1. ## DE:simple chemical conversion

a certain radioactive material has a half-life of 38 hours.find how long it takes for 90% of the radioactivity to be dissipated?

2. Hello, alderon!

A certain radioactive material has a half-life of 38 hours.
Find how long it takes for 90% of the radioactivity to be dissipated?
The "decay" formula is: .$\displaystyle A \;=\;A_oe^{-kt}$

. . where $\displaystyle A_o$ = initial amount, $\displaystyle A$ = final amount,
. . $\displaystyle t$ = time, and $\displaystyle k$ is a constant to be determined.

We are told that when $\displaystyle t = 38,\;A \,= \,\frac{1}{2}A_o$

So we have: .$\displaystyle \frac{1}{2}A_o \;=\;A_oe^{-38k}\quad\Rightarrow\quad \frac{1}{2}\;=\;e^{-38k}$

Multiply by $\displaystyle 2e^{38k}\!:\;\;e^{38t} \:=\:2\quad\Rightarrow\quad 38k \:=\:\ln(2)\quad\Rightarrow\quad k \:=\:\frac{\ln(2)}{38} \:\approx\: 0.01824$

Hence, we have: .$\displaystyle A \;=\;A_oe^{-0.01824t}$

The problem asks: .When is $\displaystyle A \,=\,\frac{1}{10}A_o$ ?

We have: .$\displaystyle \frac{1}{10}A_o \;=\;A_oe^{-0.01824t}\quad\Rightarrow\quad \frac{1}{10} \;=\;e^{-0.01824t}$

Multiply by $\displaystyle 10e^{0.01824}\!:\;\;e^{0.01824t} \:=\:10\quad\Rightarrow\quad 0.01824t \:=\:\ln10\quad\Rightarrow\quad t \:=\:\frac{\ln10}{0.01824}$

Therefore: .$\displaystyle t \:=\:126.2382178 \:\approx\:126.24\text{ hours}$

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# simple chemical conversion in differential equation

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