# Question #3594d

Apr 10, 2017

The amount left is $= 0.0625 g$

#### Explanation:

The half life of Cobalt 60 is ${t}_{\frac{1}{2}} = 5.25 y e a r s$

The radioactive decay constant is $\lambda = \ln \frac{2}{{t}_{\frac{1}{2}}}$

So,

$\lambda = \ln \frac{2}{5.25} = \frac{0.69}{5.25}$

$= 0.1320 \left(y e a r {s}^{-} 1\right)$

We apply the equation

$A = {A}_{0} \cdot {e}^{- l a m \mathrm{da} t}$

The activity is proportional to the mass.

$m = {m}_{0} \cdot {e}^{- l a m \mathrm{da} t}$

$m = 1 \cdot {e}^{-} \left(0.1314 \cdot 21\right)$

$m = 1 \cdot {e}^{-} 2.773$

$m = 1 \cdot 0.0625$

$m = 0.0625 g$

We can do this differently

Every half life, the amount is divided by $2$

$\frac{21}{5.25} = 4$times

So the initial amount is divided by

$\frac{1}{16} = 0.625$