Question

The half-life of cesium-137 is 30 years. Suppose we have a 120-mg sample. Find the mass that remains after `t` years. How much of the sample remains after 90 years? After how long will only 1 mg remain?

Answer 1

`(a) "120 mg"/2^(t/"30 yr"); (b) "15.0 mg"; (c) "199 yr"`

Explanation:

(a) Mass after t years

The formula for calculating the amount remaining after a number of half-lives, `n` is

`color(blue)(bar(ul(|color(white)(a/a)"A" = "A"_0/2^ncolor(white)(a/a)|)))" "`

where

`"A"_0` is the initial amount,

`"A"` is the amount remaining and

`n = t/t_½`

This gives

`"A" = A_0/2^( t/t_½)`

∴ `"A" = "120 mg"/2^(t/"30 yr")`

(b) Mass after 90 yr

`"A" = "120 mg"/2^(t/"30 yr") = "120 mg"/2^(("90 yr")/("30 yr")) = "120 mg"/2^3 = "120 mg"/8 = "15.0 mg"`

(c) Time for 1 mg remaining

`"A" = A_0/2^( t/t_½)`

`A_0/"A" = 2^( t/t_½)`

`(100 color(red)(cancel(color(black)("mg"))))/(1 color(red)(cancel(color(black)("mg")))) = 2^(t/"30 yr")`

`2^(t/"30 yr") = 100`

`(t/"30 yr")log2 = log100 =2`

`t/"30 yr" = 2/log2 =6.64`

`t = "6.64 × 30 yr" = "199 yr"`