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)= - 0.009
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 \(k\)

For exponential decay model \(A(t)=Ae^{kt}\), when \(t = 70\) (half - life), \(A(70)=\frac{A}{2}\).
Substitute into the formula: \(\frac{A}{2}=Ae^{70k}\).
Divide both sides by \(A\) (since \(A
eq0\)): \(\frac{1}{2}=e^{70k}\).
Take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{70k})\).
Using the property \(\ln(e^{x}) = x\), we get \(-\ln(2)=70k\).
Solve for \(k\): \(k=\frac{-\ln(2)}{70}\approx\frac{- 0.693147}{70}\approx - 0.010\).
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 \(A(t)=Ae^{-0.010t}\).
Divide both sides by \(A\) (since \(A
eq0\)): \(\frac{1}{5}=e^{-0.010t}\).
Take the natural logarithm of both sides: \(\ln(\frac{1}{5})=\ln(e^{-0.010t})\).
Using the property \(\ln(e^{x}) = x\), we get \(-\ln(5)=- 0.010t\).
Solve for \(t\): \(t=\frac{\ln(5)}{0.010}\approx\frac{1.6094}{0.010}=160.940\).

Answer:

a. \(A(t)=Ae^{-0.010t}\)
b. \(160.940\)