Question

A research assistant made 160 mg of radioactive sodium (Na^24) and found that there was only 20 mg left 45 h later, how much of the original 20 mg would be left in 12 h?

Answer 1

`=11.49` mg will be left

Explanation:

Let rate of decay be `x` per hour
So we can write
`160(x)^45=20`
or
`x^45=20/160`
or
`x^45=1/8`
or
`x=root45(1/8)`
or
`x=0.955`
Similarly after `12` hours
`20(0.955)^12`
`=20(0.57)`
`=11.49` mg will be left

Answer 2

Just to use the conventional radioactive decay model as a slight alternative method.

After 12hr we have 11.49mg

Explanation:

Let `Q(t)` denote the amount of sodium present at time `t`. At `t=0, Q = Q_0`

It's a fairly simple model to solve with ODEs but as it's not really related to the question, we end up with

`Q(t) = Q_0e^(-kt)` where `k` is a rate constant.

First we find the value of `k`

`Q_0 = 160mg, Q(45) = 20mg`

`Q(45) = 20 = 160e^(-45k)`

`therefore 1/8 = e^(-45k)`

Take natural logs of both sides:

`ln(1/8) = -ln(8) = -45k`

`k=(ln(8))/45 hr^(-1)`

`therefore Q(t) = Q_0e^(-(ln(8))/45t)`

So starting with `Q_0 = 20mg`

`Q(12) = 20e^(-(ln(8))/45*12) = 11.49mg`