# A research assistant made 160 mg of radioactive sodium (Na^24) and found that there was only 20 mg left 45 h later, what is the half-life of Na^24?

Apr 12, 2017

The half life is $= 15 h$

#### Explanation:

There are 2 ways for solving this problem

Every half life, the amount is divided by $2$

After $1 \cdot {t}_{\frac{1}{2}}$, the amount left is $\frac{160}{2} = 80$

After $2 \cdot {t}_{\frac{1}{2}}$, the amount left is $\frac{80}{2} = 40$

After $3 \cdot {t}_{\frac{1}{2}}$, the amount left is $\frac{40}{2} = 20$

The half life is

${t}_{\frac{1}{2}} = \frac{45}{3} = 15 h$

The half life of sodium 24 is ${t}_{\frac{1}{2}}$

The radioactive decay constant is $\lambda = \ln \frac{2}{{t}_{\frac{1}{2}}}$

We apply the equation

$A = {A}_{0} \cdot {e}^{- l a m \mathrm{da} t}$

The activity is proportional to the mass.

$m = {m}_{0} \cdot {e}^{- l a m \mathrm{da} t}$

$160 = 20 \cdot {e}^{-} \left(\ln \frac{2}{t} _ \left(\frac{1}{2}\right) \cdot 45\right)$

$8 = {e}^{-} \left(\ln \frac{2}{t} _ \left(\frac{1}{2}\right) \cdot 45\right)$

$\left(\ln \frac{2}{t} _ \left(\frac{1}{2}\right) \cdot 45\right) = \ln 8$

${t}_{\frac{1}{2}} = \ln \frac{2}{\ln} 8 \cdot 45$

$= \ln \frac{2}{3 \ln 2} \cdot 45$

$= 15 h$