Half life and radioactivity clarification

In summary, half-life refers to the amount of time it takes for half of a sample of unstable nuclear isotopes to undergo radioactive decay. This can result in the emission of alpha, beta, or gamma particles. When an atom decays, it can change into another element or isotope, and this process can continue until the atom becomes stable. Cesium-137, for example, has a half-life of approximately 30 years and will decay into other elements during this time. However, even before it reaches its half-life, it can emit ionizing radiation. This is because decay is a statistical process and some atoms may decay more quickly than others.
  • #1
oblong-pea
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TL;DR Summary
Just wondering if my understanding of half life's and radioactivity is correct.
As I understand, half life's are where unstable nuclear isotopes Undergo radioactive decay, where they emmit either aplha beta or gamma during this decay.

I'm looking for some clarification on whether when an atom undergoes decay, does it then change into another element/ isotope of an element, and then that has another half life where it is repeated until stable?

For example say cesium 137 has a half life of 7 years. Will this cesium 137 then decay in 7 years to another element/isotope or multiple elements and the original cesium will dissappear?

Additionally, during the 7 years of this cesium existing, how is it giving off ionising radiation if its not decaying?

Thanks
 
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  • #2
https://www.britannica.com/science/half-life-radioactivity
Radioactive decay is statistical: the half life is the time required in a large sample for half of them to have decayed. Each decay converts one Cs-137 nucleus into something else entirely which may or may not be stable. There may be more than one initial decay path. These new species then usually decay independently
 
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  • #3
Hi,

Decay processes occur in unstable nuclei. There is a certain ##\lambda {\roman d}t## probability that a nucleus decays in a period ## {\roman d}t##. See here.

[edit] ok, slow typist...

[edit2] where did you get the 7 years ? Cesium 137 half life is more like 30 years !

And yes, if you wait for a time corresponding to the half-life, half the original Cs nuclei will have decayed into Ba 137. After another half-life, half of the remainder (so one quarter of the original number of Cs nuclei) will have decayed and only one quarter of the original number of Cs nuclei are left over.

##\ ##
 
  • #4
oblong-pea said:
For example say cesium 137 has a half life of 7 years. Will this cesium 137 then decay in 7 years to another element/isotope or multiple elements and the original cesium will dissappear?

https://en.wikipedia.org/wiki/Caesium-137

CS-137 itself has a half-life of about 30 years, as BvU has already noted. Most of it decays to Ba-137m, which in turn decays via gamma emission to stable Ba-137 with a half-life of about 153 seconds.

After several half-lives (30 years each) of Cs-137, almost all of it will have decayed. Consider the sequence 1/2, 1/4, 1/8, 1/16, 1/32, 1/64... In principle, there will be some undecayed Cs-137 for a long time, but in practice it becomes negligible after some number of half-lives. The number depends on the specific application.
 
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  • #5
Thanks all that's clarified it for me now. I was getting a bit confused.

However just one more query, how does CS-137 (as an example) give off beta/gamma radiation within the 30 year period if it is not 'decaying' until it hits that 30 years time? Or am I misunderstanding something trivial.

Thanks
 
  • #6
Yes you have not understood anything. Half life is a statistical property. Compare with say rolling a dice.

Assume standard six sided dice and you want the dice to show one eye (this is "decays" in this model). You can not say in advance how many rolls are needed to make the dice show one eye (assume one roll equals one year in this model).

If we start with a huge sample of dices, lets say 60000 and we roll them all at once. Then we remove all dices that showed one eye - these have decayed. After the first roll, there will be approximetly 50000 dices left due to the law of large numbers. It could have been 49000 or 51000 left, we don't know. Repeat, after the second roll there will be approx 41700 left. And so on. This is know as exponential decrease.

Half life is a statistical property of radioactive nuclei. The more of them you have, the better this model work. Obviously it does not for single say Cs-137 or five of them.
 
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  • #7
oblong-pea said:
However just one more query, how does CS-137 (as an example) give off beta/gamma radiation within the 30 year period if it is not 'decaying' until it hits that 30 years time?
One atom of Cs137 doesn't give off anything until it decays. Then it gives off something and turns into something else. A lump of Cs137 will have some atoms that have bad luck and decay immediately, and some that decay in a year, and some that decay in ten years, and so on. The thirty year number is the average lifetime, and this means that a lump of many atoms of Cs137 emits radiation over time.

An experiment we did at school was to take a box of about a hundred dice and roll them all. Any that came up sixes "decayed" and we took them out and recorded the count, then repeated. If you do that experiment you'll find that you usually have fewer than half your initial dice left after four throws, so the half life is about four. (In fact the half life of this process is about 3.8.) It isn't a perfect simulation of radioactive decay because of the discrete time nature of it, but it's a decent first stab.
 
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  • #8
Ibix said:
The thirty year number is the average lifetime
Average life time is defined as ##\tau = \ln (2) \times T_{1/2} ## though

Ibix said:
If you do that experiment you'll find that you usually have fewer than half your initial dice left after four throws, so the half life is about four. (In fact the half life of this process is about 3.8.) It isn't a perfect simulation of radioactive decay because of the discrete time nature of it, but it's a decent first stab.

In my school, I do this "lab" with dices that have 20 sides which makes this "discrete time" make "more" sense. It is pretty fun.
 
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  • #9
malawi_glenn said:
Average life time is defined as ##\tau = \ln (2) \times T_{1/2} ## though
Yes, sloppy wording on my part. The half life is the average time needed for half the Cs137 atoms to randomly undergo decay, which is not the same as the average survival time of an atom before it decays - although they are related, as you say.
malawi_glenn said:
In my school, I do this "lab" with dices that have 20 sides which makes this "discrete time" make "more" sense. It is pretty fun.
Ours were just wooden cubes with one side painted black - may even have been made in the school workshop. But more sides would be better, I agree.

The thing I forgot to say is that (having recorded the counts) you can plot them as a function of the number of throws. It goes wonky at the end when you're down to about ten cubes because of small numbers, but until then you get a nice exponential reduction in the number of decays per unit time - exactly as you see with real radioactive material.
 
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  • #10
I also make students make a new prediction for if we remove the dices (they decay) if they show 1, 2 or 3 eyes. And then let them repeat the experiment.

Yes when there are few dices left, it get's wonky. But it also serves as a nice discussion.
 
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  • #11
I too used to do the 'dice experiment' when teaching radioactive decay. We had lots of dice.

With surprising predictability, my main challenge was preventing groups (invariably of boys) having a see-how-high-we-can-build-a-tower-of-our-dice competition.
 
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  • #12
Steve4Physics said:
I too used to do the 'dice experiment' when teaching radioactive decay. We had lots of dice.

With surprising predictability, my main challenge was preventing groups (invariably of boys) having a see-how-high-we-can-build-a-tower-of-our-dice competition.
Another reason for why using 20 sided dices
 
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  • #13
oblong-pea said:
how does CS-137 (as an example) give off beta/gamma radiation within the 30 year period if it is not 'decaying' until it hits that 30 years time?
Viewed in terms of a large collection of Cs-137 atoms, the decay process is continuous. Each second, some number of atoms decays, although the number is not fixed. It varies randomly from one second to the next, but with a well-defined average.

Viewed in terms of an individual atom, the decay process is discrete and occurs at a random time. Looking at a specific individual atom, there is no way, even in principle, as far as we know, to determine exactly when in the future it will decay. All we can do is calculate the probability that it will survive for a given length of time.

There is a quantity related to the half-life, called the decay constant: $$\lambda = \frac {\ln 2} {t_{1/2}} = \frac {\ln 2} {30 \textrm{ yr}} = \frac {\ln 2} {9.5 \times 10^8 \textrm{ s}} = 7.3 \times 10^{-10} \textrm{ s}^{-1}$$ (using the numbers for Cs-137).

To a very very good approximation (because it's such a small number in this case), this is the probability that any specific Cs-137 atom will decay during the next second. If that atom doesn't decay, then it has the same probability of decaying during the next second, and so on.

One gram of Cs-137 is 1/137 mole of Cs-137, containing ##(1/137) \left( 6.02 \times 10^{23} \right) = 4.4 \times 10^{21}## atoms. If we start with one gram of completely undecayed Cs-137, during the first second, about ##\left( 4.4 \times 10^{21} \right) \left( 7.3 \times 10^{-10} \right) = \left( 3.2 \times 10^{12} \right)## atoms will decay. This is the average number that I mentioned in my first paragraph above.

As time passes, the number of undecayed atoms decreases, and so does the number of decays per second. The decay constant ##\lambda## remains... constant!
 
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  • #14
malawi_glenn said:
Another reason for why using 20 sided dices
You can stack 20's. Its a good reason four 4's.

I don't know why we make dice out of Platonic solids.. Polygonic cylinders are much more flexible.

Steve4Physics said:
, my main challenge was preventing groups (invariably of boys) having a see-how-high-we-can-build-a-tower-of-our-dice competition.
Did you announce your own competition, that whomever builds the highest tower gets an F for the unit? Maybe witha "Muahahahah!" for emphasis.

Ibix said:
It goes wonky at the end when you're down to about ten cubes because of small numbers,
Do you combine the data from multiple groups at that point?
 
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  • #15
Vanadium 50 said:
You can stack 20's. Its a good reason four 4's.
you can in theory, but perhaps not in practice. What is your PR?
 
  • #16
malawi_glenn said:
What is your PR?
I don't know what that is.

malawi_glenn said:
you can in theory, but perhaps not in practice
I the words of the great philosopher of the previous century, L.P. Berra, "In theory there is no difference between theory and practice. In practuce there is."
 
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  • #17
Vanadium 50 said:
I don't know what that is.
I'm guessing PR in this context is Personal Record.
 
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  • #18
Vanadium 50 said:
Do you combine the data from multiple groups at that point?
I was fourteen when I did this experiment. I don't recall pooling data but, give or take your confidence in the various groups' abilities to record numbers accurately, you certainly could do so.
Steve4Physics said:
my main challenge was preventing groups (invariably of boys) having a see-how-high-we-can-build-a-tower-of-our-dice competition.
WizardTower.png
 
  • #19
The example I taught to show the random nature of decays was about a bar owner.

He gets a delivery of 200 new beer glasses, which he adds to his current stock giving him 400 in total. The glasses are strongly built but he knows that on average just over three get broken every day. So in 60 days he expects to be down to 200. But he doesn't know which ones will break. Perhaps three of the new stock will break today, perhaps three of the old stock will break today. So he has a regular order or 200 new ones every two months. Some orders have a slightly different design mark on them so just by accident he can tell which order the broken glasses came from. Some in his bar are from an order he made three years ago and still haven't been broken, while last week 50% of the broken ones were from an order that was a year old. By the end of this week he might have 15 broken ones from the new deliver and two from the previous order and one each from the three year old and one year order.

So he can calculate the half live of beer glasses (60 days) but can't say which ones will break when. And some from that three year old delivery might still be around in another two years time. But he still knows that 200 plus or minus 1 or 2 out of his stock of 400 are expected to be broken over every 60 day period.

And if he stops buying new glasses he could eventually end up with a dozen or so left, all of which break in the next seven days, or none of which get broken in the next 30 days.

Students seemed to get the random nature of things from this and understand the concept, that as the number of glasses decreased the average breakages would change but still fitted a pattern of half breaking over a 60 day period, and some surviving for a very long time.

PS I do like the multi-sided dice example, as you can do the experiment.
 
  • #20
DrJohn said:
on average just over three get broken every day
It's better if 1-1/2% get broken every day.
 
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  • #21
DrJohn said:
And if he stops buying new glasses he could eventually end up with a dozen or so left, all of which break in the next seven days, or none of which get broken in the next 30 days.
At a bar, the utilization and associated breakage of glasses should remain constant regardless of the number in stock -- at least until so few remain that the number of patrons exceeds the number of glasses. This makes breakage a memoryless process. The discipline that digs into the details of systems of random processes like this is "queuing theory". [Plenty of good queuing examples at a bar]

With a memoryless process and a rate of three per day looking for the time interval required to lose twelve glasses, the Poisson distribution is relevant. The likelihood of at least one glass of twelve surviving a week is 1.29 percent. The likelihood of at least one glass of twelve surviving a month is negligible.

https://stattrek.com/online-calculator/poisson

The "Poisson" distribution is what you get when considering a continuous random process that is memoryless and you want the probability of getting a particular number of events over a given amount of time. By "memoryless", we mean that the probability of getting an event is constant over time and does not depend on how many events have already occurred or when they had occurred.

Radioactive decay is not memoryless in the required sense. @Vanadium 50 alludes to this. Once a decay has occurred for an atom, that decay cannot happen again. The decay rate depends on how many undecayed atoms remain. Accordingly, radioactive decay is not precisely modelled with a Poisson distribution, though it may be a very good approximation in some cases. A quick trip to Google and Stackexchange speaks to a "pure death process" as a more precisely correct fit.
 
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