What does lambda in nuclear decay mean if it is big?

I am thinking about lambda, the nuclear decay constant and what it actually means.

I have built spreadsheet models looking at decay using lambda to determine the decay for each time period and then comparing the number left to that determined by using e ˆ- lamda x time.


These are my conclusions.
If lambda is very small, less than 0.00001 then it is true to say that it represents the probability of decay per time period. And that iterative calculations using lambda give results within negligible percentage difference from calculations using e.

Here is a summary of these findings. Here I used initial nuclei at 1e18

Value of lamda Percentage difference between nuclei left after 50 time periods



As lambda gets above 0.1 the maths of decay works but is no longer in agreement with A = N x lambda.

We can use any type of value for lambda including values such as 760 and the exp formula works correctly but if lambda is bigger than 0.2 and in particular bigger than one what does it mean? A = N x lambda cannot be true for values bigger than 1.

So here is my question: What precisely is lambda when it gets big?

Tibetan Blackbird said:

A = N x lambda.

Where did you get this formula and what do you think it means?

Tibetan Blackbird said:

What precisely is lambda when it gets big?

The same thing as it is when it is small: the time constant in the exponential decay equation. I.e., at the end of one time period ##\lambda##, the number of atoms of the original element decreases to ##1 / e## of what it was at the start.

Tibetan Blackbird said:

Here is a summary of these findings.

You have posted no code or anything of that sort, so I can't tell where your calculations are going wrong, but they are wrong somewhere. I'm not sure why you would try a hand-built "iterative" model (whatever that means) in any case given that it is so easy to just calculate the exponential function directly.

Tibetan Blackbird said:

We can use any type of value for lambda including values such as 760 and the exp formula works correctly

That's because the exponential formula is correct.

PeterDonis said:

Where did you get this formula and what do you think it means?


The same thing as it is when it is small: the time constant in the exponential decay equation. I.e., at the end of one time period ##\lambda##, the number of atoms of the original element decreases to ##1 / e## of what it was at the start.


You have posted no code or anything of that sort, so I can't tell where your calculations are going wrong, but they are wrong somewhere. I'm not sure why you would try a hand-built "iterative" model (whatever that means) in any case given that it is so easy to just calculate the exponential function directly.


That's because the exponential formula is correct.

Thank you, but you are not answering my question. My A level book says that lambda is the probability of decay per second. And the equation provided to A level students is here. It says that activity in Bq is equal to lambda x N (number of unstable nuclei) This makes sense for small values of lambda but not for large ones.

Tibetan Blackbird said:

My A level book

What book?

Tibetan Blackbird said:

says that lambda is the probability of decay per second.

It would be more precise to say probability density, since we are talking about a continuous function and its derivative.

Tibetan Blackbird said:

the equation provided to A level students is here.

Here, as in where? Please provide a link or reference to the book you got this from.

Tibetan Blackbird said:

It says that activity in Bq is equal to lambda x N (number of unstable nuclei)

Yes, and then it solves for the number of atoms and the activity as a function of time, which is the exponential decay law.

At any instant of time, you can say that the activity given by the exponential decay law is equal to the product of ##\lambda## and the number of atoms, ##N##, remaining at that time. That is what ##A = \lambda N## means. But if you want to evaluate the change in activity over a period of time, and how that affects the number of atoms remaining over a period of time, you need to use the exponential decay law. That's what it's for.

Tibetan Blackbird said:

This makes sense for small values of lambda but not for large ones.

No, your hand-built attempt to do different math from what is explicitly shown in the image you gave, whatever that math is (you haven't shown it), appears to be a reasonable approximation for small values of ##\lambda## but not for large ones. But why would you bother with an approximation when the correct exact equation is (a) right there in front of you, and (b) simple to compute?