# What is the rate of decay, in counts per minute per gram, of carbon-14 in the sample?

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 chemist determines that the sample of petrified wood is 4.00 X 103 years old. 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

ln([A]t/[A]o) = -kt

ln(x/13.6) = -(1.21 x 10^-4/year)(4.00 x 10^3 years)

x = 22.1

To determine k

0.693/k = halflife

0.693/k = 5730 years

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For nuclear decay the rate is ofen is expressed in units of counts per unit of time.

A chemist determines that the sample of petrified wood is 4.00 X 103 years old. 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

**Answer:**ln([A]t/[A]o) = -kt

ln(x/13.6) = -(1.21 x 10^-4/year)(4.00 x 10^3 years)

x = 22.1

To determine k

0.693/k = halflife

0.693/k = 5730 years

k = 1.21 x 10^-4/year