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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

  1. A1845 years
  2. B184.5 years
  3. C1900 years
  4. D190 years

Correct answer

A. 1845 years

Step-by-step solution

Radioactive decay follows first order kinetics. L e t [ A ] 0 = 100 ∴ [ A ] = 100 × 80 % = 80 D e c a y constant ​ ( k ) = 0.693 t 1 / 2 = 0.693 5730 t = 2.303 k log [ A o ] [ A ] = 2.303 0.693 5730 log 100 80 = 2.303 × 5730 0.693 × log 1.25 = 2.303 × 5730 0.693 × 0.0969 = 1 8 4 5 yrs

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