(a) Plot a graph showing the variation of undecayed nuclei $N$ versus time $t$. From the graph, find out how one can determine the half-life and average life of the radioactive nuclei.

(b) The air in some caves includes a significant amount of radon gas, which can lead to lung cancer if breathed over a prolonged time. In British caves, the air in the cave with the greatest amount of the gas has an activity per volume of $1.55\times 10^{5}Bq/m^3$ . Suppose that you spend two full days exploring (and sleeping in) that cave. Approximately how many ${}^{222}Rn$ atoms would you take in and out of your lungs during your two-day stay? The radionuclide ${}^{222}Rn$ in radon gas has a half-life of 3.82 days. You need to estimate your lung capacity and average breathing rate.

(a) Law of Radioactivity defines that the number of Nuclei undergoing number of Nuclei present in the sample at that Instant.

Since from the graph; we have

     $N=N_{-}{e}^{-\lambda t}$    ....1

where $\lambda$=Disintegration constant

For $\therefore$ For $T_{1/2}$ is the time at $N-\frac{1}{2}{N}_D$

$\Rightarrow$ $t=T_{1/2}$

$\frac{N_0}{2}$=$N_0{e}^{-\lambda \frac{T}{2}}$

$\Rightarrow$ $\boxed{T_{1/2}=\frac{Ln2}{\lambda }=\frac{0.693}{\lambda }}$

And for Mean life -we have to sum it over the whole Range for 

$N(t)=N_0{e}^{-\lambda t}$

for number of nuclei which decay in time t to t  t $△$;

$N(t)△t=\lambda N_0{e}^{-\lambda t}△t$

 For Integration it over the Range $T=0to\infty$

$\tau$=$\frac{\lambda N_0{\iint }_0^{\infty }t{e}^{-\lambda t}dt}{N_0}$

$=\lambda {\int }_0^{\infty }t{e}^{\lambda t}dt$

$\boxed{\tau =\frac{1}{\lambda }}$

Mean-life

(b) The equation for the activity is given by:

$R=\lambda N$

Here, R is the activity, N is the number of nuclei and $\lambda$ is the decay constant. The equation for the decay constant is given by, 

$\lambda =\frac{ln2}{T_{1/2}}$

Here, $T_{1/2}$ is the half - life

Thus, $R=\frac{ln2}{T_{1/2}}N$

$N=\frac{R{T}_{1/2}}{ln2}$

Dividing by volume, $\frac{N}{V}=\frac{R}{V}\frac{T_{1/2}}{ln2}$

Substituting the value,

$=\frac{N}{V}=(1.55\times 10^{5}Bq/m^3)\frac{3.82\times 24\times 3600s}{ln2}$

            $=7.38\times 10^{10}$  $atoms/m^3$

Thus, these are $7.38\times 10^{10}$ atoms $/m^{3}Rn$ atms per unit volume in the cave.