Question

Suppose a laboratory has a 31 g sample of polonium-210. The half-life of polonium-210 is about 138 days. How many half lives of polonium occur in 966 days? How much polonium is in the sample 966 days later?

Answer 1

`7` half-lives

`0.24g` `"^(210)Po`

Explanation:

The number of half-lives is simply the number of days given (`966d`) divided by the half-life (`138d`), which is `7` half-lives.

The amount of `"^(210)Po` remaining after `t` days is represented by the equation

`N(t) = N_0(1/2)^(t/(t_(1/2))`

where `N(t)` is the amount of substance remaining after `t` days,
`N_0` is the initial amount of substance,
`t` is the time in days, and
`t_(1/2)` is the half-life of the substance.

Plugging in known variables, we have

`N(t) = (31g)(1/2)^((966)/(138)) = 0.24g` `"_(210)Po`