A radioactive material with an original mass of $10 \mathrm{~g}$ has a mass of $8 \mathrm{~g}$ after 50 years. The material decays at a rate proportional to the amount present. Find an expression for the amount present at any time t. Also find its half-life.
A) $Q=10(0.8)^{\frac{t}{50}}$ or $Q=10 e^{-0.004463 t}$; its half- life is about 155 years.
B) $\mathrm{Q}=10(0.8)^{\frac{\mathrm{t}}{500}}$ or $\mathrm{Q}=10 \mathrm{e}^{-0.0004463 \mathrm{t}}$; its half- life is about 1553 years.
C) $\mathrm{Q}=10(8)^{\frac{\mathrm{t}}{500}}$ or $\mathrm{Q}=10 \mathrm{e}^{0.004159 t}$; its half- life is about 167 years.
D) $\mathrm{Q}=8(0.1)^{\frac{\mathrm{t}}{50}}$ or $\mathrm{Q}=8 \mathrm{e}^{-0.04605 \mathrm{t}}$; its half- life is about 15 years.

Question

Quizplus · Accepted Answer

The rate of decay is proportional to the amount present, so we have the differential equation $$\frac{dQ}{dt} = kQ$$where $k$ is a constant. Solving this equation, we get $$Q(t) = Q_0 e^{kt}$$where $Q_0$ is the initial amount. Using the given information, we have $$8 = 10 e^{50k} \Rightarrow k = \frac{1}{50} \ln \frac{8}{10} = -0.004463$$Therefore, the amount present at any time $t$ is $$Q(t) = 10 e^{-0.004463 t}$$The half-life is the time it takes for the amount to decay to half of its original value. Therefore, we have $$5 = 10 e^{-0.004463 t_{1/2}} \Rightarrow t_{1/2} = \frac{1}{-0.004463} \ln \frac{5}{10} = 155 \text{ years}$$

A Radioactive Material with an Original Mass Of 10 g10 \mathrm{~g}10 g Has a Mass Of

A Radioactive Material with an Original Mass Of $10 \mathrm{~g}$ Has a Mass Of