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So I understand that half-life $T_{1/2}$ is the time for the amount $N(0)$ to reduce by half. Basically, $$N(t)=N(0)2^{-t/T_{1/2}} $$

My question is that why can't I use this to directly pull out the the mean lifetime $\tau$, and why $\tau = ln(2)T_{1/2}$ is correct. Assume $T_{1/2}=1$ for easiness, I make the argument:

Since a particle has a 50% chance of decaying after $1$ second, it'll have a 50% chance of surviving after 1 second. Thus it'll have 25% chance of surviving for $2$ seconds, and 12.5% for $3$ seconds etc. So the average lifetime is calculated:

$$\tau = 1\cdot\frac{1}{2}+2\cdot\frac{1}{4}+3\cdot\frac{1}{8}+...=2$$

The correct answer must be $\tau = \frac{1}{ln(2)}=1.44$. What am I wrong about here? Thank you!

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    $\begingroup$ As with many things Hyperphysics has a specific page on the relationship between these quantities. $\endgroup$ – StephenG Aug 29 '18 at 10:21
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What you've calculated is the mean number of lifetimes if it's discrete, but it's continuous. The correct calculation is $\int_0^\infty t\dfrac{d}{dt}(1-2^{-t/T_{1/2}})dt$. I'll leave the rest to you.

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If you want to calculate mean lifetime by summing a series, the right way to do it is by summing $$\tau = \sum(\text{time at which particle decays})\times(\text{probability that particle decays at that time})$$ So the first issue is that you're calculating something different. Your series is computing $$\sum(\text{time at which particle hasn't decayed})\times(\text{probability that particle hasn't decayed at that time})$$ which isn't a meaningful quantity. If nothing else, this double-counts some fraction of particles. In particular, some of the particles that survive for one second are also going to survive for two seconds, or three seconds, or longer, and those get included in multiple terms in your sum.

Fixing this one issue will get you the following series: $$1\times \underbrace{\biggl(1 - \frac{1}{2}\biggr)}_{P(\text{decay},1)} + 2\times \underbrace{\frac{1}{2}}_{P(\text{survival},1)}\times\underbrace{\biggl(1 - \frac{1}{2}\biggr)}_{P(\text{decay},2)} + \cdots$$ where $P(\text{decay},1)$ represents the probability that the particle decays after $1\ \mathrm{s}$, and so on.

At this point you might realize that the other thing you're doing is restricting the particles to only decaying at integral numbers of seconds. If you think about it, your calculation makes no distinction between a particle that decays after $1.1\ \mathrm{s}$ and a particle that decays after $1.9\ \mathrm{s}$, but that should make a difference because it changing $1.1\ \mathrm{s}$-lifetime particles for $1.9\ \mathrm{s}$-lifetime particles will bump up the average lifetime.

You can actually extrapolate your logic to work out the right solution, just by shrinking the time interval. For example, from your first equation $N(t) = N(0)\ 2^{-t/T_{1/2}}$, you know the particle has probability $\frac{1}{2}$ of surviving for the first second. But what about the first half-second? $$\begin{align} P(\text{survival},1/2) &= \frac{N(1/2)}{N(0)} = 2^{-1/2} = \frac{1}{\sqrt{2}} \\ P(\text{decay},1/2) &= 1 - P(\text{survival},1/2) = 1 - \frac{1}{\sqrt{2}} \end{align}$$ And the second half-second? $$\begin{align} P(\text{survival},1) &= \frac{1}{2} \\ P(\text{decay},1) &= 1 - P(\text{survival},1) = \frac{1}{2} \end{align}$$ And so on. So if you allow particles to decay at half-second increments instead of one-second increments, you'd get $$\frac{1}{2}\times\biggl(1 - \frac{1}{\sqrt{2}}\biggr) + 1\times\frac{1}{\sqrt{2}}\times\biggl(1 - \frac{1}{\sqrt{2}}\biggr) + \frac{3}{2}\times\biggl[1 - \biggl(1 - \frac{1}{\sqrt{2}}\biggr) - \frac{1}{\sqrt{2}}\biggl(1 - \frac{1}{\sqrt{2}}\biggr)\biggr]\times\frac{1}{\sqrt{2}} + \cdots$$ which simplifies to $$\frac{1}{2}\times\biggl(1 - \frac{1}{\sqrt{2}}\biggr) + 1\times\frac{1}{\sqrt{2}}\times\biggl(1 - \frac{1}{\sqrt{2}}\biggr) + \frac{3}{2}\times\biggl(\frac{1}{\sqrt{2}}\biggr)^2\biggl(1 - \frac{1}{\sqrt{2}}\biggr) + \cdots = \sum_n \frac{n}{2}\biggl(\frac{1}{\sqrt{2}}\biggr)^{n - 1}\biggl(1 - \frac{1}{\sqrt{2}}\biggr) \approx 1.707$$ Doing the same thing with quarter-second intervals gets you to $$\sum_n \frac{n}{4}\biggl(\frac{1}{2^{1/4}}\biggr)^{n - 1}\biggl(1 - \frac{1}{2^{1/4}}\biggr) \approx 1.571$$ Maybe you can see the pattern here: if $\Delta$ is the time interval in seconds, it's $$\sum_n n\Delta\frac{1}{2^{\Delta(n - 1)}}\biggl(1 - \frac{1}{2^{\Delta}}\biggr)$$ Taking the limit as $\Delta\to 0$ gives you $\frac{1}{\ln 2}$.

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