Question

(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.

Answer 1

(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 $N-\frac{1}{2}{N}_D$

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

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

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

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

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

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

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

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

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

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

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

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.