Question

The half-life of an element is `5.8 x 10^11`. How long does it take a sample of the element to decay to `2/5` its original mass?

Answer 1

The expression for the first-order decay of a population is
`A/A_0=e^(-kt)`
where the rate constant `k` is related to the half-life by
`k=ln2/t_(1/2)`

Explanation:

In the question, the half-life should have units of time. Let's assume that the half-life is `5.8 times 10^11 s`

In this case, the value of the rate constant is
`k=ln2/t_(1/2)=ln2/(5.8times10^11 s)=1.20times 10^-12 s^-1`

Using the first equation, we can find the time, `t` at which the fraction of remaining atoms is `2/5`.

`2/5 = e^(-(1.20times10^-12 s^-1)(t))`

Solve for `t` by first taking the natural logarithm of both sides:

`-0.92=-(1.2 times 10^-12 s^-1) t`

`t=0.92/(1.2times10^-12 s^-1)=7.67times10^11 s`

(or about 24,300 years)