# What is average life in radioactivity and what is its significance?

By definition, average life of radioactive sample is the amount of time required for it to get decayed to 36.8% of its original amount.

But what is the significance of 36.8% and why has that value been chosen?

• I do not think that for radioactivity one uses the term "average life" .It is called the half life and it is the time it takes for the radioactive material to be reduced by 1/2. Have a look at this article. en.wikipedia.org/wiki/Half-life – anna v Jan 30 '14 at 17:46
• Anna, you are confused about the bases of the logarithms just like Kyle. After one half-life, the percentage of the material left is, by definition, 50% and not 36.8%. We use the lifetime, the average life, which is exactly the e-folding time after which the percentage drops to 36.8%. This lifetime (average life) is more natural exactly because e is a more natural base of logarithms than 2 or 10. The (average) lifetime with the base e is also literally the expectation value of the lifetime of a large ensemble of nuclei of the same kind. – Luboš Motl Jan 30 '14 at 17:47
• @LubošMotl true, i am not discussing the 36.8% I am just saying that the terminology "average life" is strange to me for radioactive materials, have not used it in any lab context. – anna v Jan 30 '14 at 17:51
• Dear anna v, the term "lifetime" is the standard term for the base-e "average" lifetime. But even the term "average lifetime" inevitably occurs often, see e.g. these papers scholar.google.com/… , simply because you don't need to consider this a "new term" in the jargon. It is literally the average of the lifetime of a large enough number of the decaying nuclei of the same kind. See my answer for some more details why it's so. – Luboš Motl Jan 30 '14 at 18:02

The number 36.8% is literally $$36.8 \approx 100 \exp(-1) =\frac{100}{2.71828\dots}$$ Moreover, it is right to call this quantity "average lifetime" or just "lifetime" because it is literally the average value of the time for which a nucleus (or something else) from the ensemble lives.
If the initial number is $N_0$, they decrease to $$N(t) = N_0 \cdot \exp (-t/t_0)$$ at time $t$ where $t_0$ is what we want to call the (average) lifetime. How many nuclei $dN\lt 0$ die (decay) in the short interval $(t,t+dt)$? Well, it's given by the derivative $$dN = dt\cdot \frac{dN(t)}{dt} = N_0\cdot dt\cdot \exp(-t/t_0)\cdot \left(-\frac{1}{t_0}\right)$$ To calculate the average "age at death" (a statistical expectation value), we must integrate $$\langle t \rangle = \int_0^\infty dt\cdot t\cdot P({\rm lifetime}=t) =\\ = -\int_0^\infty t\cdot \frac{1}{N_0} \cdot dN/dt \cdot dt = \int_0^\infty dt\cdot t\cdot \exp(-t/t_0)\frac{1}{t_0} = t_0$$ where $P$ refers to the probability density that the lifetime was $t$ which can be calculated by integration by parts. So the average "age at death" for a large ensemble of nuclei will really be equal to the $t_0$ that appears in the exponent of $\exp(-t/t_0)$.
When radioactive element A decays to produce element B, the (infinitesimal) number of decayed elements A, $dN$, that occurs in a small time interval, $dt$, is proportional to the initial population of A, $N$: $$-\frac{dN}{dt}\propto N$$ Assuming the proportionality is a constant, then the above becomes $$-\frac{dN}{dt}=\lambda N$$ which has a known solution: $$N(t)=N(0)e^{-\lambda t}=N(0)e^{-t/\tau}$$ where $\tau$ is the mean lifetime. Now when 36.8% of the initial material remains then, $N(t)/N(0)=0.368$ and we get $$0.368=e^{-t/\tau}$$ Taking the natural logarithm of this to eliminate the exponential, we find that $$\ln0.368=-1=-\frac{t}{\tau}$$ Thus, the value 36.8% signifies the amount of material left over after one mean lifetime has passed.