# Phosphorus-32 is radioactive and has a half life of 14.3 days. How long would it take a sample to decay from 2.20 mg to 600 mug?

Nov 1, 2016

26.8 days.

#### Explanation:

The equation for 1st order decay is:

$\textsf{{N}_{t} = {N}_{0} {e}^{- \lambda t}}$

We can get the value of the decay constant $\textsf{\lambda}$ from the 1/2 life:

$\textsf{\lambda = \frac{0.693}{t} _ \left(\frac{1}{2}\right) = \frac{0.693}{14.3} = 0.0484 \textcolor{w h i t e}{x} {d}^{- 1}}$

Taking natural logs of both sides:

$\textsf{\ln {N}_{t} = \ln {N}_{0} - \lambda t}$

Putting in the numbers and converting to milligrams:

$\textsf{\ln 0.6 = \ln 2.2 - 0.0484 t}$

$\therefore$$\textsf{- 0.5108 = 0.7884 - 0.0484 t}$

$\therefore$$\textsf{t = \frac{1.30}{0.0484} = 26.8 \textcolor{w h i t e}{x} \text{days}}$