# Nuclear decay of a small number of atoms calculation

1. Jan 23, 2016

### resurgance2001

I came across a rather dubious question that a teacher had put in a power point. It said something like,"Given a sample of 100 atoms of isotope x, after one half life of the said isotope, how many atoms of the original isotope will be left?"

My answer was that it was a trick question because you cannot know exactly how many atoms will be left. Nuclear decay is a random process and it is only possible to make predictions about the probable mass of a macroscopic sample of an isotope that will decay in a given length of time. Is that correct?

Is it actually possible to calculate the probability that there will be a specific number of the original isotope's atoms left? Say, for example, one says that the expected number is 50, how could one calculated the probability that it would be exactly 50?

I apologize if this question is rather banal, but I am just trying to understand better what is going on. Thanks

2. Jan 23, 2016

### Orodruin

Staff Emeritus
Yes it is possible to do this computation. Start by computing the probability p of a single nucleus decaying. The number of total decays is then Bin(N,p) distributed, where N is the total number of nuclei.

3. Jan 23, 2016

### Staff: Mentor

I do not pretend to understand decay. The standard use of Bin(n,p) is most usually in modeling the number of successes/failures in a sample of size n drawn, with replacement, from a population of size n.
n does not vary.
The original problem as stated appears to be without replacement. Please correct me.

Assumption: start with the condition where n=1, Bernoulli distribution, multiply to get n=100, and just go on forward with new n from there as each decay occurs? If so, it does not quite follow other applications of Bin().

4. Jan 23, 2016

### resurgance2001

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Thanks but sorry I can't see how to do that calculation. We are given that the time that has passed in one half life, so for each nucleus the probability is 50% but I can't see how to carry on the calculation from there.

5. Jan 23, 2016

### resurgance2001

Ok - thanks. I searched for an online binomial probability calculator. So with 100 atoms, each of which has a probability of 0.5 to have decayed after one half life-time, the calculator worked out that there is about a 7.9% probability that you would have 50 atoms of the original isotope at the end. It would be fairly easy to find out the probability of there being 49 atoms, 48, etc. and going the other way to get 51, 52 etc. and then add up the probabilities. I am guessing the number might have to be somewhere between 45 and 55 to get an 80% probability

6. Jan 23, 2016

### resurgance2001

Well I have now found if this is correct that there is a probability of something like 72% that the remaining number of atoms is somewhere between 45 and 55. That sounds reasonable, and lends more weight to the argument that you can't specify a particular number of atoms that will be left.

7. Jan 23, 2016

### Staff: Mentor

Right.

I get 72.9% for the probability that 45 to 55 (inclusive) atoms are left. You cannot be sure how many atoms are left, but you can give the expectation value - 50.

8. Jan 23, 2016

### Orodruin

Staff Emeritus
The replacement is done so that the probability of each draw is independent. This is exactly the case here. The decay of each nucleus is independent from the decay of the others. If you will, you can see each nucleus as its own draw pile with the same (but independent) drawing probabilities. The distribution is binomial.

9. Jan 23, 2016

### Orodruin

Staff Emeritus
Also this is wrong. The binomial distribution says nothing about the population size which samples are drawn from. This is encoded in the probability p of an individual draw being successful. The replacement ensures p is not changing between draws and that subsequent draws are independent.

10. Jan 23, 2016

### Staff: Mentor

@Orodruin thanks for the clarification. We can debate the 'with replacement' clause another time. However, this is done elsewhere commonly in other contexts, obviously not in this model.

11. Jan 23, 2016

### Orodruin

Staff Emeritus
For once, Wikipedia puts it quite neatly:
Note that there is no mention of draws and replacement. Only n independent yes/no experiments, which is exactly what we have here - each nucleus decaying constituting one experiment.

It is later mentioned that:
However, this is an application, not the definition.