# 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?

Aug 6, 2016

$= 11.49$ mg will be left

#### Explanation:

Let rate of decay be $x$ per hour
So we can write
$160 {\left(x\right)}^{45} = 20$
or
${x}^{45} = \frac{20}{160}$
or
${x}^{45} = \frac{1}{8}$
or
$x = \sqrt[45]{\frac{1}{8}}$
or
$x = 0.955$
Similarly after $12$ hours
$20 {\left(0.955\right)}^{12}$
$= 20 \left(0.57\right)$
$= 11.49$ mg will be left

Aug 6, 2016

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

After 12hr we have 11.49mg

#### Explanation:

Let $Q \left(t\right)$ 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

$Q \left(t\right) = {Q}_{0} {e}^{- k t}$ where $k$ is a rate constant.

First we find the value of $k$

${Q}_{0} = 160 m g , Q \left(45\right) = 20 m g$

$Q \left(45\right) = 20 = 160 {e}^{- 45 k}$

$\therefore \frac{1}{8} = {e}^{- 45 k}$

Take natural logs of both sides:

$\ln \left(\frac{1}{8}\right) = - \ln \left(8\right) = - 45 k$

$k = \frac{\ln \left(8\right)}{45} h {r}^{- 1}$

$\therefore Q \left(t\right) = {Q}_{0} {e}^{- \frac{\ln \left(8\right)}{45} t}$

So starting with ${Q}_{0} = 20 m g$

$Q \left(12\right) = 20 {e}^{- \frac{\ln \left(8\right)}{45} \cdot 12} = 11.49 m g$