Question 29.4: Radon Gas Goal Calculate the number of nuclei after an arbit...
Radon Gas
Goal Calculate the number of nuclei after an arbitrary time and the time required for a given number of nuclei to decay.
Problem Radon { }_{86}^{222} \mathrm{Rn} is a radioactive gas that can be trapped in the basements of homes, and its presence in high concentrations is a known health hazard. Radon has a half-life of 3.83 days. A gas sample contains 4.00 \times 10^{8} radon atoms initially. (a) How many atoms will remain after 14.0 days have passed if no more radon leaks in? (b) What is the activity of the radon sample after 14.0 days? (c) How long before 99 \% of the sample has decayed?
Strategy The activity can be found by substitution into Equation 29.5,
T_{1/2}=\frac{ln\,2}{\lambda}={\frac{0.693}{\lambda}} (29.5)
as before. Equation 29.4a (or Eq. 29.4b)
N=N_{0}e^{-\lambda t} (29.4a)
N=N_{0}{\bigg(}{\frac{1}{2}}{\bigg)}^{n} (29.4b)
must be used to find the number of particles remaining after 14.0 days. To obtain the time asked for in part (c), Equation 29.4a must be solved for time.
(a) How many atoms will remain after 14.0 days have passed?
Determine the decay constant from Equation 29.5:
\lambda=\frac{0.693}{T_{1 / 2}}=\frac{0.693}{3.83 \text { days }}=0.181 \text { day }^{-1}
Now use Equation 29.4a, taking N_{0}=4.0 \times 10^{8}, and the value of \lambda just found to obtain the number N remaining after 14 days:
\begin{aligned} N & =N_{0} e^{-\lambda t}=\left(4.00 \times 10^{8} \text { atoms }\right) e^{-\left(0.181 \text { day }^{-1}\right)(14.0 \text { days })} \\ & =3.17 \times 10^{7} \text { atoms } \end{aligned}
(b) What is the activity of the radon sample after 14.0 days?
Express the decay constant in units of \mathrm{s}^{-1} :
\lambda=\left(0.181 \text { day }^{-1}\right)\left(\frac{1 \text { day }}{8.64 \times 10^{4} \mathrm{~s}}\right)=2.09 \times 10^{-6} \mathrm{~s}^{-1}
From Equation 29.3
R=\left|{\frac{\Delta N}{\Delta t}}\right|=\lambda N (29.3)
and this value of \lambda, compute the activity R :
\begin{aligned} R & =\lambda N=\left(2.09 \times 10^{-6} \mathrm{~s}^{-1}\right)\left(3.17 \times 10^{7} \text { atoms }\right) \\ & =66.3 \text { decays } / \mathrm{s}=66.3 \mathrm{~Bq} \end{aligned}
(c) How much time must pass before 99 \% of the sample has decayed?
Solve Equation 29.4a for t, using natural logarithms:
\begin{aligned} \ln (N)=\ln \left(N_{0} e^{-\lambda t}\right) & =\ln \left(N_{0}\right)+\ln \left(e^{-\lambda t}\right) \\ \ln (N)-\ln \left(N_{0}\right) & =\ln \left(e^{-\lambda t}\right)=-\lambda t \\ t=\frac{\ln \left(N_{0}\right)-\ln (N)}{\lambda} & =\frac{\ln \left(N_{0} / N\right)}{\lambda} \end{aligned}
Substitute values:
t=\frac{\ln \left(N_{0} / 0.01 N_{0}\right)}{2.09 \times 10^{-6} \mathrm{~s}^{-1}}=2.20 \times 10^{6} \mathrm{~s}=25.5 \text { days }
Remarks This kind of calculation is useful in determining how long you would have to wait for radioactivity at a given location to fall to safe levels.