Worked exercise — radon decay in a cellar

Problem

Radon $^{222}\text{Rn}$ has a half-life $\tau_{1/2} = 3{,}82$ days. A sealed cellar initially contains $N_0 = 10^{10}$ radon atoms. How many atoms remain after $20$ days? What is the initial activity of the sample?

Solution

Decay constant. The decay constant is obtained from the half-life:

$\lambda = \frac{\ln 2}{\tau_{1/2}} = \frac{0{,}693}{3{,}82\ \text{days}} \approx 0{,}181\ \text{days}^{-1}$

Number of atoms after 20 days. The exponential decay law gives:

$N = N_0\,e^{-\lambda t} = 10^{10}\cdot e^{-0{,}181\cdot 20} = 10^{10}\cdot e^{-3{,}63} \approx \ev{N \approx 2{,}7\cdot 10^{8}\ \text{atoms}}$

In twenty days (a little more than five half-lives) the radon drops to under $3\%$ of its initial value.

Initial activity. The activity is $A_0 = \lambda N_0$. To get a result in becquerel (decays per second), $\lambda$ must be converted to $\text{s}^{-1}$:

$\lambda = \frac{0{,}181}{86400\ \text{s}} \approx 2{,}09\cdot 10^{-6}\ \text{s}^{-1}$

$A_0 = \lambda N_0 = 2{,}09\cdot 10^{-6}\cdot 10^{10} \approx \ev{A_0 \approx 2{,}1\cdot 10^{4}\ \text{Bq}}$

That is, roughly twenty thousand decays per second.

Why it matters $200\ \text{Bq/m}^3$). Radon is a radioactive gas that builds up in closed, poorly ventilated spaces: it is worth airing cellars and basement rooms often, especially in areas with granitic subsoil, where the natural emission is higher.

The value is well above the recommended limit for dwellings (

Connections

Topics: Nuclear physics Concepts: Radioactive decay Skills: Using logarithmic scales