Question

The age of a piece of wood from an archeological site is to be determined using the Carbon-14 method. The activity of the sample is measured to be 0.459 times the Carbon-14 activity of living plants. What is the age of the sample in years? (The half-life of the Carbon-14 isotope is 5730 years.)

Tries 0/20

Answer 1

The relation of activity is given by A = Ao*e^-*t

given

A/Ao = 0.459

T_1/2 = 5730 years

we have

T_1/2 = 0.693/

= 0.693/T_1/2 = 0.693/5730 = 1.2094*10^-4 /year

Hence age of the sample t can be found by the relation.

0.459 = e^{-(1.2094*10^-4)*(t)}

e^{(1.2094*10^-4)*(t)} = 1/0.459 = 2.17

1.2094*10^-4 *t = ln(2.17)

1.2094*10^-4 *t = 0.774

t = 6405.88 years