# 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?

Dec 10, 2016

$\left(a\right) \text{120 mg"/2^(t/"30 yr"); (b) "15.0 mg"; (c) "199 yr}$

#### 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

${\text{A}}_{0}$ is the initial amount,

$\text{A}$ 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

$\text{A" = "120 mg"/2^(t/"30 yr") = "120 mg"/2^(("90 yr")/("30 yr")) = "120 mg"/2^3 = "120 mg"/8 = "15.0 mg}$

(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")

${2}^{\frac{t}{\text{30 yr}}} = 100$

$\left(\frac{t}{\text{30 yr}}\right) \log 2 = \log 100 = 2$

$\frac{t}{\text{30 yr}} = \frac{2}{\log} 2 = 6.64$

$t = \text{6.64 × 30 yr" = "199 yr}$