- #1
Rectifier
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The problem
The isotope ## ^{14}C ## has a half-life of ##5730## years. The ratio between the amount of ## ^{14}C ## nuclei and ## ^{12}C ## nuclei in a living matter is ## 1.3 \cdot 10 ^{-12} ##. When an organism dies the amount of ## ^{14}C ## atoms starts dropping and thus the ratio decreases.
a) What is the activity of ## ^{14}C ## -nuclei in a sample from a living matter that contains 1g Carbon.
b) Assume that an archaeological finding (from a dead animal) that contains 200g carbon has an activity of 400 decays per minute. How old is the finding?
The attempt
a)
The activity is given by:
##N=N_0 \cdot e^{-\lambda t}##
##\lambda = \frac{ln(2)}{T_{\frac{1}{2}}} ##
##T_{\frac{1}{2}}=5730 \ years =
2.091 \cdot 10^6 \ days ##
There is 1g of carbon. That means that
## m_{^{14}C} + m_{^{12}C} = 1 ##
the ration of ## ^{14}C ## atoms and ## ^{12}C ## atoms in a living matter is ## 1.3 \cdot 10 ^{-12} ## thus
## \frac{ ^{14}C n }{ ^{12}C n} = 1.3 \cdot 10 ^{-12} ##
To calculate N i need N_0 which is initial activity.
The activity N halves then t=half-time
Therefore:
## \frac{1}{2} = N_0e^{- \lambda t} \\ \frac{1}{2} = N_0e^{- \frac{ln(2)}{t_{\frac{1}{2}}} t} ## and for ##t = t_{\frac{1}{2}}## ##N_0 = 1##
Aaaand I am stuck :,(
Can someone please help?
The isotope ## ^{14}C ## has a half-life of ##5730## years. The ratio between the amount of ## ^{14}C ## nuclei and ## ^{12}C ## nuclei in a living matter is ## 1.3 \cdot 10 ^{-12} ##. When an organism dies the amount of ## ^{14}C ## atoms starts dropping and thus the ratio decreases.
a) What is the activity of ## ^{14}C ## -nuclei in a sample from a living matter that contains 1g Carbon.
b) Assume that an archaeological finding (from a dead animal) that contains 200g carbon has an activity of 400 decays per minute. How old is the finding?
This problem was translated from Swedish. Sorry for my weak English ;(
The attempt
a)
The activity is given by:
##N=N_0 \cdot e^{-\lambda t}##
##\lambda = \frac{ln(2)}{T_{\frac{1}{2}}} ##
##T_{\frac{1}{2}}=5730 \ years =
2.091 \cdot 10^6 \ days ##
There is 1g of carbon. That means that
## m_{^{14}C} + m_{^{12}C} = 1 ##
the ration of ## ^{14}C ## atoms and ## ^{12}C ## atoms in a living matter is ## 1.3 \cdot 10 ^{-12} ## thus
## \frac{ ^{14}C n }{ ^{12}C n} = 1.3 \cdot 10 ^{-12} ##
I am not entirely sure whether I should have n (amount of mols) or another unit. But I guess that it does not matter since units cancel out in the division.
To calculate N i need N_0 which is initial activity.
The activity N halves then t=half-time
Therefore:
## \frac{1}{2} = N_0e^{- \lambda t} \\ \frac{1}{2} = N_0e^{- \frac{ln(2)}{t_{\frac{1}{2}}} t} ## and for ##t = t_{\frac{1}{2}}## ##N_0 = 1##
Aaaand I am stuck :,(
Can someone please help?