Question

100 points possible answered: 1/20 question 2 an initial sample of 658 grams was found in a radioactive substance. the equation, $a = 658e^{-0.0226t}$, gives the amount left in the substance after $t$ years. a. how many grams of the substance will remain after 43 years? (round your answer to one decimal place.) grams b. what is the half-life of the substance, that is, how long will it take for the substance to decrease to half its initial amount? (round your answer to one decimal place.) years c. in what year will the sample decay to 7 grams? (round your answer to one decimal place) next question

Explanation:

Part a:

Step1: Identify the formula and values

The formula for the amount of radioactive substance remaining is \( A = 658e^{-0.0226t} \), where \( t = 43 \) years.

Step2: Substitute \( t = 43 \) into the formula

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Step3: Calculate the exponent and then the value

First, calculate \( e^{-0.9718}\approx0.386 \) (using a calculator for the exponential function). Then, \( A = 658\times0.386\approx254.0 \) (rounded to one decimal place).

Part b:

Step1: Recall the half - life condition

Half - life is the time \( t \) when \( A=\frac{658}{2}=329 \) grams. So we set up the equation \( 329 = 658e^{-0.0226t} \).

Step2: Solve for \( t \)

Divide both sides by 658: \( \frac{329}{658}=e^{-0.0226t} \), which simplifies to \( 0.5 = e^{-0.0226t} \).
Take the natural logarithm of both sides: \( \ln(0.5)=\ln(e^{-0.0226t}) \).
Since \( \ln(e^{x}) = x \), we have \( \ln(0.5)=- 0.0226t \).
Then, \( t=\frac{\ln(0.5)}{-0.0226} \).

Step3: Calculate the value of \( t \)

We know that \( \ln(0.5)\approx - 0.6931 \), so \( t=\frac{- 0.6931}{-0.0226}\approx30.7 \) years (rounded to one decimal place).

Part c:

Step1: Set up the equation for \( A = 7 \)

We set \( A = 7 \) in the formula \( A = 658e^{-0.0226t} \), so \( 7 = 658e^{-0.0226t} \).

Step2: Solve for \( t \)

Divide both sides by 658: \( \frac{7}{658}=e^{-0.0226t} \), or \( 0.01064 = e^{-0.0226t} \).
Take the natural logarithm of both sides: \( \ln(0.01064)=\ln(e^{-0.0226t}) \).
Since \( \ln(e^{x})=x \), we get \( \ln(0.01064)=-0.0226t \).

Step3: Calculate \( t \)

We know that \( \ln(0.01064)\approx - 4.547 \), so \( t=\frac{-4.547}{-0.0226}\approx201.2 \) years (rounded to one decimal place).

Answer:

a. \(\boxed{254.0}\) grams
b. \(\boxed{30.7}\) years
c. \(\boxed{201.2}\) years