question 3 of 10\nthe function $q(t)=q_0e^{-kt}$ may be used to model radioactive decay. $q$ represents the…

question 3 of 10\nthe function $q(t)=q_0e^{-kt}$ 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?\na. 3,150 years\nb. 6,301 years\nc. 0.076 years\nd. 1,512,321 years

question 3 of 10\nthe function $q(t)=q_0e^{-kt}$ 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?\na. 3,150 years\nb. 6,301 years\nc. 0.076 years\nd. 1,512,321 years

Answer

Explanation:

Step1: Set up the half - life equation

At half - life, $Q(t)=\frac{Q_0}{2}$. Substitute into $Q(t)=Q_0e^{-kt}$: $\frac{Q_0}{2}=Q_0e^{-kt}$ Since $Q_0\neq0$, we can divide both sides by $Q_0$ to get $\frac{1}{2}=e^{-kt}$.

Step2: Take the natural logarithm of both sides

$\ln(\frac{1}{2})=\ln(e^{-kt})$. 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 $k = 0.00011$. So $t=\frac{\ln(2)}{k}$. Substitute $k = 0.00011$ into the formula: $t=\frac{\ln(2)}{0.00011}\approx\frac{0.693147}{0.00011}\approx6301$.

Answer:

B. 6,301 years