# If a 12.0-g sample of radon-222 decays so that after 19 days only 0.375 g remains, what is the half-life of radon-222?

Jun 26, 2017

#### Answer:

The half life is $= 3.8$ days

#### Explanation:

The original mass is ${m}_{0} = 12.0 g$

After $19$ days, amount remaining is ${m}_{19} = 0.375 g$

The ratio of ${m}_{19} / {m}_{0} = \frac{0.375}{12} = \frac{1}{32}$

That is

${m}_{0} / {m}_{19} = 32$

${m}_{19} = \frac{1}{32} {m}_{0}$

The half life is $= {t}_{\frac{1}{2}}$

${m}_{19}$ corresponds to $5 {t}_{\frac{1}{2}}$

So,

$5 {t}_{\frac{1}{2}} = 19$

${t}_{\frac{1}{2}} = \frac{19}{5} = 3.8$ days

We can perform the calculation

${m}_{19} = {m}_{0} {e}^{- 19 \lambda}$

${e}^{- 19 \lambda} = {m}_{19} / {m}_{0} = \frac{1}{32}$

$19 \lambda = \ln \left(32\right)$

$\lambda = \frac{1}{19} \ln \left(32\right) = \frac{5}{19} \ln 2$

${t}_{\frac{1}{2}} = \ln \frac{2}{\lambda} = \ln \frac{2}{\frac{5}{19} \ln 2} = \frac{19}{5} = 3.8$days