Question

a scientist begins with 250 grams of a radioactive substance. after 250 minutes, the sample has decayed to 14 grams. write an exponential equation ( f(t) ) representing this situation. (let ( f ) be the amount of radioactive substance in grams and ( t ) be the time in minutes.)

( f(t)=250left(\frac{1}{2}
ight)^{left(\frac{t}{125}
ight)} )

to the nearest minute, what is the half - life of this substance?

min

use the model for continuous exponential decay, ( y = a_0e^{-kt} ), where ( y ) is the amount of radioactive substance left after ( t ) minutes. how can the coefficient ( k ) be determined from the amount left after 250 minutes? how can ( a_0 ) be determined from the given information? what is the relationship between the half - life and the constant ( k )?

Explanation:

Step1: Recall the formula for exponential decay

The general formula for exponential decay is \(f(t)=A_0e^{-kt}\), where \(A_0\) is the initial amount, \(k\) is the decay constant, and \(t\) is time. Given \(A_0 = 250\) (initial amount of the radioactive substance), and when \(t = 250\), \(f(250)=14\).

Step2: Substitute the values into the formula

Substitute \(A_0 = 250\), \(t = 250\), and \(f(t)=14\) into \(f(t)=A_0e^{-kt}\). We get \(14 = 250e^{-250k}\).

Step3: Solve for \(k\)

First, divide both sides by 250: \(\frac{14}{250}=e^{-250k}\). Then, take the natural logarithm of both sides: \(\ln(\frac{14}{250})=- 250k\). So, \(k=-\frac{\ln(\frac{14}{250})}{250}\).

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Step4: Recall the formula for half - life

The formula for half - life \(T\) is \(T=\frac{\ln(2)}{k}\).

Step5: Calculate the half - life

Substitute \(k = 0.01152\) into \(T=\frac{\ln(2)}{k}\).

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Answer:

The half - life of the substance is approximately \(60\) minutes.