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Exponential decay

  • Thread starter chroot
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chroot
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From a member called "Banana":

Now I'm trying to do one with radioactive decay. Do you think I would use those same formulas? The only confusing thing is that it's not presented the same. It says that a material decays so that it is 99 percent gone in 6.65 half-lives (so would you double that?). After how many half-lives is the material 99.9 percent gone? I tried to piece together some ideas I found in other text books, but I don't think it makes sense. Here's what I tried:

- ln (N/No)/ln2
- ln (.001/.01)/.693 = 3.3 (I don't think this is right because I didn't even find anywhere to insert the 6.65)

Is that wrong? Thanks for your help, I REALLY appreciate it. Banana
 

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chroot
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The key equation is the same. In this case, after one half-life has elapsed, the population of remaining radioactive atoms is one-half what it was at time t=0.

Thus

[tex]0.5 N_0 = N_0 e^{k t_{hl}}[/tex]

Where [itex]t_{hl}[/itex] is the half-life.

If one-half the population remains after one half-life, you get this equation:

[tex]0.5 = e^{k \cdot 1}[/tex]

Thus, k always equals [itex]\ln{0.5}[/itex], or about -0.69, in problems dealing with half-life.

Now, using this k, we can solve for the situation where 99.9% of the original population is gone (therefore 0.1% remains):

[tex]0.001 N_0 = N_0 e^{\ln{0.5} t}[/tex]

Solve for t. Let me know what you get.

- Warren
 
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[tex] \frac{a}{a-x}=2^{\frac{t}{\tau}} [/tex]

tau : half life

And a is initial amt and a-x is the remaining
from the data

[tex] log(1000) =log2*\frac{t}{\tau}[/tex]

So u get t={log1000/log2}*tau =9.96 [tex]\tau[/tex]
 
Last edited:

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