# Question #21f9a

Oct 28, 2016

1.2 grams

#### Explanation:

Let n = the number of half lives

Let Q(n) = the quantity after n half lives.

Let Q(0) = the original quantity

The exponential decay is
$Q \left(n\right) = Q \left(0\right) {e}^{\alpha \left(n\right)}$

when $n = 1$:

$\frac{Q \left(1\right)}{Q \left(0\right)} = \frac{1}{2}$

This helps us to find the value of $\alpha$

$\frac{1}{2} = {e}^{\alpha \left(1\right)}$

$\ln \left(\frac{1}{2}\right) = \alpha$

$\alpha = - \ln \left(2\right)$

$Q \left(n\right) = Q \left(0\right) {e}^{- \ln \left(2\right) n}$

Your conditions are Q(0) = 75.0 g and n = 6

$1.2 g = \left(75.0 g\right) {e}^{- \ln \left(2\right) 6}$