# If the radon concentration has become 1/20th of what it started as, and its half-life is 3.8 days, what amount of time in days has passed?

Dec 24, 2017

The time is $\text{16.4 days}$

#### Explanation:

The half life of Radon is ${t}_{1 / 2} = \text{3.8 days}$

The radioactive constant is $\lambda = \ln \frac{2}{t} _ \left(1 / 2\right)$

The equation for radioactive decay is

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

Therefore,

${e}^{- \lambda t} = \frac{N \left(t\right)}{N} _ 0 = \frac{1}{20}$.

Taking natural logs of both sides,

$- \lambda t = \ln \left(\frac{1}{20}\right) = - \ln 20$

$t = \frac{\ln 20}{\lambda}$

$= \ln \frac{20}{\ln 2 / {t}_{1 / 2}}$

$= \ln \frac{20}{\ln} 2 \cdot {t}_{1 / 2}$

$= \ln \frac{20}{\ln} 2 \cdot \text{3.8 days}$

$=$ $\text{16.4 days}$