Question 29.3: The Activity of Radium Goal Calculate the activity of a radi...
The Activity of Radium
Goal Calculate the activity of a radioactive substance at different times.
Problem The half-life of the radioactive nucleus { }_{88}^{226} \mathrm{Ra} is 1.6 \times 10^{3} \mathrm{yr}. If a sample initially contains 3.00 \times 10^{16} such nuclei, determine (a) the initial activity in curies, (b) the number of radium nuclei remaining after 4.8 \times 10^{3} \mathrm{yr}, and (c) the activity at this later time.
Strategy For parts (a) and (c), find the decay constant and multiply it by the number of nuclei. Part (b) requires multiplying the initial number of nuclei by one-half for every elapsed half-life. (Essentially, this is an application of Equation 29.4b.)
N=N_{0}{\bigg(}{\frac{1}{2}}{\bigg)}^{n} (29.4b)
(a) Determine the initial activity in curies.
Convert the half-life to seconds:
T_{1 / 2}=\left(1.6 \times 10^{3} \mathrm{yr}\right)\left(3.156 \times 10^{7} \mathrm{~s} / \mathrm{yr}\right)=5.0 \times 10^{10} \mathrm{~s}
Substitute this value into Equation 29.5
T_{1/2}={\frac{\ln2}{\lambda}}={\frac{0.693}{\lambda}} (29.5)
to get the decay constant:
\lambda=\frac{0.693}{T_{1 / 2}}=\frac{0.693}{5.0 \times 10^{10} \mathrm{~s}}=1.4 \times 10^{-11} \mathrm{~s}^{-1}
Calculate the activity of the sample at t=0, using R_{0}=\lambda N_{0}, where R_{0} is the decay rate at t=0 and N_{0} is the number of radioactive nuclei present at t=0 :
\begin{aligned} R_{0} & =\lambda N_{0}=\left(1.4 \times 10^{-11} \mathrm{~s}^{-1}\right)\left(3.0 \times 10^{16} \text { nuclei }\right) \\ & =4.2 \times 10^{5} \text { decays } / \mathrm{s} \end{aligned}
Convert to curies to obtain the activity at t=0, using the fact that 1 \mathrm{Ci}=3.7 \times 10^{10} decays / \mathrm{s} :
\begin{aligned} R_{0} & =\left(4.2 \times 10^{5} \text { decays } / \mathrm{s}\right)\left(\frac{1 \mathrm{Ci}}{3.7 \times 10^{10} \text { decays } / \mathrm{s}}\right) \\ & =1.1 \times 10^{-5} \mathrm{Ci}=11 \mu \mathrm{Ci} \end{aligned}
(b) How many radium nuclei remain after 4.8 \times 10^{3} years?
Calculate the number of half-lives, n :
n=\frac{4.8 \times 10^{3} \mathrm{yr}}{1.6 \times 10^{3} \mathrm{yr} / \text { half-life }}=3.0 \text { half-lives }
Multiply the initial number of nuclei by the number of factors of one-half:
\quad N=N_{0}\left(\frac{1}{2}\right)^{n} (1)
Substitute N_{0}=3.0 \times 10^{16} and n=3.0:
N=\left(3.0 \times 10^{16} \text { nuclei }\right)\left(\frac{1}{2}\right)^{3.0}=3.8 \times 10^{15} \text { nuclei }
(c) Calculate the activity after 4.8 \times 10^{3} \mathrm{yr}.
Multiply the number of remaining nuclei by the decay constant to find the activity R :
\begin{aligned} R & =\lambda N=\left(1.4 \times 10^{-11} \mathrm{~s}^{-1}\right)\left(3.8 \times 10^{15} \text { nuclei }\right) \\ & =5.3 \times 10^{4} \text { decays } / \mathrm{s} \\ & =1.4 \mu \mathrm{Ci} \end{aligned}
Remarks The activity is reduced by half every half-life, which is naturally the case because activity is proportional to the number of remaining nuclei. The precise number of nuclei at any time is never truly exact, because particles decay according to a probability. The larger the sample, however, the more accurate are the predictions from Equation 29.4.