Question

In 2002, a wooden beam from the Wupatki Indian ruin in Arizona was found to contain 235 mg of carbon-14. The same mass of wood from a tree-cut in 2002 contained 264 mg of carbon-14. In what year was the beam from Wupatki cut if the half-life of carbon-14 is 5730 years?

Answer 1

Radio active decay is a first order reaction.

For first order recation,

half life t_1/2 = 0.693 /k where k is rate constant

k = 0.693/ t_{1/2 --- Eq (1)}

k = 1/t ln { [A]_o/[A]_t} -----Eq (2)

From Eqs (1) and (2),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t} ------Eq (3)

Given that

half life of C-14 = 5730 yrs

time t = ? hrs

Initial amount of C-14  = 264 mg

Final amount of C-14  [A]_t = 235 mg

Substitute all the values in Eq (3),

0.693/ t_1/2 = (1/t) ln {[A]_o/ [A]_t}

[(0.693)/(5730 yrs)] = (1/t )  ln{ 264 mg/ 235 mg}

t =ln{ 264 mg/ 235 mg} x (5730 yrs / 0.693)

= 962 yrs

t = 962 yrs

Therefore,

the year in which the beam from Wupatki cut = 2002 + 962 = 2964