Curium has a half life of 1.5 hours, how much of a 1000g sample remains after: 1.5 hrs? 3hrs? 4.5 hrs? 6hrs?

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Answer 1 · 2018-06-19T13:02:08

Let's assume the radioactive decay of curium follows first-order kinetics.

Given,

$t_{\frac{1}{2}} = 1.5 hr$ $t_(1/2) = 1.5"hr"$

${[A]}_{0} = 1000 g$ $[A]_0 = 1000"g"$

Recall,

$t_{\frac{1}{2}} = \frac{ln (2)}{k} (1)$ $t_(1/2) = ln(2)/k" "(1)$

$ln (\frac{{[A]}_{t}}{{[A]}_{0}}) = - k t (2)$ $ln(([A]_"t")/([A]_0)) = -kt" "(2)$

Let's derive the rate constant,

$\Rightarrow k = \frac{ln (2)}{t_{\frac{1}{2}}} \approx 0.462 {hr}^{- 1}$ $=> k = ln(2)/t_(1/2) approx 0.462"hr"^-1$ ,

and rearrange (2),

${[A]}_{t} = {[A]}_{0} e^{- k t}$ $[A]_"t" = [A]_0e^(-kt)$

Hence,

puu.sh

is how your data would work out.

Note: I used a relatively advanced method to solve this, lots of beginning chemistry students use simple arithmetic,

${(\frac{1}{2})}^{n}$ $(1/2)^n$ , where $n = \frac{t}{t_{\frac{1}{2}}}$ $n = t/t_(1/2)$