A sample of a radioactive substance decayed to 94.5% of its original amount after a year. a) What is the half-life of the substance? b) How long would...

Denote
the relative amount of the substance remaining after `t` years as `R( t ).` We know that after
`H` years (the half-life period) `1/2` will remain: `R(H) = 1/2.` Then after `H` more years the
remaining amount will be `1/2` of `1/2,` i.e. `1/4,` then `1/8` and so on.

It
is a logarithmic decay function with the base 2, `R(t) = 1/2^(t/H).`

a) First, we want to find the half-life H given that `R(1)
= 94.5%:`

`1/2^(1/H) = 94.5% = 0.945, 2^(1/H) = 1/0.945.`

Now take natural logarithm `ln` on both parts: `(1/H)ln(2) = ln(1/0.945),` `H =
ln(1/0.945) / ln(2) approx 12.25 (years).`

We used `ln` because it is present
on most calculators.

b) Now we know `R(t) approx
1/2^(t/12.25)` and our goal is to find for what t it will be 15%. It is the same as solve the
equation `1/2^(t/12.25) = 0.15` for t. It is not hard:

`2^(t/12.25) =
1/0.15,`

and now repeat the same steps we performed at a), apply ln and get a
linear equation. I got about 33.5 years.