Question

suppose ( w(t) ) denotes the amount of a radioactive material left after time ( t ). assume that ( w(0)=9 ) and ( w(1)=3 ). complete parts (a) through (c).
(a) find the differential equation that describes this situation.
the differential equation is ( \frac{dw}{dt}=(-ln 3)w(t) )
(b) how much material is left at time ( t = 5 )?
the amount of material left after ( t = 5 ) is ( w(5)=\frac{1}{27} ).
(type an exact answer. use integers or fractions for any numbers in the expression.)
(c) what is the half - life of the material?
half - life ( =square )
(type an exact answer. use integers or fractions for any numbers in the expression.)

Explanation:

Step1: Recall the general form of radioactive decay

The general solution of the differential equation \(\frac{dW}{dt}=kW\) is \(W(t) = W(0)e^{kt}\). Given \(W(0) = 9\) and \(W(1)=3\), we substitute into \(W(t)=W(0)e^{kt}\). So \(3 = 9e^{k\times1}\).

Step2: Solve for \(k\) in the equation \(3 = 9e^{k}\)

Divide both sides of \(3 = 9e^{k}\) by \(9\), we get \(\frac{3}{9}=e^{k}\), i.e., \(\frac{1}{3}=e^{k}\). Taking the natural - logarithm of both sides, \(\ln(\frac{1}{3})=k\), and since \(\ln(\frac{1}{3})=-\ln(3)\), the differential equation is \(\frac{dW}{dt}=-\ln(3)W(t)\) and \(W(t)=9e^{-\ln(3)t}\).

Step3: Use the formula for half - life

The formula for half - life \(T\) is \(W(T)=\frac{W(0)}{2}\). Substitute \(W(t)=W(0)e^{kt}\) into \(W(T)=\frac{W(0)}{2}\), we have \(W(0)e^{kT}=\frac{W(0)}{2}\). Since \(W(0)
eq0\), we can cancel \(W(0)\) on both sides, getting \(e^{kT}=\frac{1}{2}\).
Take the natural logarithm of both sides: \(kT=\ln(\frac{1}{2})\). We know \(k =-\ln(3)\), so \(T=\frac{\ln(\frac{1}{2})}{k}\). Substitute \(k =-\ln(3)\) into the formula, \(T=\frac{-\ln(2)}{-\ln(3)}=\frac{\ln(2)}{\ln(3)}\).

Answer:

\(\frac{\ln(2)}{\ln(3)}\)