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About LinkBacksok i got the first part to be k = log 2 / 5730
but i dont understand the disintegrations per minute part.
Any help would be greatly appreciated. Thanks
The half life of naturally occuring radioactive carbon C14 is 5730 years. In a living organism, the rate of C14 decay is typically constant at around 15.3 disintegrations per minute. When the organism dies the level of C14 decays away. Let C be the amount of radioactive carbon measured in disintegrations per minute at time t years.
a) assuming natural decay, C = Co * e^(-kt) Find the values of k and Co
b) Bones are found which exhibit 11 disintegrations per minute. How old are the bones to the nearest year?
c) Carbon dating is only deemed reliable for ages between 1000 and 10000 years. In another archaeological expedition, artefacts are found in a cave which exhibit 2.25 disintegrations per minute
(i) How old do these artefacts seem to be?
(ii) Is this figure reliable or should other tests be carried out to confirm the age of the artefacts.
Sorry its so long but i have no idea how to work this out. :S
The "disintegrations per minute" is just the value of C: " Let C be the amount of radioactive carbon measured in disintegrations per minute at time t years. "
You were told that the a living creature has "around 15.3 disintegrations per minute" so when t= 0 (the instant it dies) C(0)= C0 e^{-k0}= C0= 15.3. "The half life of naturally occuring radioactive carbon C14 is 5730 years." So in 5730 year, that level will have decreased to 15.3/2= 7.65. 15.3e^{-k(5730)}= 7.65 which gives e^{-k(5730)}= 1/2. Taking the logarithm of both sides, -k(5730)= log(1/2)= -log(2) so that k= log(2)/5730, as you have.
Now solve 11= 15.2e^{-kt} and 2.25= 15.2e^{-kt}, with that value of k, for t.
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. Ive got it now. You're explanation was very clear.
