-life ?

The answer is "nothing," because a single nucleus doesn't have an internal clock.

Here's a better model for what's happening in an unstable nucleus. Think of it as an enclosed container with a small hole, or a tunnel, somewhere near the top.

a drink pitcher, closed except at the spout

Somewhere down in the bottom of the container is a rubber ball. That's the alpha particle. But the whole thing isn't just sitting there, static, because oscillators in quantum mechanics don't sit still even in their lowest energy states. The container is jumping and jiggling around, like it's on a paint shaker, and the ball is constantly bouncing off the bottom and off the walls. Importantly, the hole in the container always stays up near the top.

What happens to this system over the long term? Most of the time the ball only bounces partway up the walls of the container and falls back down. Sometimes it goes higher than the opening — but because the opening is small, the ball usually misses and falls back to the floor again. Eventually, by chance, the ball will find just the right trajectory so that it hops right out the tiny exit hole and leaves the container for ever. But the vibrations that take the ball near the hole are chaotic, and there's no way for you know whether the ball is "about" to come out or how long it's been bouncing around. Every time the ball hits the bottom, it gets a new random trajectory. The dice roll again.

If you have an unstable system, there's 50% chance it'll still be intact after one half-life. If you check on it later, and it's still there, there's still a 50% chance that it'll be intact after another half-life. But there's no memory, and no "almost decayed," for a single nucleus.


The key fact you don't seem to realize is that the radioactive nuclei have no memory. Each second is a new second, and the probability the nucleus decays in the new second does not depend on how long it has been around. The only reason the decay rate of a material decreases with time is that after time a fraction of it has already decayed.

Thus when the fraction of uranium that lasted long enough to mix with the zircon does in fact mix, you can treat that uranium as having just been created, since, after all, it doesn't have any memory and so it doesn't know to behave as if it had just been created or if it had been around more millions of years.

What is unclear from your question, though, is why the lead decay product from the uranium that had decayed before forming the molten mixture with the zircon would not also be in this mixture. But taking for granted that the mixture consists solely of uranium and zircon initially, you can use the half life of uranium to tell the age of the object.

  • $\begingroup$ Thak you for your answer. To clarify things: I was quoting @JohnRennie (see the 3rd comment, made by himself to his own answer, here: physics.stackexchange.com/questions/55898/…). From what I understand it seems like a key ingredient in whole radiometric dating activity $\endgroup$ Apr 25, 2014 at 4:52

It's probably better to talk about a number $N_{0}$ of uranium nuculei travelling through space after getting created in the supernova. Once you do this, it should be pretty obvious that the amount left is going to be very close to $N = N_{0}2^{-t/t_{h}}$, where $t_{h}$ is the half-life. The key point is that the whole planet will be made from the same, approximately mixed, cloud, so the ratios of the atomic isotopes should be relatively constant across the planet at the time of formation.

Also, note that the half-life of U-238 is 4.5 billion years. There have only been a little more than three U-238 half lives since the beginning of the universe. So, even if you assumed (incorrectly) that all of the initial Uranium supply was made in the big bang, we'd still have 1/8 of the initial amount today.

  • $\begingroup$ Thanks for the almost instantaneous answer :). It was very helpful $\endgroup$ Apr 25, 2014 at 4:48

To answer what seems to be your primary question first:

Is that clock reset because of external electromagnetic reasons the moment the nucleus becomes part of the crystal ?

No, the uranium atom's "internal clock" is not reset. To understand this, let's go back and correct one misunderstanding earlier in your question:

does one uranium nucleus start the radioactive decaying process the instant it is born inside a supernova ?

The nucleus doesn't really start the decay process. Rather, over any given length of time, the nucleus has a probability of decaying given by a probability distribution function with the half-life as the only free parameter. That probability is independent of time, meaning that the nucleus has no memory, and thus no internal clock. Having a constant decay probability is what gives the exponential decay pattern. The half-life is length of time over which the probability of the nucleus to decay is 50% (not 100%).


We use the word 'lifetime' for both, but nuclei are not like people. Don't be misled by the comparison.

Obviously, a 60 year old person is not the same as a 16 year old person.

But a 60 million year old uranium nucleus is exactly the same as a 16 million year old nucleus, or indeed a 16 seconds old nucleus. It does not age. The probability of its decaying is the same as it always was, and always will be. It does not have an internal clock - or, if you like, that clock is getting reset, not just just when it enters a crystal but continuously.


When we talk about half life, we talk about amount of substance. Suppose 2Kg Uranium was fused with Zircon crystal. After 99% or 0.9 Half Life, 0.9Kg Uranium would be no more. After 1 Half Life, there would be 1Kg Uranium and 1Kg Lead.

When talking about single Uranium atom, one atom can bleed out its instability at this point and it's neighbor atom would after billions of years. We don't know it's reason. The atomic clock you're talking about is also based on macroscopic amount of substance. The probabilistic behavior of atoms inside a macroscopic heap is periodic making the clock tick (See SI definition of Second).

  • $\begingroup$ "After 1 Half Life, there would be 1Kg Uranium and 1Kg Lead." Well, strictly speaking, there would be 1kg of U and 1kg of various daughter elements. Most (radium, bismuth, etc.) have considerably shorter half-lives than U-238. It's likely that most of the daughter elements have already decayed all the way down to Pb-206, but not all of them. $\endgroup$
    – Phil Perry
    Apr 25, 2014 at 17:10
  • $\begingroup$ @PhilPerry Technically, yes. But, amount of all those intermediate elements would be negligible. $\endgroup$ Apr 25, 2014 at 17:13

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