Question

question 9 of 10
plutonium - 240 decays according to the function $q(t)=q_0e^{-kt}$ where q represents the quantity remaining after t years and k is the decay constant, 0.00011... to the nearest 10 years, how long will it take 24 grams of plutonium - 240 to decay to 20 grams?
a. 1,660 years
b. 1.06 years
c. 80.11 years
d..00048 years

Explanation:

Step1: Substitute values into decay formula

We know that $Q(t) = 20$, $Q_0=24$ and $k = 0.00011$. Substitute into $Q(t)=Q_0e^{-kt}$:
$20 = 24e^{-0.00011t}$

Step2: Solve for $e^{-0.00011t}$

Divide both sides by 24: $\frac{20}{24}=e^{-0.00011t}$, so $e^{-0.00011t}=\frac{5}{6}$

Step3: Take natural - logarithm of both sides

$\ln(e^{-0.00011t})=\ln(\frac{5}{6})$. Since $\ln(e^x)=x$, we have $- 0.00011t=\ln(\frac{5}{6})$

Step4: Solve for $t$

$t=\frac{\ln(\frac{5}{6})}{-0.00011}$. We know that $\ln(\frac{5}{6})=\ln(5)-\ln(6)\approx1.6094 - 1.7918=- 0.1824$. Then $t=\frac{-0.1824}{-0.00011}\approx1658.18\approx1660$ years

Answer:

A. 1,660 years