For a first order reaction the rate R = −k N?
so the ratio of rates at two different times is given by R(t1)/R(t2) = N(t1)/N(t2).
For nuclear decay the rate is ofen is expressed in units of counts per unit of time.
A chemists determines that a sample of petrified wood is 5.00 X 103 years old. What is the rate of decay, in counts per minute per gram, of carbon-14 in the sample? The decay rate of carbon-14 in wood today is 13.6 counts per minute per gram, and the half life of carbon-14 is 5730 years.
Enter a numeric answer only do not include units
Hint: Assume decay rate of wood hasn't changed over a 1000 years
i.e., No = 13.6
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Since radioactive decay is a first order reaction, use the equation:
ln(N/No)= -kt (you're looking for N since No is already given as 13.6.)
First, find the decay constant : 0.693/(half-life of C-14)
0.693/5730=
1.20942E-4
Next, multiply the decay constant (negative as stated by first order reaction equation) by the age of the tree:
-(1.20942E-4)(5000)=
-.6047120419
Now you are left with:
ln(N/13.6) = -.6047120419
Negate the natural log by "e-ing" it:
e^(ln (N/13.6) = e^(-.6047120419)
This will leave:
N/13.6 = .5462316958
Finally, solve for N which should be : 7.428751063