Theorem

Let a radioactive element $S$ decay with a rate constant $k$.

Then its half-life $T$ is given by:

$T = \dfrac {\ln 2} k$

Proof

Let $x_0$ be the quantity of $S$ at time $t = 0$.

At time $t = T$ the quantity of $S$ has been reduced to $x = \dfrac {x_0} 2$.

This gives:

 $\displaystyle x_0 e^{-k T}$ $=$ $\displaystyle \frac {x_0} 2$ First-Order Reaction $\displaystyle \leadsto \ \$ $\displaystyle e^{k T}$ $=$ $\displaystyle 2$ $\displaystyle \leadsto \ \$ $\displaystyle k T$ $=$ $\displaystyle \ln 2$ $\displaystyle \leadsto \ \$ $\displaystyle T$ $=$ $\displaystyle \frac {\ln 2} k$

$\blacksquare$