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Re: Simulating radioactive decay.



On Thu, 13 May 1999, Ludwik Kowalski wrote:

This morning, while responding to a message, I wrote:

An activity with coins (or m%m candies) can be supplemented
with a similar activity with pencils. Just declare that "a label
up means a decay". This will give students an opportunity to
compare two decay curves, one whose probability of decay
per unit time (one throw) is 1/2 and another with p=1/6.

[snip]

The well known formula, T=ln(2)/lambda, would produce
wrong results if p was used instead of lambda, except when
p<<1. In that sense simulations with pencils (p=1/6) are better
than simulations with coins (p=1/2). The decay curves are
exactly exponential for all constant values of p but the formula
for calculating T is valid only when very short dt makes p<<1.

[snip]

The curve N=f(t) can be plotted to determine the "experimental"
value of T. It will NOT be equal to 4.16 units. (1 unit =1 throw).
By the way, one does not need a program to realize that for p=1/2
the answer 1.38 units is larger than the experimental half-life T.
The discrepancy for p=1/6 will be considerably smaller than 38 %,
perhaps only 5 or so. Can somebody predict the exact value of the
discrepancy without using a program (when p=1/6, for example)?
I can not do this.

After one throw, there are (1-p) left. After n throws, there are (1-p)^n
left. We want to know the number of throws, n, that leave us with 1/2
left: (1-p)^n = 1/2

Solve for n to get n = - ln(2) / ln (1-p)

So, what is the discrepancy caused by using p rather than ln(1-p)?

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