Question

the function $q(t) = q_oe^{-kt}$ 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?

a. 1,512,321 years
b. 6,301 years
c. 3,150 years
d. 0.076 years

Explanation:

Step1: Recall half - life definition

For radioactive decay, at half - life, the quantity remaining \(Q(t)=\frac{Q_0}{2}\). We start with the formula \(Q(t) = Q_0e^{-kt}\). Substitute \(Q(t)=\frac{Q_0}{2}\) into the formula: \(\frac{Q_0}{2}=Q_0e^{-kt}\).

Step2: Simplify the equation

Divide both sides of the equation \(\frac{Q_0}{2}=Q_0e^{-kt}\) by \(Q_0\) (assuming \(Q_0
eq0\)). We get \(\frac{1}{2}=e^{-kt}\).

Step3: Take the natural logarithm of both sides

Take the natural log of both sides: \(\ln(\frac{1}{2})=\ln(e^{-kt})\). Since \(\ln(e^x) = x\), the right - hand side simplifies to \(-kt\). And \(\ln(\frac{1}{2})=-\ln(2)\), so we have \(-\ln(2)=-kt\).

Step4: Solve for t

Divide both sides by \(-k\) to solve for \(t\). We get \(t=\frac{\ln(2)}{k}\).

Step5: Substitute the value of k

We know that \(k = 0.00011\) and \(\ln(2)\approx0.6931\). Then \(t=\frac{0.6931}{0.00011}\approx6301\) years.

Answer:

B. 6,301 years