I have a question that in short reads:

If some element decays exponentially with a half life of 25,000 years, how long will it take for 99.9% of it to decay away?

Using the exponential decay formula: $P(t)=P_0e^{-kt}$ where $k$ is the half life, $t$ is time, $P_0$ is the initial quantity of the element (at $t=0$) and $P(t)$ is the quantity at time $t$. And $k=ln(2)\div T$ where $T$ is the half life.

This is what I have tried:

First get the decay rate:

$k = ln(2)\div T$

$k=ln(2)\div 25000$

$k \approx 0.000027725$

Now I substitute the decay rate and solve for $t$:

$P(t) = P_0e^{-kt}$

$0.1=1\times e^{-0.000027725t}$

$ln(0.1) = ln(e^{-0.000027725t})$

$ln(0.1) = -0.000027725t$

$t= \frac {ln(0.1)}{-0.000027725}$

$t \approx 83,050.85998$

However, the answer from my class solutions is $t \approx 249,144.61$

Can anyone point me in the right direction or let me know where my mistake is?

  • 1
    $\begingroup$ You were doing things basically right. However, $99.9\%$ gone means that $0.001$ remains, not $0.1$. $\endgroup$ – André Nicolas Nov 17 '15 at 23:35
  • $\begingroup$ oh oops. :/ thanks for the catch $\endgroup$ – 121c Nov 17 '15 at 23:36
  • $\begingroup$ Either $k$ is the decay rate, and not half life, or you are using the wrong formula. $\endgroup$ – 5xum Nov 17 '15 at 23:45
  • $\begingroup$ You are right. Supposed to be decay rate $\endgroup$ – 121c Nov 17 '15 at 23:47

Your original formula is wrong. The decay must be such that when $t=k$, that is, when one half life has passed, the remaining value is exactly half.

The formula is $$P(t)=P_0\cdot e^{-\frac{t}{k}\cdot \ln 2}$$ or, avoiding $\ln$ and $e$, it is $$P(t) = P_0 2^{-\frac tk}$$ (which is the same thing).

You can easily check that if $t=k$, then $P(t)=\frac12$.

  • $\begingroup$ Actually, OP has the formula right, but has difficulty interpreting it. $\endgroup$ – André Nicolas Nov 18 '15 at 4:15

You made a miscalculation. It wants to know when 99.9 percent of it is gone, which means 0.001 of it is left. You put percent on the left-hand side of your equation by mistake, so you should be taking the natural log of 0.001, not 0.1.

  • $\begingroup$ But that's not where the mistake was made. The mistake was in the beginning. $\endgroup$ – 5xum Nov 17 '15 at 23:39
  • $\begingroup$ Using the formula I stated and changing 0.1 to 0.001 I get the correct answer $\endgroup$ – 121c Nov 17 '15 at 23:40
  • $\begingroup$ @Cryptic Well then $k$ is not the half life. $\endgroup$ – 5xum Nov 17 '15 at 23:44
  • $\begingroup$ @5xum, his formula is correct (it is the same as yours as he states k = ln(2) % T, where T is the half life, where in your formula you use k for the half life). He simply mis-stated k to be the half life in his text. $\endgroup$ – Dave Nov 17 '15 at 23:50
  • $\begingroup$ @Dave Yeah, I understand that now. Thanks. $\endgroup$ – 5xum Nov 17 '15 at 23:52

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