Calculations Involving Radioactive
Decay and Time
If we start with \(1.000 \mathrm{~g}\) of strontium- \(90,0.953 \mathrm{~g}\) will
remain after \(2.00 \mathrm{yr}\). (a) What is the half-life of strontium- 90 ?
(b) How much strontium- 90 will remain after \(5.00 \mathrm{yr}\) ? (c) What is
the initial activity of the sample in becquerels and curies?
SOLUTION
(a) Analyze We are asked to calculate a half-life, \(t_{1 / 2}\) based on data
that tell us how much of a radioactive nucleus has decayed in a time interval
\(t=2.00 \mathrm{yr}\) and the information \(N_{0}=1.000 \mathrm{~g}, N_{t}=0.953
\mathrm{~g}\).
Plan We first calculate the rate constant for the decay, \(k\), and then use
that to compute \(t_{1 / 2}\).
Solve Equation \(21.19\) is solved for the decay constant, \(k\), and then
Equation \(21.20\) is used to calculate half-life, \(t_{1 / 2}\) :
$$
\begin{aligned}
k &=-\frac{1}{t} \ln \frac{N_{\mathrm{f}}}{N_{0}}=-\frac{1}{2.00 \mathrm{yr}}
\ln \frac{0.953 \mathrm{~g}}{1.000 \mathrm{~g}} \\
&=-\frac{1}{2.00 \mathrm{yr}}(-0.0481)=0.0241 \mathrm{yr}^{-1} \\
t_{1 / 2} &=\frac{0.693}{k}=\frac{0.693}{0.0241 \mathrm{yr}^{-1}}=28.8
\mathrm{yr}
\end{aligned}
$$
(b) Analyze We are asked to calculate the amount of a radionuclide remaining
after a given period of time.
Plan We need to calculate \(N_{b}\) the amount of strontium present at time \(t\),
using the initial quantity, \(N_{0}\), and the rate constant for decay, \(k\),
calculated in part (a).
Solve Again using Equation \(21.19\), with \(k=0.0241 \mathrm{yr}^{-1}\), we have
$$
\ln \frac{N_{t}}{N_{0}}=-k t=-\left(0.0241 \mathrm{yr}^{-1}\right)(5.00
\mathrm{yr})=-0.120
$$
\(N_{t} / N_{0}\) is calculated from \(\ln \left(N_{t} / N_{0}\right)=-0.120\)
using the \(e^{x}\) or INV LN function of a calculator:
$$
\frac{N_{t}}{N_{0}}=e^{-1.120}=0.887
$$
Because \(N_{0}=1.000 \mathrm{~g}\), we have
$$
N_{t}=(0.887) N_{0}=(0.887)(1.000 \mathrm{~g})=0.887 \mathrm{~g}
$$
(c) Analyze We are asked to calculate the activity of the sample in becquerels
and curies.
Plan We must calculate the number of disintegrations per atom per second and
then multiply by the number of atoms in the sample.
Solve The number of disintegrations per atom per second is given by the decay
constant, \(k\) :
$$
k=\left(\frac{0.0241}{\mathrm{yr}}\right)\left(\frac{1 \mathrm{yr}}{365 \text
{ days }}\right)\left(\frac{1 \text { day }}{24
\mathrm{~h}}\right)\left(\frac{\mathrm{lh}}{3600 \mathrm{~s}}\right)=7.64
\times 10^{-10} \mathrm{~s}^{-1}
$$
To obtain the total number of disintegrations per second, we calculate the
number of atoms in the sample. We multiply this quantity by \(k\), where we
express \(k\) as the number of disintegrations per atom per second, to obtain
the number of disintegrations per second:
$$
\begin{aligned}
\left(1.000 \mathrm{~g}^{90} \mathrm{Sr}\right)\left(\frac{1
\mathrm{~mol}^{90} \mathrm{Sr}}{90 \mathrm{~g}^{90}
\mathrm{Sr}}\right)\left(\frac{6.022 \times 10^{23} \mathrm{atoms}
\mathrm{Sr}}{1 \mathrm{~mol}^{90} \mathrm{Sr}}\right)=6.7 \times 10^{21} \text
{ atoms }^{90} \mathrm{Sr} \\
\text { Total disintegrations/s } &=\left(\frac{7.64 \times 10^{-10} \text {
disintegrations }}{\text { atom }{ }^{2} \mathrm{~s}}\right)\left(6.7 \times
10^{21} \text { atoms }\right) \\
&=5.1 \times 10^{12} \text { disintegrations/s }
\end{aligned}
$$
Because \(1 \mathrm{~Bq}\) is one disintegration per second, the activity is
\(5.1 \times 10^{12} \mathrm{~Bq}\). The activity in curies is given by
$$
\left(5.1 \times 10^{12} \text { disintegrations/s }\right)\left(\frac{1
\mathrm{Ci}}{3.7 \times 10^{10} \text { disintegrations/s }}\right)=1.4 \times
10^{2} \mathrm{Ci}
$$
We have used only two significant figures in products of these calculations
because we do not know the atomic weight of \({ }^{90} \mathrm{Sr}\) to more
than two significant figures without looking it up in a special source.
