The half life of bromine-73 is 20 min. How much 10 mg sample is still active after 60 min?

Jul 19, 2016

1.3 mg

Explanation:

Let me start by saying that the half life of bromine-73 is not 20 minutes but 3.33 minutes . But lets assume that the given numbers are correct.

Half life means that half of the sample you start with has decayed in the given period of time. It doesn't matter whether that is given in grams, number of atoms or the activity, the calculation is the same!

The simple way
The example is fairly easy because exactly 3 half times have past (60 min / 20 min = 3). How much is active after:

• 1 half life: 10 mg/2 = 5 mg
• 2 half lifes: 5 mg/2 = 2.5 mg
• 3 half lifes: 2.5 mg/ 2 = $\textcolor{red}{1.25 \text{mg}}$ (=1.3 mg taking into account the number of significant figures in the example)

The less simple way
When the example wouldn't have been as simple you can use the following equation:

$N \left(t\right) = N \left(0\right) \cdot {0.5}^{\frac{t}{T}}$

In which $N \left(0\right)$ is the amount you start with, and $N \left(t\right)$ the amount left after a certain time $t$ for a nuclide with a half life of $T$.

Let's do the calculation for the example with the actual half life of 3.33 minutes:

$N \left(t\right) = 10 m g \cdot {0.5}^{\frac{60 \min}{3.33 \min}} = 3.77 \cdot {10}^{-} 5 m g$

Always make sure the half life (T) and the time (t) have the same units!

Note: bromine-73 decays to selenium-73, this nuclide is also radioactive and will emit radiation. The half life of selenium-73 is longer (about 7 hours) so in this case it doesn't influence the result much. Otherwise you would measure more radiation than you expect based solely on decay of bromine-73.