Question

a certain radioactive isotope has leaked into a small stream. four hundred days after the leak, 5% of the original amount of the substance remained. determine the half - life of this radioactive isotope. the half - life is about \\(\square\\) days (do not round until the final answer. then round to the nearest whole number as needed.)

Explanation:

Step1: Recall radioactive decay formula

Radioactive decay: $A(t) = A_0 e^{-kt}$, where $A(t)$=remaining amount, $A_0$=initial amount, $t$=time, $k$=decay constant.

Step2: Use given data to find k

At $t=400$, $A(400)=0.05A_0$. So $0.05A_0 = A_0 e^{-400k}$. Cancel $A_0$: $0.05 = e^{-400k}$. Take ln: $\ln(0.05) = -400k$. Calculate $\ln(0.05) \approx -2.9957$, so $k = \frac{2.9957}{400} \approx 0.007489$.

Step3: Find half-life ($t_{1/2}$)

Half-life: $A(t_{1/2}) = 0.5A_0$. So $0.5 = e^{-kt_{1/2}}$. Take ln: $\ln(0.5) = -kt_{1/2}$. $t_{1/2} = \frac{-\ln(0.5)}{k} \approx \frac{0.6931}{0.007489} \approx 92.55$.

Step4: Round to nearest whole number

92.55 rounds to 93.

Answer:

93