The half-life of a radioactive material is 4.5 days.How many days are required for a sample,with an initial activity of 1.0 × 10⁵ Bq,to decay to an activity of 100 Bq?

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The half-life formula is used to calculate the time required for a quantity to reduce to half its initial value. The formula is $t = \frac{\ln(\frac{N_0}{N})}{\lambda}$, where $t$ is the time, $N_0$ is the initial quantity, $N$ is the final quantity, and $\lambda$ is the decay constant. The decay constant can be found using the half-life formula $\lambda = \frac{\ln(2)}{T_{1/2}}$, where $T_{1/2}$ is the half-life. For a half-life of 4.5 days, to decay from $1.0 \times 10^{15}$ Bq to 100 Bq, we can use the concept of half-lives directly without delving into the formula's specifics. Every 4.5 days, the activity halves. To go from $1.0 \times 10^{15}$ Bq to 100 Bq, we halve the amount repeatedly until we reach 100 Bq. This process takes 10 half-lives (since $2^{10} = 1024$, and $1.0 \times 10^{15} / 1024 \approx 100$). Therefore, $10 \times 4.5 = 45$ days are required.

The Half-Life of a Radioactive Material Is 4