# The half-life of P-32 is 14.30 days. How many milligrams of a 20.00 mg sample of P-32 will remain after 85.80 days?

Aug 4, 2017

The amount remaining is $= 0.3125 m g$

#### Explanation:

The half life is ${t}_{\frac{1}{2}} = 14.3 d$

This means that after $14.3$ days 50% of the sample will remain

The time is $T = 85.8$days

This is equal to $= \frac{85.8}{14.3} = 6$ half lives

Therefore,

The amount remaining is $= \frac{1}{2} ^ 6 \cdot 20 = 0.3125 m g$

We can solve this problem with the equation

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

The radioactive constant is $\lambda = \ln \frac{2}{t} _ \left(\frac{1}{2}\right) = \ln \frac{2}{14.3}$

Therefore,

$m = 20 \cdot {e}^{- \ln 2 \cdot \frac{85.3}{14.3}} = 20 \cdot {e}^{- 6 \ln 2} = 20 \cdot {e}^{-} \ln \left({2}^{6}\right) = 20 \cdot {e}^{-} \ln 64$

$m = \frac{20}{64} = 0.3125 m g$