Radioactivity
The activity,\(A\), of a radioactive nuclide describes how many decay occurs. The unit Becquerel (Bq) is a measure of decays per second. As nuclide decay, there are fewer left avaible to decay. Thus, the activity decreases with time:
\(A(t) = A_0 \cdot e^{-\lambda \cdot t}\)
\(A(t) = A_0 \cdot e^{-\ln(2) \cdot t/t_{1/2}}\)
Given activity is number of decays per second (or the rate of decays) is can be expressed as:
\(A = \lambda \cdot N\)
Another common unit is the Curie (Ci) with the conversion:
\(1 \; [\text{Ci}] = 3.7 \cdot 10^{10} \; [\text{Bq}]\)
A nonillion (\(10^{30}\)) nuclide of element X decay with a half life of \(10\) years. What is the activity in \(\text{Bq}\) and \(\text{Ci}\)?
First calculate the decay constant (remember to use units of \(s^{-1}\) if we want to calculate the activity using \(A=\lambda N\):
\(\lambda = \cfrac{\ln(2)}{t_{1/2}}\)
\(\lambda = \cfrac{\ln(2)}{10 \times 365.25 \times 24 \times 60 \times 60}\)
\(\lambda = \cfrac{\ln(2)}{3.16 \cdot 10^8 \; [\text{s}] }\)
\(\lambda = 2.20 \cdot 10^{-9} [\text{s}^{-1}]\)
Next, we can calculate the activity in Becquerels:
\( A = \lambda N \)
\( A = (2.20 \cdot 10^{-9}) (1 \cdot 10^30) \)
\( A = 2.20 \cdot 10^{21} \; [\text{Bq}]\)
Finally, we can convert this value to Curies:
\(A = 2.20 \cdot 10^{21} \; [\text{Bq}] \times \cfrac{1 \; [\text{Ci}]}{3.7 \cdot 10^{10} \; [\text{Bq}]}\)
\(A = 5.94 \cdot 10^{10} \; [\text{Ci}]\)
Therefore, we can determine the activity can be expressed as either \(\boldsymbol{2.20 \cdot 10^{21} \; [\textbf{Bq}]}\) or \(\boldsymbol{5.94 \cdot 10^{10} \; [\textbf{Ci}]}\).
Review these lessons:
Try these questions:
Branching Decays
Radioactive elements may decay through different reactions. A decay scheme shows the path to a stable element. Originally, Vanadium decays into titanium which is also radioactive. Titanium has two decay processes and eventually ends up at stable Potassium.
We can add up the decay constants to obtain the total:
\(\lambda_{\text{total}} = \sum_i^n {\lambda_i}\)
A radioactive sample has measured activity of \(2 \; [\text{mCi}]\) and two decay modes \(\lambda_1 = 0.005 \; [\text{s}^{-1}]\) and \(\lambda_2 = 0.0005 \; [\text{min}^{-1}]\). What is the activity after a minute?
We know that the activity will decrease according to the radioactive nuclide decay formula:
\(A(t) = A_0 \cdot e^{-\lambda \cdot t} \)
Next, we can calculate the total decay constant:
\(\lambda_{tot} = \sum_i^n {\lambda_i}\)
\(\lambda_{tot} = 0.0005 + 0.0005 \times 60\)
\(\lambda_{tot} = 0.0305 \; [\text{s}^{-1}]\)
Then, we can substitute the pertinent value(s) into the radiaoctive nuclide decay formula and solve:
\(A(t = 60s) = 2 \cdot e^{-0.0305[\text{s}^{-1}] \cdot 60 [\text{s}]}\)
\(A(t = 60s) = 0.32 \; [\text{mCi}]\)
Therefore, we can determine the activity after a minute is \(\boldsymbol{0.32 \; [\textbf{mCi}]}\).