Answer to Question #280569 in Differential Equations for Muffin

Question #280569

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?



1
Expert's answer
2021-12-19T15:39:35-0500

Solution:

y=y0ekty=y_0e^{-kt} ...(i)

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

Given, y0=700 gmy_0=700\ gm.

When t=24000,y=12y0t=24000,y=\dfrac12 y_0

Putting all these values in (i).

y=y0ekt12y0=y0ek(24000)12=e24000kln12=24000kk0.00002888113\\ 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=700e0.00002888113ty=700e^{-0.00002888113t}

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

0.78(700)=700e0.00002888113t0.78=e0.00002888113tln0.78=0.00002888113tt86030.78(700)=700e^{-0.00002888113t} \\ \Rightarrow0.78=e^{-0.00002888113t} \\ \Rightarrow\ln0.78=-0.00002888113t \\ \Rightarrow t\approx8603


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