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

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Answer 1 · 2016-12-10T23:46:20

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

(a) Mass after t years

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

$"A" = A_0/2^( t/t_½)$

$A_0/"A" = 2^( t/t_½)$

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

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

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

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

$t = "6.64 × 30 yr" = "199 yr"$

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