Step 1: Determine the fraction of the original atoms remaining. The problem states that the radioactive element decayed to $\frac{1}{8}$ of its original atoms. This means the fraction remaining ($N/N_0$) is $\frac{1}{8}$. Step 2: Calculate the number of half-lives ($n$). The formula for radioactive decay is: $$ \frac{N}{N_0} = \left(\frac{1}{2}\right)^n $$ Substitute the given fraction: $$ \frac{1}{8} = \left(\frac{1}{2}\right)^n $$ Since $2^3 = 8$, we can write $\frac{1}{8}$ as $\left(\frac{1}{2}\right)^3$. $$ \left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^n $$ Therefore, the number of half-lives ($n$) is 3. Step 3: Calculate the half-life ($T_{1/2}$). The total time elapsed ($t$) is given as 6 minutes. The relationship between total time, number of half-lives, and half-life period is: $$ t = n \times T_{1/2} $$ Substitute the known values: $$ 6 \text{ minutes} = 3 \times T_{1/2} $$ Now, solve for $T_{1/2}$: $$ T_{1/2} = \frac{6 \text{ minutes}}{3} $$ $$ T_{1/2} = 2 \text{ minutes} $$ The final answer is $\boxed{\text{2 minutes}}$. 3 done, 2 left today. You're making progress.