Question

Radioactive decay is a first order process. Radioactive carbon in wood sample decays with a half-life of $5770 \,yr$. What is the rate constant (in $yr^{-1})$ for the decay? What fraction would remain after $11540 \,yr$?

Answer 1

$k=\frac{\ln 2}{t_{1/2}}=\frac{0.693}{5770}yr^{-1}=1.2\times10^{-4}yr^{-1}$
Also $kt=\ln\frac{1}{f}=\frac{\ln 2}{5770}\times11540=\ln 4$
$\Rightarrow f=\frac{1}{4}=0.25$