The decay constant of a radioactive isotope is 1.60 × 10^-3 s^-1.At a given instant,the number of atoms of the radioactive isotope is 1.85 × 10¹².How many atoms of the isotope remain after a time interval of 30.0 minutes?

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Quizplus · Accepted Answer

The number of atoms remaining after a time interval can be calculated using the formula $N = N_0 e^{-\lambda t}$, where $N_0$ is the initial number of atoms, $\lambda$ is the decay constant, and $t$ is the time interval. Given $\lambda = 1.60 \times 10^{-3} \, \text{s}^{-1}$ and $N_0 = 1.85 \times 10^{12}$, and converting 30.0 minutes to seconds ($30 \times 60 = 1800 \, \text{s}$), we can calculate $N$. Plugging these values into the formula gives the result closest to $1.04 \times 10^{11}$, which corresponds to choice A.

The Decay Constant of a Radioactive Isotope Is 1