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Question
Read the passage carefully and answer the questions that follow.
A plant or any other living being maintains a reasonable balance of C-14 (radioactive carbon) in its tissue during its lifetime. C-14 is used to determine the age of fossils. In the upper atmosphere, neutrons present in the cosmic rays are captured to produce the following nuclear reaction. \[\ce{_7N^14 + _0n^1 -> _6C^14 + _1H^1}\] C-14 isotope circulates in the atmosphere and is absorbed by living organisms during photosynthesis. The ratio of C-14 to C-12 in living beings is 1 : 1012. Once a living being dies, the level of C-14 in the dead being decreases due to the following reaction. \[\ce{_6C^14 -> _7N^14 + _-1e^0 + γ rays}\] The death of the plant brings an end to its tendency to take up C-14. The half-life period of C-14 is 5770 years. By knowing the concentration of C-14 in the living plant and the piece of dead material at a particular time, the age of the material (fossil) can be determined. |
- Write the relation between the decay constant and half-life period.
- In a piece of dead wood, the activity or concentration of C-14 is found to be one third of its initial activity. Calculate the age of the old wood.
- The half-life period (t1/2) for a first order reaction is 30 minutes. Calculate the time taken to complete 87·5% of the reaction.
Solution
- k = `0.693/t^(1/2)`
- 9146.5 years
- 90.03 minutes
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