# The half-life of sodium-24 is 14.9h. A medical facility buys a 50mg sample. How do you determine the exponential function that models the amount, A(t), of sodium-24 that will be present in the sample after t hours?

Mar 1, 2018

$\textcolor{b l u e}{A \left(t\right) = \left(50\right) {2}^{- \frac{t}{14.9}}}$

#### Explanation:

We need to find an equation of the form:

$A \left(t\right) = a {e}^{k t}$

Where:

$A \left(t\right) =$ the amount after time $t$ ( in milligrams )

$a =$ the initial amount.

$k =$ the growth/decay factor.

$t =$ time ( in this case hours )

From the given information. If the half life is $14.9 \textcolor{w h i t e}{8}$hrs, then we would expect the mass of sodium to have halved after this period.

Plugging this into our equation:

$25 = 50 {e}^{14.9 k}$

Dividing both sides by $50$

$\frac{25}{50} = \frac{50}{50} {e}^{14.9 k}$

$\frac{1}{2} = {e}^{14.9 k}$

Taking natural logs of both sides:

$\ln \left(\frac{1}{2}\right) = 14.9 k \ln \left(e\right)$

$\ln \left(e\right) = 1 \textcolor{w h i t e}{888}$ ( the logarithm of the base is always 1 )

$\ln \left(\frac{1}{2}\right) = 14.9 k$

By the laws of logarithms:

$\ln \left(\frac{a}{b}\right) = \ln \left(a\right) - \ln \left(b\right)$

So:

$\ln \left(\frac{1}{2}\right) = \ln \left(1\right) - \ln \left(2\right) = 0 - \ln \left(2\right)$

$- \ln \left(2\right) = 14.9 k$

$k = - \ln \frac{2}{14.9}$

$A \left(t\right) = 50 {e}^{\left(- \ln \frac{2}{14.9}\right) t}$

Exploiting the laws of indices:

$A \left(t\right) = 50 {e}^{\left(\left(\ln \left(2\right)\right) t\right) \left(- \frac{1}{14.9}\right)}$

$A \left(t\right) = \left(50\right) {2}^{- \frac{1}{14.9} t}$

$\textcolor{b l u e}{A \left(t\right) = \left(50\right) {2}^{- \frac{t}{14.9}}}$

Check:

$25 \text{mg}$ after 14.9 hours.

$A \left(t\right) = \left(50\right) {2}^{- \frac{14.9}{14.9}}$

$A \left(t\right) = \left(50\right) {2}^{-} 1 = 25 \text{mg} \textcolor{w h i t e}{88}$ as expected.

Note:

If you leave this with $\boldsymbol{e}$ as a base, the results will not be exact. This is because $\boldsymbol{e}$ is an irrational number.

Mar 1, 2018

The function is $A \left(t\right) = 50 {e}^{- 0.0465 t}$

#### Explanation:

The equation for the radioactive decay is

$A \left(t\right) = {A}_{0} {e}^{- \lambda t}$

$\lambda$ is the radioactive constant

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

The half life of $\text{Sodium-24}$ is ${t}_{\frac{1}{2}} = 14.9 h$

The initial mass is ${A}_{0} = 50 m g$

Therefore,

$A \left(t\right) = 50 {e}^{- \frac{\ln 2}{14.9}} t$

$A \left(t\right) = 50 {e}^{- 0.0465 t}$

graph{50e^(-0.0465x) [-11.27, 61.8, -0.3, 36.22]}