How do I determine the age of a sample using radiocarbon dating?

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SUMMARY

The discussion focuses on determining the age of a carbon sample using radiocarbon dating, specifically carbon-14. The user presents a problem involving a 9.2g carbon sample with an activity of 1.6Bq but struggles due to the unspecified isotope and unknown half-life. Participants clarify that the half-life of carbon-14 is essential for calculating the decay constant and solving for the initial quantity of carbon-14 (No) and time (t). The equations A = -(lambda)*N and N=No * e ^ -(lambda)*t are highlighted as critical for the calculations.

PREREQUISITES
  • Understanding of radiocarbon dating principles
  • Familiarity with the decay constant and half-life concepts
  • Knowledge of the equations A = -(lambda)*N and N=No * e ^ -(lambda)*t
  • Basic grasp of isotopes, specifically carbon-14
NEXT STEPS
  • Research the half-life of carbon-14 (approximately 5730 years)
  • Learn how to calculate the decay constant (lambda) for carbon-14
  • Explore methods for determining the initial quantity of carbon-14 in a sample
  • Study practical applications of radiocarbon dating in archaeology and geology
USEFUL FOR

Students in chemistry or geology, researchers in archaeology, and anyone interested in understanding radiocarbon dating techniques and calculations.

mss90
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Homework Statement


9.2g of carbon got an activity of 1.6Bq, how old is it?

Homework Equations


A = -(lambda)*N
N=No * e ^ -(lambda)*t
(lambda) = ln2/t1/2

The Attempt at a Solution


The question doesn't specify which isotope of carbon it is so halflife remains unknown.
I can't see how to solve this by only knowing the mass and activity.
Could anyone give me pointers?

R
 
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They are most likely referring to carbon-14 dating.
 
Alright in that case we know the half life and can find the decay constant.
Using A = -(lambda)*N we can find N as well (not sure If I need to divide the activity on 9.2g to get it per grams?)
However It still leaves No and t unsolved.

An alternative method is to; N=No * e ^ -(lambda)*t |*(lambda) => A=Ao * e ^ -(lambda)*t
But it seems to get me to the same "deadend" where Ao and t is unsolved.
I thought I might be able to set t=0 to get Ao but that only makes the expression " e ^ -(lambda)*t" equal 1.
Any tips?

R

PS: thanks for fast reply
 
Atoms per mole.
The sample starts out with a certain fraction of it being carbon-14.
Over time the amount of carbon-14 decays.
 

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