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Re: Radioactive decay



POST SCRIPTUM:

Program Simulation ! In True Basic computer language
randomize ! to avoid exact reproducibilities
let lambda=1/6 ! probability of "decay" per throw
let t=0 ! initialize counter of throws (timer)
let N=10000 ! initialize number of atoms (pencils)
print"At t=0 the number of radiactive atoms was: ";N
for i=1 to 20 ! perform 20 throws
let decays=0 ! initialize counter of decays
for k=1 to N ! for every pencil in a throw
let try=rnd ! a random number (between 0 and 1)
if try<lambda then let decays=decays+1
next k
let N=N-decays ! remaining number of pencils (atoms)
print "after t=";i;" only ";N; "atoms remained."
next i ! ready for the next throw
end
! ********************************************
! The output is different each time. For example:
! 10000 8319 6446 5776 4807 3363 etc.; only
! 253 atoms were left after 20 time units.
!
! I believe that programs of that kinds should be
! composed by students. Programming is a good place
! to learn how to solve multiple-steps problems.
! All high school students should learn programming.
! ********************************************
Ludwik Kowalski wrote:

Dan Macisaaac wrote:

... I also hope to ... simulate radioactive decay with dice
... and I'm seeking your suggestions, particularly when
suggestions are from experience, and can be performed
cheaply in HS classrooms.

Simulating radioactive decay on a computer is easy when a
code for generating random numbers (uniformly distributed
between 0 and 1) is available, and when students know how
to program. But I just tried a hardware simulation using ten
common pencils (hexagonal cross sections). Each pencil
has a black spot labeled 2 HB on one of its sides. I threw
all pencils on the flat table randomly and counted black
labels facing up. (It is like throwing ten dice and counting
one's). The procedure was repeated ten times. The result were:

1, 1, 4, 0, 2, 1, 1, 2, 1, 3

On the average 16 black spots or 1.6 per throw. I rounded
this to 2 and removed 2 pencils. They represented decayed
atoms per unit time. Thus starting with 10 pencils I had 8
after one time unit. Using 8 pencils I repeated the procedure.
The result was:

2, 2, 0, 0, 2, 0, 0, 2, 1, 1

This times the average was 1.0 black spots and I removed
one pencil. Thus starting with 10 "radioactive atoms" I had
only 7 left after two time units. Then, using 7 pencils, I
repeated the procedure (ten throws again) and got:

0, 1, 3, 1, 2, 2, 0, 1, 1, 2

This times the average was 1.4 black spots. I rounded it
to 1 and removed one pencil. Thus starting with 10
"radioactive atoms" I had only 6 left after three time units.
Using 6 pencils I repeated the procedure and got:

1, 0, 2, 0, 0, 1, 0, 2, 0, 2

This times the average was 0.8 black spots. I rounded it
to 1 and removed one pencil. Thus starting with 10
"radioactive atoms" I had only 5 left after four time units.
Using 5 pencils I repeated the procedure and ...

Working with only ten pencils is far from ideal. I had no
choice but to round fractional outcomes because pencils
(atoms) are indivisible. But suppose the same is done
independently by twenty or thirty students. Then the
number of remaining pencils, after i throws, could be
obtained by averaging individual results. This would
show that the number of remaining pencils decreases,
more or less, exponentially.

Why is decay exponential? Because the probability of
a decay per unit time (i.e. per throw) is a constant (1/6
for each pencil). In a computer code the number of
pencils (atoms) would be, for example, 100,000 and
there would be no need for averaging. The effect of
rounding would be minimal till only ~0.1% of the
original pencils remain. The law of exponential
decay depends on large numbers.

How many throws are needed to eliminate ~50% of
pencils (atoms)? A good mini-exercise for Excel? Yes,
but the question can also be answered without using
a mighty computer. [Hint: N=No*exp(-lambda*t)]

If you have a very very strong, and rapidly decaying,
source try to follow its decay with the a counter
for ten or more half-lives. What for? To show that,
unlike cars or people, atoms have no age. If very old
atoms were more likely to die than young atoms then
the value of lambda would increase in time and the
curve would deviate from exponential. It would be
a good exercise in hypothesis testing. Deviations from
the perfect exponential would be observed by students
near the end. This could provide a chance to discuss
what is and what is not significant. Nobody should
say that "atoms have no age" is only a speculative
theory; the law of radioactive decay is real, as far as
we can determine experimentally.
Ludwik Kowalski