Answer to Question #280569 in Differential Equations for Muffin

Question

A radioactive substance plutonium-239 has a half-life of 24,000 years. Initially, there is 700 mg of the substance. How long for the substance to decay 78% of its initial mass?

Answer 1

Solution:

$y=y_0e^{-kt}$ ...(i)

Here, y is the amount of element at time t, $y_o$ is the initial amount, k is decay constant.

Given, $y_0=700\ gm$.

When $t=24000,y=\dfrac12 y_0$

Putting all these values in (i).

$\\ y=y_0e^{-kt} \\ \Rightarrow\dfrac12 y_0=y_0e^{-k(24000)} \\ \Rightarrow \dfrac12=e^{-24000k} \\ \Rightarrow \ln\dfrac12=-24000k \\ \Rightarrow k\approx0.00002888113$

So, $y=700e^{-0.00002888113t}$

Now, $y=0.78(700),t=?$

$0.78(700)=700e^{-0.00002888113t} \\ \Rightarrow0.78=e^{-0.00002888113t} \\ \Rightarrow\ln0.78=-0.00002888113t \\ \Rightarrow t\approx8603$