Calculating Atomic Fraction of C in Carbon Dioxide at STP

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iamalexalright
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Homework Statement



In a sample of one litre of carbon dioxide at STP an average of 5 disintegrations

C -> N + e + v

are observed per minute. Calculate the atomic fraction of C present if the mean lifetime of this nucleus is 8267 years.


Homework Equations



[tex]\tau[/tex] = [tex]\omega[/tex]-1 = 8267 years

N(t) = N0 * [tex]e[/tex]-[tex](\omega[/tex] * t)


The Attempt at a Solution



First I found [tex]\tau[/tex] = 2.607 * 1011 seconds.

And really I'm at a loss of where to go next. I figure N(t)/N0 is what I am looking for but I cannot solve for this without knowing the time t and I do not know where the 5 disintegrations per minute comes in. I'm not looking for an answer(already know it), i just want a little push in the right direction.
 
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iamalexalright said:

Homework Statement



In a sample of one litre of carbon dioxide at STP an average of 5 disintegrations

C -> N + e + v

are observed per minute. Calculate the atomic fraction of C present if the mean lifetime of this nucleus is 8267 years.

Homework Equations



[tex]\tau[/tex] = [tex]\omega[/tex]-1 = 8267 years

N(t) = N0 * [tex]e[/tex]-[tex](\omega[/tex] * t)

The Attempt at a Solution



First I found [tex]\tau[/tex] = 2.607 * 1011 seconds.

And really I'm at a loss of where to go next. I figure N(t)/N0 is what I am looking for but I cannot solve for this without knowing the time t and I do not know where the 5 disintegrations per minute comes in. I'm not looking for an answer(already know it), i just want a little push in the right direction.

Since there are an average of 5 disintegrations per minute, you would expect:

[tex]N(t)-N(t+60s)=5[/tex]...can you take it from there?
 
iamalexalright said:

Homework Statement



In a sample of one litre of carbon dioxide at STP an average of 5 disintegrations

C -> N + e + v

are observed per minute. Calculate the atomic fraction of C present if the mean lifetime of this nucleus is 8267 years.


Homework Equations



[tex]\tau[/tex] = [tex]\omega[/tex]-1 = 8267 years

N(t) = N0 * [tex]e[/tex]-[tex](\omega[/tex] * t)


The Attempt at a Solution



First I found [tex]\tau[/tex] = 2.607 * 1011 seconds.

And really I'm at a loss of where to go next. I figure N(t)/N0 is what I am looking for but I cannot solve for this without knowing the time t and I do not know where the 5 disintegrations per minute comes in. I'm not looking for an answer(already know it), i just want a little push in the right direction.

One can find the total amount of carbon from "one litre of carbon dioxide" - simply find the moles of CO2 in 1 liter at STP. It would appear that one is to find the atomic fraction of radiocarbon or C-14.

The activity, decays per unit time, is the product of the decay constant [itex]\lambda[/itex] and the number of atoms of the radionuclide present, i.e. A(t) = [itex]\lambda[/itex] N(t). One is given A, so find N.