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If I know that the sample originally has an activity of $A_0$ and after a time $t$ has an activity of $A_1$, how can I use this to determine the half-life of the material?

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  • $\begingroup$ What would the half-life be (in terms of $t$) if $A_1=A_0/4$ ? $\endgroup$
    – PM 2Ring
    Jan 16 at 12:58
  • $\begingroup$ HL = t / 2... so half life is just $2t \times \frac{A_1}{A_0}$? $\endgroup$
    – physBa
    Jan 16 at 13:41
  • $\begingroup$ $t/2$ is correct for my example, but your equation isn't right. Eg, if $A_1=A_0/8$ then the half-life is $t/3$. Think about exponents and logarithms... $\endgroup$
    – PM 2Ring
    Jan 16 at 13:57

2 Answers 2

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When you are aiming at the half-life $t_{1/2}$ it is actually easier to write the decay law with power of $2$, rather than with power of $e$. $$A(t)=A_0 2^{-t/t_{1/2}}$$ or $$\frac{A(t)}{A_0}= 2^{-t/t_{1/2}}$$

Now take the logarithm ($\ln$) on both sides (remember $\ln(a^b)=b\ln(a)$): $$\ln\frac{A(t)}{A_0}=-\frac{t}{t_{1/2}} \ln(2)$$

Then it is easy to solve this for $t_{1/2}$.

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In the general case it is not a simple question. But for the simplest case it is quite simple.

For a single radioactive species that decays into a stable species, and has only one decay path, you get a simple equation. Given that the activity is $A_0$ at $t=0$ the activity at other times is given by the following.

$$ A(t) = A_0 \exp\left(-\frac{t\ln(2)}{t_{1/2}}\right) $$

Here $t_{1/2}$ is the half life. So at $t=t_{1/2}$ you have just half the activity you started with. So if you are given the activity at $t=0$ (that's the $A_0$) and at some measurement time $t_m$, you have to solve this equation for $t_{1/2}$.

It's a lot more fun if you have multiple decay chains so that you have multiple species of radioactive materials with different half lives. A fairly simple one is I-135 decaying to Xe-135 (a Beta decay with a half life of about 6.57 hours) and then Xe-135 decaying to Cs-135 with a half life of about 9.14 hours. The Cs-135 has a half life of 2.3 million years, so you can treat it as stable for most purposes. So, when you get comfortable with the previous equation, try writing down the equation for activity of this combined system. This is important in nuclear reactors because fission produces (among a bunch of other stuff) both I-135 and Xe-135, and Xe-135 is a strong neutron absorber.

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