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Radioactive Decay

Date: 10/26/2004 at 19:48:21
From: Alicia 
Subject: Radioactive decay

The half-life of a radioactive element is 131 days, but your sample 
will not be useful to you after 90% of the radioactive nuclei 
originally present have disintegrated.  For about how many days can 
you use the sample?



Date: 10/27/2004 at 11:30:43
From: Doctor Ian
Subject: Re: Radioactive decay

Hi Alicia,

In radioactive decay, you have atoms of one elements changing into 
atoms of another element because some of its neutrons turn into 
protons, changing the atomic number.  (For example, if we start with a
Carbon-14 atom, with 6 protons and 8 neutrons, and one of its neutrons
becomes a proton, we end up with a Nitrogen-14 atom, with 7 protons 
and 7 neutrons.)  

The rate at which this happens isn't constant.  Rather, in some given
period, half of the elements will decay; then in the same period, half
of the remaining elements will decay; and so on. 

Suppose we have an imaginary element Bobium, and it decays into the
element Stanium with a half-life of two days.  Then if we start with
128 grams of Bobium, after two days we'll have 64 grams of Bobium, and
64 grams of Stanium.  After another two days, we'll have 32 grams of
Bobium, and 96 grams of Stanium.  And so on:

  Days     Bobium    Stanium
  ----     ------    -------
    0        128          0
    2         64         64
    4         32         96
    6         16        112
    8          8        120
   10          4        124
   12          2        126
   14          1        127
   16        1/2        127 1/2
   18        1/4        127 3/4
   20        1/8        127 7/8

Does this make sense? 

Now, in this case, suppose I need have at least 10% Bobium in order to
use it.  I can compute the percentage of Bobium:

  Days     Bobium    Stanium         Percent Bobium
  ----     ------    -------         --------------
    0        128          0               100
    2         64         64                50
    4         32         96                25
    6         16        112                12.5
    8          8        120                 6.25
   10          4        124
   12          2        126
   14          1        127
   16        1/2        127 1/2
   18        1/4        127 3/4
   20        1/8        127 7/8

So sometime between 6 and 8 days, my sample will become unusable. 

Could I figure this out without making a table?  Well, note that after
n days, the amount of Bobium I have left is 

  128 * (1/2)^(n/2)

To check, 

   n (days)    128 * (1/2)^(n/2)
   --------    -----------------
      0        128 * (1/2)^(0/2) = 128 * 1     = 128
      2        128 * (1/2)^(2/2) = 128 * (1/2) =  64
      4        128 * (1/2)^(4/2) = 128 * (1/4) =  32

So it looks like this captures the data in the table.  What percentage
of my sample will still be Bobium after n days?  It will be 

   amount of Bobium
   ---------------- * 100
   amount of sample

So that's just 

  128 * (1/2)^(n/2)
  ----------------- 
        128

and I need for that be more than 10%:

  128 * (1/2)^(n/2)
  ----------------- > 0.10 
        128

Note that the initial amount (128 grams) isn't really doing anything
here: I multiply by it, and then divide by it.  Which just means that
the length of time the sample will be useful doesn't depend on the
size of the sample!  Which is interesting, but makes sense when you
think about what's going on.  (If I cut something in half, and then in
half again, I have 25% of what I started with, regardless of whether I 
started with one ounce or a million tons.) 

Anyway, the initial amount cancels out, and I'm left with

  (1/2)^(n/2) > 0.10

And now I can find the largest value of n for which this is true by
trying different values (which is what I did in the table); or I can
use logarithms--which is probably what you're studying now, if you've 
been assigned a question like this.  :^D

Does this help?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Logs
High School Physics/Chemistry

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