Chemical Kinetics - Result Question 79
####78. 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 $\left(\right.$ in $yr^{-1}$ ) for the decay? What fraction would remain after $11540 yr$ ?
(1984, 3M)
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Solution:
- $k=\frac{\ln 2}{t _{1 / 2}}=\frac{0.693}{5770} yr^{-1}=1.2 \times 10^{-4} yr^{-1}$
Also $k t=\ln \frac{1}{f}=\frac{\ln 2}{5770} \times 11540=\ln 4 \Rightarrow f=\frac{1}{4}=0.25$
