(a) (i) Define radioactive decay constant. (ii) Show that the decay constant λ is related to the half-life t_½ of a radioactive isotope by the expression λt_½ = In2 (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.
- (a)(i)Define radioactive decay constant.[2]
- (a)(ii)Show that the decay constant λ is related to the half-life t_½ of a radioactive isotope by the expression λt_½ = In2[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]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2018). The mark or mark type is shown in brackets.
- [M1] probability of decay (of a nucleus)
- [A1] per unit time
- [M1] A = A₀ e⁻ᵞᵗ
- [A1] after one half-life, ½A₀ = A₀ e⁻ᵞᵗ¹/²
- [A1] ½ = exp(−λt_½) and hence taking logs, ln2 = λt_½
- [C1] activity = 3.8 × 10⁴ exp(−ln2 × 36/15)
- [C1] = 7200 Bq
- [(C1)] or activity = 3.8 × 10⁴ / 2²․⁴
- [(C1)] = 7200 Bq
- [C1] volume = (7200/1.2) × 5.0
- [A1] = 3.0 × 10⁴ cm³
- [(C1)] OR activity of 5.0 cm³ = 1.2 x 2²․⁴
- [(C1)] = 6.3336 Bq
- [(C1)] volume = (3.8 × 10⁴ / 6.3336) × 5.0
- [(A1)] = 3.0 × 10⁴ cm³
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