Question

cesium-137 has a half - life of about 30 years. given this half - life, we can represent its decay with the exponential decay function

$a = a_0e^{(\frac{ln(0.5)}{30})t}$

if we begin with 200 mg of cesium - 137, how long will it take for the cesium to decay to the point where there is only 1 milligram remaining? round to the closest year.

provide your answer below:

Explanation:

Step1: Substitute values into formula

We know $A = 1$, $A_0=200$ and the formula $A = A_0e^{(\frac{\ln(0.5)}{30})t}$. So, $1 = 200e^{(\frac{\ln(0.5)}{30})t}$.

Step2: Isolate the exponential term

Divide both sides by 200: $\frac{1}{200}=e^{(\frac{\ln(0.5)}{30})t}$.

Step3: Take natural - logarithm of both sides

$\ln(\frac{1}{200})=\ln(e^{(\frac{\ln(0.5)}{30})t})$. Since $\ln(e^x)=x$, we have $\ln(\frac{1}{200}) = (\frac{\ln(0.5)}{30})t$.

Step4: Solve for t

First, $\ln(\frac{1}{200})=-\ln(200)\approx - 5.2983$. And $\ln(0.5)\approx - 0.6931$. Then $t=\frac{- \ln(200)}{\frac{\ln(0.5)}{30}}=\frac{-5.2983}{\frac{- 0.6931}{30}}$.
$t=\frac{-5.2983\times30}{- 0.6931}=\frac{-158.949}{-0.6931}\approx229$.

Answer:

229