# If the half-life of uranium-232 is 70 years, how many half-lives will it take for 10 g of it to be reduced to 1.25 g?

##### 1 Answer
Nov 13, 2016

It will take 210 years.

#### Explanation:

The formula for radioactive decay is

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \frac{N}{N} _ 0 = {\left(\frac{1}{2}\right)}^{n} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

${N}_{0} = \text{original amount of isotope}$
$N = \text{amount of isotope remaining}$
$n = \text{number of half-lives}$

and

n = t/t_½

where

$t = \text{the time for the decay}$
t_½ = "the half-life"

In your problem,

${N}_{0} = \text{10 g}$
$N = \text{1.25 g}$
t_½ = "70 years"

$\frac{N}{N} _ 0 = {\left(\frac{1}{2}\right)}^{n}$

$\left(1.25 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))))/(10 color(red)(cancel(color(black)("g}}}}\right) = {\left(\frac{1}{2}\right)}^{n}$

$\frac{1}{8} = \frac{1}{2} ^ n$

$n = 3$

So, the uranium has decayed for 3 half-lives.

n = t/t_½

t = nt_½ = "3 × 70 years" = "210 years"