# The half-life of an element is 5.8 x 10^11. How long does it take a sample of the element to decay to 2/5 its original mass?

Dec 1, 2016

The expression for the first-order decay of a population is
$\frac{A}{A} _ 0 = {e}^{- k t}$
where the rate constant $k$ is related to the half-life by
$k = \ln \frac{2}{t} _ \left(\frac{1}{2}\right)$

#### Explanation:

In the question, the half-life should have units of time. Let's assume that the half-life is $5.8 \times {10}^{11} s$

In this case, the value of the rate constant is
$k = \ln \frac{2}{t} _ \left(\frac{1}{2}\right) = \ln \frac{2}{5.8 \times {10}^{11} s} = 1.20 \times {10}^{-} 12 {s}^{-} 1$

Using the first equation, we can find the time, $t$ at which the fraction of remaining atoms is $\frac{2}{5}$.

$\frac{2}{5} = {e}^{- \left(1.20 \times {10}^{-} 12 {s}^{-} 1\right) \left(t\right)}$

Solve for $t$ by first taking the natural logarithm of both sides:

$- 0.92 = - \left(1.2 \times {10}^{-} 12 {s}^{-} 1\right) t$

$t = \frac{0.92}{1.2 \times {10}^{-} 12 {s}^{-} 1} = 7.67 \times {10}^{11} s$