Question

the function ( q(t)=q_{0}e^{-kt} ) may be used to model radioactive decay. ( q ) represents the quantity remaining after ( t ) years; ( k ) is the decay constant. the decay constant for plutonium - 240 is ( k = 0.00011 ). what is the half - life, in years?
0.076 years
3,150 years
6,301 years
1,512,321 years

Explanation:

Step1: Substitute \( Q(t)=\frac{Q_0}{2} \) into the formula

We know that \( Q(t) = Q_0e^{-kt} \). When it is half - life, \( Q(t)=\frac{Q_0}{2} \). So, \(\frac{Q_0}{2}=Q_0e^{-kt}\). Divide both sides by \( Q_0 \) (since \( Q_0
eq0 \)), we get \(\frac{1}{2}=e^{-kt}\).

Step2: Take the natural logarithm of both sides

Take the natural logarithm of \(\frac{1}{2}=e^{-kt}\). Using the property \(\ln(e^{x}) = x\), we have \(\ln(\frac{1}{2})=-kt\). Since \(\ln(\frac{1}{2})=-\ln(2)\), the equation becomes \(-\ln(2)=-kt\).

Step3: Solve for \( t \)

We know \( k = 0.00011 \). Substitute \( k \) into the equation \(t=\frac{\ln(2)}{k}\). Substitute \(\ln(2)\approx0.693\) and \(k = 0.00011\) into \(t=\frac{\ln(2)}{k}\), then \(t=\frac{0.693}{0.00011}\approx6300.91\).

Answer:

6,301 years