Question

In a living organism, the decay of C-14 produces 15.3 disintegrations per minute per gram of carbon. The half-life of C-14 is 5730 years. A bone sample with 2.8 g of carbon has 36.7 disintegrations per minute. How old is the bone sample in years?

Answer 1

Answer:

The bone sample is 1278.86 years.

Explanation:

Using the integrated rate law:

N=$N_{o} e^{-kt}$

where:

k is the rate constant, N is the number of atoms at time t, No is the number of atoms at time t=0.

The rate constant can be determined using:

k = 0.693/$t_{1/2}$

where $t_{1/2}$ is the half life which is equal to 5730 years

Thus:

k = 0.693/5730 = 1.210*$10^{-4}$ (1/year)

Then, we calculate the current rate. The organism produces 36.7 disintegrations per minute per 2.8 g. However, the rate constant is in years. Therefore, we convert the unit to disintegrations per year per gram.

$\frac{36.7}{2.8*g*min} *\frac{60 min}{1 h} * \frac{24 h}{1 d} *\frac{356.25 d}{1 y} = 6.724*10^{6} g^{-1}y^{-1}$

To determine N, we use:

r=kN

where r is the current rate which is 6.724*$10^{6} g^{-1}y^{-1}$

Thus:

$6.724*10^{6} = 1.210*10^{-4}*N$

N = 5.557*$10^{10}$

Similarly, we have 15.3 disintegrations per minute per gram of carbon. Therefore,

$\frac{15.3}{g*min}*\frac{60 min}{1 h}*\frac{24 h}{1 d}*\frac{356.25 d}{1 yr} = 7.849*10^{6} g^{-1}y^{-1}$

To determine No, we use:

r=kNo

Thus:

$7.849*10^{6} = 1.210*10^{-4}*N_{0}$

No = 6.487*$10^{10}$

Therefore using the integrated rate law:

N=$N_{o}*e^{-kt}$

5.557*$10^{10} = 6.487*10^{10}*e^{-1.210*10^{-4}*t }$

0.8567 = $e^{-1.210*10^{-4}*t }$

Taking the natural log (ln) of both sides which is the inverse of exponential, we have:

ln (0.8567) = ln ($e^{-1.210*10^{-4}*t }$

-0.1547 = -1.210*$10^{-4}*t$

t = 1278.86 years. Therefore, the bone sample is 1278.86 years old.