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Nuclear decay and half-life

  • Thread starter plexus0208
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  • #1
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Homework Statement


Background info: The first order rate of nuclear decay of an isotope depends only upon the isotope, not its chemical form or temperature. The half-life for decay of carbon-14 is 5730 years. Assume that the amount of C-14 present in the atmosphere as CO2 and therefore in a living organism has been constant for the last 50,000 years. An ancient sample containing C-14 will show fewer disintegrations of the C-14 that is present than a modern sample because the concentration of C-14 is lower in the ancient sample.

If a 1.00 gram sample of wood found in an archaelogical site in Arizona underwent 7.90x103
disintegrations in a given time period (e.g., 20 h) and a modern sample underwent 1.84x104 disintegrations in the same time period, how old is the ancient sample?

Homework Equations



First order:
ln[A]t = -kt + ln[A]o
[A]t = e-kt[A]o

ln(([A]o/2)/[A]o) = -kt1/2 = ln(1/2)
or ln2 = kt1/2 = 0.693

The Attempt at a Solution


kt1/2 = 0.693
k = 0.693/5730 = 1.21x10-4

ln[A]t = -kt + ln[A]o
ln[A]t = ?
ln[A]o = ?
Solve for t?
Is this the right equation to use?
 
Last edited:

Answers and Replies

  • #2
Borek
Mentor
28,448
2,843
kt1/2 = 0.693
t1/2 = 0.693/5730 = 1.21x10-4
t1/2 is given, simple mistake here.

Other than that go for

[tex]\frac {A_t} {A_0} = e^{-kt} [/tex]

and it becomes almost simple plug and chug.

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  • #3
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What do I use for At and A0?
The number of disintegrations?
 
  • #4
Borek
Mentor
28,448
2,843
Yes. You are interested in ratio of activities.

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methods
 

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