Half-Life of Radioactive Substance
Jump to navigation
Jump to search
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