Question

the half - life of a certain isotope is 70 years.

a. given an initial amount of ( a ) grams of this isotope, at ( t = 0 ), find an exponential decay model,
( a(t)=ae^{kt} ) that gives the amount of the isotope at time ( t,t > 0 ). round your ( k ) value to 3

decimal places.

( a(t)= )

b. use the function you found in part ( a ) to calculate the time required (in years) for the isotope to

decay to ( \frac{1}{5}a ). round to 3 decimal places.

Explanation:

Step1: Find the value of \(k\)

The half - life formula for \(A(t)=Ae^{kt}\) is when \(t = T\) (half - life), \(A(T)=\frac{A}{2}\).
Substitute into the formula: \(\frac{A}{2}=Ae^{kT}\).
Since \(T = 70\), we have \(\frac{1}{2}=e^{70k}\).
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{70k})\).
Using the property \(\ln(e^{x})=x\), we get \(\ln(\frac{1}{2}) = 70k\).
Since \(\ln(\frac{1}{2})=-\ln(2)\approx - 0.693\), then \(k=\frac{-\ln(2)}{70}\approx\frac{- 0.693}{70}\approx - 0.010\).
So the model is \(A(t)=Ae^{-0.010t}\).

Step2: Calculate the time \(t\) when \(A(t)=\frac{1}{5}A\)

Substitute \(A(t)=\frac{1}{5}A\) into \(A(t)=Ae^{-0.010t}\).
We get \(\frac{1}{5}A=Ae^{-0.010t}\).
Cancel out \(A\) (since \(A
eq0\)), so \(\frac{1}{5}=e^{-0.010t}\).
Take the natural logarithm of both sides: \(\ln(\frac{1}{5})=\ln(e^{-0.010t})\).
Using \(\ln(e^{x}) = x\), we have \(\ln(\frac{1}{5})=-0.010t\).
Since \(\ln(\frac{1}{5})=-\ln(5)\approx - 1.609\), then \(t=\frac{-\ln(5)}{-0.010}=\frac{1.609}{0.010}=160.944\).

Answer:

a. \(A(t)=Ae^{-0.010t}\)
b. \(160.944\)