For a first order reaction the rate R = 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 3.50 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.

For a first order reaction the rate R = 鈭択 N?

Since radioactive decay is a first order reaction, use the equation:

ln(N/No)= -kt (you%26#039;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)(3500)=

-.4232984293

Now you are left with:

ln(N/13.6) = -.4232984293

Negate the natural log by %26quot;e-ing%26quot; it:

e^(ln (N/13.6) = e^(-.4232984293)

This will leave:

N/13.6 = .6548831676

Finally, solve for N which should be : 8.906411079

For a first order reaction the rate R = 鈭択 N?

R/13.6 = 3500/5730. Solve for R.