Half-Life of Radioactive Substance

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:

\(\ds x_0 e^{-k T}\)

\(=\)

\(\ds \frac {x_0} 2\)

First-Order Reaction

\(\ds \leadsto \ \ \)

\(\ds e^{k T}\)

\(=\)

\(\ds 2\)

\(\ds \leadsto \ \ \)

\(\ds k T\)

\(=\)

\(\ds \ln 2\)

\(\ds \leadsto \ \ \)

\(\ds T\)

\(=\)

\(\ds \frac {\ln 2} k\)

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $1$: The Nature of Differential Equations: $\S 4$: Growth, Decay and Chemical Reactions