Question

Carbon-11 has a half-life of 20. min. A sample has the initial activity of 64 mCi. What is the activity of the sample after 1.0 h?

Answer 1

Ans. Ans. Radioactive decay follows first order kinetics.

First order kinetics-

ln ([A]_t / [A]₀) = -kt                                -equation 1

where, [A]_t = Final amount of isotope

            [A]₀ = Initial amount of isotope

            k = rate constant

            t = time of reaction

Also, Half-life of first order reaction, t_½ = 0.693 / k

            Or, k = 0.693 / t_½                                        - equation 2

Combining equation 1 and 2 –

            ln ([A]_t / [A]₀) = -(0.693 / t_½) t               

            or, ln [A]_t – ln [A]₀ = -(0.693 / t_½) t                    - equation 3

Given, [A]₀ = Initial activity = 64 mCi

            Time, t = 1.0 hr = 60 min

            t_½ = 20.0 min

Putting the values in equation 3-

            ln[A]_t – ln (64) = - (0.693 / 20.0 min) x 60.0 min

            Or, ln[A]_t – 4.159 = -2.079

            Or, ln[A]_t = -2.079 + 4.159 = 2.080

            Or, 2.303 log [A]_t = 2.080

            Or, log [A]_t = 2.080 / 2.303 = 0.90317

            Or, [A]_t = antilog (0.90317) = 8.00

Hence amount of C-11 after 1.0 hr = 8.00 mCi