Question

230Th and 226Ra are intermediate nuclides in the decay series of 238U to 206Pb. The half-lives of 238U, 230Th, and 226Ra are 4.47 billion years, 75,000 years, and 1600 years, respectively. At secular equilibrium, when activities are equal, 10 billion atoms of 238U are present. The number of atoms of 226Ra present at equilibrium is ________

Hint:

At secular equilibrium in a decay series, the number of atoms of the daughter isotopes can be calculated by using the ratio of their decay constants.

Answer 1

Step 1: Understanding the relationship in the decay series.
At secular equilibrium, the activities (rate of decay) of parent and daughter isotopes are equal. The activity of a radioactive isotope is proportional to the number of atoms and its decay constant (\(\lambda\)). Therefore, at secular equilibrium: \[ {Activity of } 238U = {Activity of } 230Th = {Activity of } 226Ra \] At equilibrium, the number of atoms of the daughter isotopes can be related to the number of atoms of the parent isotope using the ratio of their decay constants.

Step 2: Decay constants and activity relations.
The decay constant \(\lambda\) is related to the half-life (\(t_{1/2}\)) by the equation: \[ \lambda = \frac{\ln 2}{t_{1/2}} \] For each isotope: \[ \lambda_{238U} = \frac{\ln 2}{4.47 \times 10^9 \, {years}} \approx 1.55 \times 10^{-10} \, {years}^{-1} \] \[ \lambda_{230Th} = \frac{\ln 2}{75,000 \, {years}} \approx 9.24 \times 10^{-6} \, {years}^{-1} \] \[ \lambda_{226Ra} = \frac{\ln 2}{1600 \, {years}} \approx 4.35 \times 10^{-4} \, {years}^{-1} \]
Step 3: Relating the number of atoms of 226Ra to 238U.
Since the activities are equal at equilibrium, the number of atoms of 226Ra (\(N_{226Ra}\)) can be calculated using the ratio of the decay constants: \[ \frac{N_{226Ra}}{N_{238U}} = \frac{\lambda_{238U}}{\lambda_{226Ra}} \] Given that the number of atoms of 238U is 10 billion (i.e., \(N_{238U} = 10 \times 10^9\)), we can substitute the decay constants into the equation: \[ \frac{N_{226Ra}}{10 \times 10^9} = \frac{1.55 \times 10^{-10}}{4.35 \times 10^{-4}} \] \[ N_{226Ra} = 10 \times 10^9 \times \frac{1.55 \times 10^{-10}}{4.35 \times 10^{-4}} \approx 3.5 \times 10^3 \] Therefore, the number of atoms of 226Ra present at equilibrium is approximately 3500 atoms.