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?

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Answer 1 · 2016-11-13T01:34:57

It will take 210 years.

The formula for radioactive decay is

$color(blue)(bar(ul(|color(white)(a/a) N/N_0 = (1/2)^n color(white)(a/a)|)))" "$

where

$N_0 = "original amount of isotope"$
$N = "amount of isotope remaining"$
$n = "number of half-lives"$

and

$n = t/t_½$

where

$t = "the time for the decay"$
$t_½ = "the half-life"$

In your problem,

$N_0 = "10 g"$
$N = "1.25 g"$
$t_½ = "70 years"$

$N/N_0 = (1/2)^n$

$(1.25 color(red)(cancel(color(black)("g"))))/(10 color(red)(cancel(color(black)("g")))) = (1/2)^n$

$1/8 = 1/2^n$

$n = 3$

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

$n = t/t_½$

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