# Derivation of mean life of a radioactive nucleus

How can the mean life of a radioactive nucleus be derived? Consider R.dt number of nuclei decaying in the time interval t and t+dt. Then, isn't the lifetime of those R.dt number of nuclei is t? But, I couldn't understand why in many derivations, lifetime is mentioned as tR.dt instead of t and we integrate tR.dt from 0 to infinity?

Average values are calculated by integrating over a probability distribution: $$\langle x\rangle=\int_a^b xP(x)\;dx.$$ Thus to solve the problem one must find the probability that a given nuclei does not decay, so that the mean lifetime is given by: $$\langle t\rangle=\int_0^\infty tP_{not decayed}(t)\; dt.$$ Since we know the probability that a given nucleus will decay is given by the distribution: $$P_{decayed}(t)=\int_0^t \lambda e^{-\lambda t^\prime}\; dt^\prime=1-e^{-\lambda t},$$ where $$\lambda$$ is the decay constant of the given radioactive species, and is included for normalization so that: $$P_{decayed} (t)=\int_0^\infty \lambda{N(t)\over N_0}\; dt=1.$$ So the probability that the given nucleus has not decayed is given by: $$P_{not decayed}(t)=1-P_{decayed}(t)=e^{-\lambda t}.$$ So that we have the mean life-time as: $$\langle t\rangle=\int_0^{\infty} te^{-\lambda t}\; dt.$$ Using integration by parts, we find that the mean life time is : $$\langle t\rangle={1\over \lambda}$$. Why must we integrate over infinity, because the sample life time is such that $$N(t)\rightarrow 0$$ asymptoticaly as $$t\rightarrow \infty$$.

The basic assumption is that the rate of decay is proportional to the number of unstable nuclei,

ie $$\dfrac{dN}{dt}\propto N \Rightarrow \dfrac{dN}{dt}= -\lambda N\Rightarrow N = N_0e^{-\lambda t}$$

where $$\lambda$$ is the decay constant and $$N_0$$ is the number of unstable nuclei at time $$t=0$$.

As an aside suppose that you started with $$15$$ and $$8$$ nuclei decayed in the first two second interval, $$4$$ nuclei decayed in the second two second interval, $$2$$ nuclei decayed in the third two second interval and the last nucleus decayed in the fourth two second interval.

How might you estimate the mean lifetime of the unstable nuclei?

How about $$\langle t\rangle=\dfrac{(8\times 1) + (4 \times 3)+(2\times 5)+(1\times7)}{8+4+2+1} \approx 2.47$$.

Where did the $$(8\times 1)$$ come from?
It comes from the fact that between time $$= 0$$ and time $$= 2$$, mid interval time $$1$$ seconds, eight nuclei decayed and so on for $$(\underbrace{4}_{\rm number}\times \underbrace{3}_{\rm mid\,interval})$$ etc, with $$(8+4+2+1)$$ being the total number.

Coming back to the radioactive decay.

The number of unstable nuclei which decay in infinitesimal intervals of time from $$t$$ to $$t+dt$$ is

$$dN= -\lambda N(t) \,dt=-\lambda N_0e^{-\lambda t} \,dt$$.

and the mean lifetime is $$\langle t\rangle=\dfrac{\displaystyle\int_{t=0}^\infty \left(-\lambda N_0e^{-\lambda t} \,dt\right) \times t}{N_0}=\dfrac{1}{\lambda}$$