Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

The half-life for decay of radioactive $^{14} C$ is 5730 years. An archaeological artefact containing wood has only 80% of the $^{14} C$ activity as found in living trees. The age of the artefact would be:
[Given: log 1.25 = 0.0969]

Options:

Answer:

Solution:

Radioactive decay follows first order kinetics. $L e t {[A]}_{0} = 100$ $\begin{array}{l} ∴ [A] = 100 \times 80 % = 80 \\ D e c a y constant (k) = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{5730} \\ t = \frac{2.303}{k} log \frac{[A o]}{[A]} \\ = \frac{2.303}{\frac{0.693}{5730}} log \frac{100}{80} \\ = \frac{2.303 \times 5730}{0.693} \times log 1.25 \\ = \frac{2.303 \times 5730}{0.693} \times 0.0969 = 1845 yrs \end{array}$

Stream:NTA_ABHYASSubject:ChemistryTopic:Chemical KineticsSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...