The half-life of C-14 is 5730 years. The abundance of C-14 in materials taking p

Question

The half-life of C-14 is 5730 years. The abundance of C-14 in materials taking part in the C-N cycles [living plants, for example) is such that 14 disintegrations per minute [dpm] occur per gram of pure carbon. A lump of charcoal from a firepit presumably from the period 600 BCE is tested. What level of radiation (dpm) is expected? What could be the impact on the measurement if the sample was handled by the discoverers? What countermeasures could be taken to assure that the sample tested is not contaminated?

Explanation / Answer

(a): Given half-life, t1/2 = 5730 years

=> disintigration constant, k = 0.693 / t1/2 = 0.693 / 5730 yr = 1.21x10-4 yr-1.

Initial activity, Ao = 14 dpm per gram of pure C.

Given age of the charcoal, t = 600 + 2015 = 2615 yr

Let the activity be, 'At'

Applying the integrated rate equation for 1st order reaction:

kt = ln(Ao / At)

=> 1.21x10-4 yr-1 x 2615 yr = ln(14 dpm / At)

=>  ln(14 dpm / At) = 0.316264

=> 14 dpm / At = exp (0.316264) = 1.372

=> At = 10.2 dpm per gram of pure C (answer)

(b): If the sample is handled by the discoverers, it gets contaminated and fresh carbon is added. This causes an increase in the activity of the sample and the calculated age of the sample decreases.

(c): The sample should not be come in contact with contaminants like paper, cigerettee ash, card board, cotton, polythine glycol, polyethene, hydrocarbons, etc.

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