Question #21f9a

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Answer 1 · 2016-10-28T02:20:36

1.2 grams

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
$Q (n) = Q (0) e^{α (n)}$ $Q(n) = Q(0)e^(alpha(n))$

when $n = 1$ $n = 1$ :

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

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

$\frac{1}{2} = e^{α (1)}$ $1/2 = e^(alpha(1))$

$ln (\frac{1}{2}) = α$ $ln(1/2) = alpha$

$α = - ln (2)$ $alpha = -ln(2)$

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

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

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