Finding the decay rate of some element

Question

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?

Answer 1

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 $P(t)=\frac12$.

Answer 2

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.