# Question #dbed0

Aug 11, 2017

The mass left after $12000 y$ is $= 1.81 g$ and the time to reach $1 g$ is $= 16800 y$

#### Explanation:

The half life is ${t}_{\frac{1}{2}} = 5600 y$

This means that half the amount present at time $t = 0$ will decay after $5600 y$

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

We can solve this problem with the equation

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

The radioactive constant is $\lambda = \ln \frac{2}{5600}$

The initial mass is ${m}_{0} = 8 g$

Therefore, the mass left after $12000 y$

$m = 8 \cdot {e}^{- \ln 2 \cdot \frac{12000}{5600}} = 8 \cdot {e}^{- 2.14 \ln 2} = 1.81 g$

Now,

The mass left is $m = 1 g$

We apply the same equation

$1 = 8 \cdot {e}^{- \ln \frac{2}{5600} \cdot t}$

${e}^{- \ln \frac{2}{5600} \cdot t} = \frac{1}{8}$

$\ln \frac{2}{5600} \cdot t = \ln 8$

$t = \ln 8 \cdot \frac{5600}{\ln} 2 = 16800 y$

The time is $= 16800 y$