Question

The half life of ¹⁴C is 5730 years. What fraction of the ¹⁴C nuclei in a sample are left after 8140 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 14C t_1/2= 5730 y

time t = 8140 years

Initial amount of14C = [A]_o

Final amount of 14C = [A]_t

Substitute all the values in Eq (3),

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

[(0.693)/ 5730] = (1/8140)  ln {[A]_o / [A]_t}

ln {[A]_o / [A]_t} = [(0.693)/5730]x 8140

-  ln {[A]_t/ [A]_o } = [(0.693)/5730]x 8140

ln {[A]_t/ [A]_o } = - [(0.693)/5730]x 8140

  [A]_t/ [A]_o = e ^-[(0.693)/5730]x 8140

   [A]_{t =} ([A]_o) . e^{-[(0.693)/5730]x 8140}

= [A]_o x 0.37

= 37 % of [A]_o

[A]_t = 37 % of [A]_o

Final amount of 14C = 37% of initial amount

Therefore, 37% of 14C nuclei in a sample are left after 8140 years.