# Exponential Decay

howsockgothap

## Homework Statement

A radioactive substance has a half-life of 20 years. If 8 mg of the substance remains after
100 years, find how much of the substance was initially present.

A=A0ekt

## The Attempt at a Solution

I set the equation up so that 8=A0e100k and figured I could solve from there taking the ln but that gives me ln8/A0=100k... I still have two variables so I can't solve. How do I fix this? Can I plug in 20 for k?

Gold Member
Just a thought: an easier way to do this might be to recognize that the object has gone through half-life five times, namely that the object has been halved five times. So if you multiply by 2^5, you'll get the answer you're looking for.

howsockgothap
Yes, I do realize that, but my prof will give 0 marks for an answer that doesn't use the equation.

Gold Member
I assume that k here is the decay constant. Just remember that one of the definitions for k is...

$$k=\frac{log(2)}{t_{\frac{1}{2}}}$$

Since you know the half-life, finding the decay constant should be easy.

howsockgothap
Thanks!

Gold Member
No problem. Have a great day!

Homework Helper
What Char Limit said is excellent: 100= 5(20) so the substance must have been halved 5 times. Work backwards- what is the opposite of "halving"?

As for your method, you have more variables than equations because you did not use all of the information- in particular you made no use of the fact that the half-life is 20 years.

If you start with any amount C, after 20 years, you will have C/2 left:
$$Ce^{20k}= C/2$$
so
$$e^{20k}= \frac{1}{2}$$
You could solve that for t and then use
$$Ce^{100k}= 8$$

But using $Ce^{kt}$ at all is the "hard way". All "exponentials", to any base, are interchangeable.
$$Ca^{\alpha t}= Ce^{ln(a^{\alpha t}}= Ce^{(\alpha ln(a))}$$
which is the same as $Ce^{kt}$ with $k= \alpha ln(a)$.

Knowing that the decay is exponential and that the half life is 20 years tells you that you can use
$$C\left(\frac{1}{2}\right)^{t/20}$$
where dividing t by 20 tells you how many "20 year periods" there are in t years.

But, again, the way Char Limit suggested is simplest and best.