The decay constant of a radioactive nuclide is $3.1 \times 10 ^ { - 3 } \mathrm {~s} ^ { - 1 }$. At a given instant, the activity of a specimen of the nuclide is $70 \mathrm { kBq }$. How much later has the activity declined to $10 \mathrm { kBq }$ ?

Question

Quizplus · Accepted Answer

The activity of a radioactive sample decreases exponentially according to the formula $A = A_0 e^{-\lambda t}$, where $A$ is the final activity, $A_0$ is the initial activity, $\lambda$ is the decay constant, and $t$ is the time. Rearranging to solve for $t$, we get $t = \frac{\ln(A_0/A)}{\lambda}$. Substituting $A_0 = 70 \, \text{kBq}$, $A = 10 \, \text{kBq}$, and $\lambda = 3.1 \times 10^{-3} \, \text{s}^{-1}$, we find $t = \frac{\ln(70/10)}{3.1 \times 10^{-3}} \approx 630 \, \text{s}$.

The Decay Constant of a Radioactive Nuclide Is 3.1×10−3 s−13.1 \times 10 ^ { - 3 } \mathrm {~s} ^ { - 1 }3.1×10−3 s−1

The Decay Constant of a Radioactive Nuclide Is $3.1 \times 10 ^ { - 3 } \mathrm {~s} ^ { - 1 }$