Question

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?

Answer 1

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

Given,

`t_(1/2) = 1.5"hr"`

`[A]_0 = 1000"g"`

Recall,

`t_(1/2) = ln(2)/k" "(1)`

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

Let's derive the rate constant,

`=> k = ln(2)/t_(1/2) approx 0.462"hr"^-1`,

and rearrange (2),

`[A]_"t" = [A]_0e^(-kt)`

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,

`(1/2)^n`, where `n = t/t_(1/2)`