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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Answer:
Correct Answer: 78. $(0.25)$
Solution:
- $k=\dfrac{\ln 2}{t _{1 / 2}}=\dfrac{0.693}{5770} yr^{-1}=1.2 \times 10^{-4} $ $yr^{-1}$
Also $k t=\ln \dfrac{1}{f}=\dfrac{\ln 2}{5770} \times 11540=\ln 4 \Rightarrow f=\dfrac{1}{4}=0.25$
BETA
