A sample contains $1.000 \times 10^{-3} \mathrm{~g}$ of $40 \mathrm{~K}$. The half-life of $40 \mathrm{~K}$ is $1.250 \times 10^{9}$ years. What is the activity of the sample?

Question

Quizplus · Accepted Answer

The activity of a radioactive sample is given by the formula $A = \lambda N$, where $A$ is the activity in decays per second (Bq), $\lambda$ is the decay constant, and $N$ is the number of radioactive nuclei. The decay constant $\lambda$ can be calculated from the half-life $T_{1/2}$ using the formula $\lambda = \ln(2)/T_{1/2}$. Given the half-life of $40\mathrm{K}$ is $1.250 \times 10^9$ years, we first convert the half-life into seconds ($T_{1/2} = 1.250 \times 10^9 \times 365.25 \times 24 \times 3600$) and then calculate $\lambda$. To find $N$, we use the mass of the sample and the molar mass of $40\mathrm{K}$ ($39.964 \, \mathrm{g/mol}$) to calculate the number of moles and then multiply by Avogadro's number ($6.022 \times 10^{23} \, \mathrm{mol}^{-1}$). Plugging these values into the formula for $A$, we find that the activity is closest to $265.0 \, \mathrm{Bq}$, making option A correct.

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A Sample Contains $1.000 \times 10^{-3} \mathrm{~g}$ Of $40 \mathrm{~K}$ the Half-Life Of \times