Living wood takes in radioactive carbon 14 from the atmosphere during the process of photosynthesis; the proportion of carbon 14 to carbon 12 being 1.25×10−12. When the wood dies the carbon 14 decays, its half life being 5600 years. 4.00 g of carbon from a piece of dead wood gave a total count of 20.0 disintegrations per minute, with an uncertainty of 0.04 disintegrations per minute. The uncertainty in the age (in years) is. Take: ln 2 = 0.7

Question

Living wood takes in radioactive carbon 14 from the atmosphere during the process of photosynthesis; the proportion of carbon 14 to carbon 12 being 1.25×10−12. When the wood dies the carbon 14 decays, its half life being 5600 years. 4.00 g of carbon from a piece of dead wood gave a total count of 20.0 disintegrations per minute, with an uncertainty of 0.04 disintegrations per minute. The uncertainty in the age (in years) is. Take:  ln 2 = 0.7

Accepted Answer

Explanation:Given Data:Proportion of $ C^{14} $ to $ C^{12} $: $ 1.25 \times 10^{-12} $Half-life of $ C^{14} $: $ 5600 \, \text{years} $Mass of carbon: $ 4.00 \, \text{g} $Total count rate: $ 20.0 \, \text{disintegrations per minute} $Uncertainty in count rate: $ 0.04 \, \text{disintegrations per minute} $$ \ln 2 = 0.7 $Step-by-Step Calculation:Decay constant ($ \lambda $): $ \lambda = \frac{\ln 2}{T_{1/2}} = 1.25 \times 10^{-4} \, \text{year}^{-1} $.Number of $ C^{14} $ atoms in 4 g of carbon: $ N_0 = 2.51 \times 10^{11} $.Activity of living wood: $ R_0 = 59.76 \, \text{disintegrations per minute} $.Age of dead wood: $ t = 8752 \, \text{years} $.Uncertainty in age: $ \Delta t = 16 \, \text{years} $.Final Answer: The uncertainty in the age is 16 years.

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