Question

lead - 202 has a half - life of 53,000 years. how long will it take for 15/16 of a sample of lead - 202 to decay?
106,000 years
159,000 years
212,000 years
265,000 years

Explanation:

Step1: Determine remaining fraction

If 15/16 of the sample decays, the remaining fraction of the sample is $1-\frac{15}{16}=\frac{1}{16}$.

Step2: Relate remaining fraction to half - life formula

The formula for radioactive decay in terms of half - life $t_{1/2}$ is $N = N_0(\frac{1}{2})^n$, where $N$ is the final amount, $N_0$ is the initial amount, and $n$ is the number of half - lives. We know $\frac{N}{N_0}=\frac{1}{16}$, and $\frac{1}{16}=(\frac{1}{2})^4$, so $n = 4$.

Step3: Calculate total time

Given $t_{1/2}=53000$ years and $n = 4$, the total time $t=nt_{1/2}$. So $t = 4\times53000=212000$ years.

Answer:

212,000 years