A small volume of solution containing the radioactive isotope sodium-24 (2411Na) has an initial activity of 3.8 × 104 Bq.

Question 16

(a) (i) Define radioactive decay constant. [2]

(ii) Show that the decay constant λ is related to the half-life t_1/2  of a radioactive isotope by the expression

λ t_1/2 = ln2

[2]

(b) A small volume of solution containing the radioactive isotope sodium-24 (²⁴₁₁Na) has an initial activity of 3.8 × 10⁴ Bq. Sodium-24, of half-life 15 hours, decays to form a stable daughter isotope.

                                                          

All of the solution is poured into a container of water. After 36 hours, a sample of water of volume 5.0 cm³, taken from the container, is found to have an activity of 1.2 Bq.

Assuming that the solution of the radioactive isotope is distributed uniformly throughout the container of water, calculate the volume of water in the container. [4]

 [Total: 8]

Reference: Past Exam Paper – March 2018 Paper 42 Q13

Solution:

(a) (i) Radioactive decay constant is defined as the probability of decay (of a nucleus) per unit time.

(ii)

A = A0 exp(-λt)

{After one half-life, the initial activity would be halved.}

after one half-life, ½A0 = A0 exp(-λt)

½ = exp(–λt½) and hence taking logs, ln2 = λt_1/2

(b)

{Activity A at time t = A₀ exp (-λt)

Note: ln 2 = λt_1/2

Decay constant λ = ln 2 / t_1/2

Activity A at time t = A₀ exp (-λt)

Activity A at time t = A₀ exp (- ln 2 × t / t_1/2)}

Activity = 3.8 × 10⁴ exp(–ln2 × 36 / 15)

Activity = 7200 Bq                             

{An activity of 1.2 Bq corresponds to a volume of 5.0 cm³.

1.2 Bq - - > 5.0 cm³

7200 Bq - - > (7200 / 1.2) × 5.0}

volume = (7200 / 1.2) × 5.0

volume = 3.0 × 10⁴ cm³