Question

Question 37 2.56 pts A piece of fossilized wood has a carbon-14 radioactivity that is 1/4 that of new wood. The half- life of carbon-14 is 5730 years. How old is the cloth? O 3 x 5730 = 17,190 years O2 5730 - 11,460 years O 1 x 5730 - 5730 years a 0.25 x 5730 - 1432 years

Answer 1

Radioactive deacy follows first order kinetics. The equation for radioactive decay is:

$A_t=A_0e^{\frac{-ln(2)\times t}{T_{{1/2}}}}$

where:

A_t - activity of sample at time t

A₀ - initial activity of sample

T_1/2 - half life

t - time or age

This equation can be re-arranged to find the age t, as follows:

$A_t=A_0e^{\frac{-ln(2)\times t}{T_{{1/2}}}}$

OR

In(2)xt A = Ate T1/2

OR

$\frac{A_0}{A_t}=e^{\frac{ln(2)\times t}{T_{{1/2}}}}$

OR

Ао In(2) xt In = At T1/2

OR

$t=\frac{T_{{1/2}}}{ln(2)}\times ln\left (\frac{A_0}{A_t} \right )$

Here the activity of fossilized wood is given to be 1/4 that of new wood.

The activity of new wood is denoted by A₀ and the activity of fossilized wood is denoted by A_t.

Therefore we have:

$A_t=\frac{1}{4}A_0$

OR

$\frac{A_0}{A_t}=4$

Substituting this in the equation for t, we get:

$t=\frac{T_{{1/2}}}{ln(2)}\times ln\left (\frac{A_0}{A_t} \right )$

$t=\frac{T_{{1/2}}}{ln(2)}\times ln\left (4 \right )=T_{{1/2}}\times 2$

Given:

T_1/2 = 5730 years

Therefore we have:

$t=2\times T_{{1/2}}=2\times 5730\: years=11460\: years$

Therefore the correct answer is the second option, that is:

2x5730 = 11460 years