# What is "average life" in terms of nuclear physics?

I am having a doubt after reading on so many sites. I am not able to understand the clear meaning of average life. Whether it is 36.18 or something else.

• Do you mean half life? Dec 27 '19 at 9:56
• @OfekGillon Half-life is not the same as the average lifetime.
– user59991
Dec 27 '19 at 10:03
• I know, but average life isn't common Jargon. Second, they are the same up to a constant factor (if he does mean what I think he means) Dec 27 '19 at 10:05
• @Ofek "Average life" isn't common jargon, but "mean lifetime" is. There's a simple mathematical relationship between the mean lifetime and the half-life, see en.wikipedia.org/wiki/Exponential_decay Dec 27 '19 at 15:30

Each unstable nucleus, in a given sample of radioactive matter consisting of nuclei of the same type, decays after a certain period of time.

If you add up all those times and then divide the sum by the number of unstable nuclei in the beginning you obtain the average time for an unstable nucleus to decay.
This average time is often called the (average) lifetime of the decay and can be shown to equal the reciprocal of the decay constant of the decay.

Considering a radioactive nucleus, let the initial number of nuclei be $$N_o$$ with its disintegration constant $$\lambda$$.

From it's first order decay/radioactive decay law, $$N(t)=N_o e^{-\lambda t}$$( let this be equation $$\boxed{1}$$) is the number of nuclei left at a general time t.

The lifetime of each $$dN$$ nuclei which decays at a given instant is NOT the same here. For example, at t=0 the lifetime of dN nuclei which decays is 0, whereas at t=10s(say) the lifetime of dN nuclei which has just decayed is 10s.

So we it makes sense to only talk about average lifetime of all the nuclei.

Average lifetime of all the nuclei

$$=\frac{\int tdN}{\int dN}$$(by definition)

$$=\frac{\int_{0}^{\infty} tNo(-\lambda)e^{-\lambda t}dt}{No}$$. (from equation $$\boxed {1}$$)

$$=\frac{1}{\lambda}$$ after doing the simple math.

mean lifetime (sometimes called just the lifetime) $$\tau$$, which is defined as the average time that a nucleus is likely to survive before it decays. The number that survive to time $$t$$ is just $$N(t)$$, and the number that decay between $$t$$ and $$t+dt$$ is $$\left |\frac{dN}{dt} \right |dt$$ . The mean lifetime is then [1]:

$$\tau=\frac{\int_{0}^{\infty}t\left |\frac{dN}{dt} \right |dt}{\int_{0}^{\infty}\left |\frac{dN}{dt} \right |dt}$$ let's try to evaluate this integral, we start by using the decay constant relation (the probability per unit time for the decay of an atom): $$\lambda =\frac{1}{N}\left |\frac{dN}{dt} \right |$$ and the exponential law of radioactive decay (which is its solution): $$N=N_{0}e^{-\lambda t}\\$$ then: $$\left |\frac{dN}{dt} \right |=N_{0}e^{-\lambda t}\lambda$$ substituting in $$\tau$$: $$\tau=\frac{\int_{0}^{\infty}tN_{0}e^{-\lambda t}\lambda dt}{\int_{0}^{\infty}N_{0}e^{-\lambda t}\lambda dt}\\$$ $$\tau=\frac{\int_{0}^{\infty}te^{-\lambda t}dt}{\int_{0}^{\infty}e^{-\lambda t} dt}\\$$ the integral in the nominator is equal to (integration by parts): $$\int_{0}^{\infty}te^{-\lambda t}dt=\frac{1}{\lambda^{2}}$$ the integral in the denominator is equal to: $$\int_{0}^{\infty}e^{-\lambda t} dt=\frac{1}{\lambda}$$ then: $$\tau=\frac{\frac{1}{\lambda^{2}}}{\frac{1}{\lambda}}=\frac{1}{\lambda}$$ Thus the mean lifetime is simply the inverse of the decay constant.

[1] introductory to nuclear physics by krane