24.123 CP - Chemistry: Molecular Nature Of Matter And Change

Question

An earthquake in the area of present-day San Francisco is to be dated by measuring the ¹⁴C activity (t = 5730 yr) of parts of a tree uprooted during the event. The tree parts have an activity of 12.9 d/min-g C, and a living tree has an activity of 15.3 d/min C. How long ago did the earthquake occur?

Step-by-Step Solution

Answer

1410.5002 years ago, there was an earthquake

Step 1: Definition of Earthquake

An earthquake (also called a quake, tremor, or temblor) is the shaking of the Earth's surface caused by a sudden release of energy in the lithosphere, which results in seismic waves.

Step 2: Calculating time at which earthquake occur

The change in the number of nuclei (N) divided by the change in time is the decay rate, or activity (A), of a radioactive sample (t).

$A = \frac{-4N}{t}$

We put N₁ equal to $\frac{1}{2}{N}_0$and solve for to get the half-life t_1/2.

$\ln \frac{N_0}{\frac{1}{2}{N}_0} = k{t}_{\frac{1}{2}}$

Rearranging the formula that was derived,

$t_{\frac{1}{2}}= \frac{\ln 2}{k}$

$N_t= N_0{e}^{-kt}$or $\ln \frac{N_t}{N_0} = -kt$is an expression for calculating the number of nuclei left after a given period t.

Step 3: Calculating time at which earthquake occur

We must use the half-life formula to find the constant (k) in this situation.

$t_{\frac{1}{2}} = \frac{\ln 2}{k}\to k = \frac{\ln 2}{t_{\frac{1}{2}}}$

$k = \frac{\ln 2}{t_{\frac{1}{2}}} = \frac{\ln 2}{5730yr} = \frac{1.2096809x{10}^{-4}}{yr}$

We can solve for the value of t by using the expression for calculating the number of nuclei left. 

$\ln \frac{N_0}{N_t} = kt$

$\ln \frac{N_0}{N_t} = kt$

t = 1410.5002 yr