Question

What is the half-life of an isotope that decays to 6.25% of its original activity in 18.9 hours?

Answer 1

Radioactive material obeys 1st order decay kinetics,
For 1st order reaction, we have 
k = $\frac{2.303}{t}Xlog \frac{\text{initial conc.}}{\text{final conc.}}$
where, k = rate constant of reaction

Given: Initial conc. 100, Final conc. = 6.25, t = 18.9 hours

∴ k = $\frac{2.303}{18.9} X log \frac{100}{6.25}$ = 0.1467 hours^(-1)

Now, for 1st order reactions: half life = $\frac{0.693}{k} = \frac{0.693}{0.1467}$ = 4.723 hours.