# 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?

##### 1 Answer
Jun 19, 2018

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

Given,

${t}_{\frac{1}{2}} = 1.5 \text{hr}$

${\left[A\right]}_{0} = 1000 \text{g}$

Recall,

${t}_{\frac{1}{2}} = \ln \frac{2}{k} \text{ } \left(1\right)$

ln(([A]_"t")/([A]_0)) = -kt" "(2)

Let's derive the rate constant,

$\implies k = \ln \frac{2}{t} _ \left(\frac{1}{2}\right) \approx 0.462 {\text{hr}}^{-} 1$,

and rearrange (2),

${\left[A\right]}_{\text{t}} = {\left[A\right]}_{0} {e}^{- k t}$

Hence,

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,

${\left(\frac{1}{2}\right)}^{n}$, where $n = \frac{t}{t} _ \left(\frac{1}{2}\right)$