Question

A nuclear reaction involves the first order decay from and has a half-life of 5730 yr. Living organisms have a 14C decay rate of 0.85 disintegrations per minute per gram of carbon. What is the value for the rate constant for this decay? If the initial concentration of 14C in a sample were 8.82 x 10-3 M, how much would remain after 936 years?

Answer 1

Given data

Half life = 5730 yr

Rate constant k= ?

Process is first order

Lets first calculate the rate constant for the process

t_1/2 = 0.693 / k

therefore k = 0.693 / t_1/2

k = 0.693 / 5730 yr

k = 0.000121 yr-1

Therefore rate constant for the process is k = 0.000121 yr-1

now lets use this rate constant to calculate the concentration after 936 years

the integrates equation for the first order decay process is as follows

ln[[A]t /[A]o] = - k*t

where [A]t = concentration after time t

[A]o = initial concentration

K= rate constant

T= time

Now lets put the values in the formula

ln[[A]t /[8.82*10^-3]] = - 0.000121 yr-1 * 936 yr

ln[[A]t /[8.82*10^-3]] = -0.1132

[A]t /[8.82*10^-3] = antiln -0.1132

[A]t /[8.82*10^-3] = 0.8929

[A]t = 0.8929 * 8.82*10^-3

[A]t = 7.88*10^-3 M

So the amount of the sample remain = 7.88*10^-3 M