The half-life of Au¹⁹⁸ is 2.7 days. The activity of 1.50 mg of Au¹⁹⁸ if its atomic weight is 198 g mol⁻¹ is, (Nₐ = 6 × 10²³/mol).

Solution

Physics sir · 4 steps

Activity is λ N: count the nuclei, find λ in s⁻¹, then convert to curie.

1. Count the nuclei. Moles of gold = (1.50 × 10⁻³ g)/(198 g mol⁻¹) = 7.58 × 10⁻⁶ mol. So N = 7.58 × 10⁻⁶ × 6 × 10²³ = 4.55 × 10¹⁸ nuclei.
2. Decay constant. Convert the half-life to seconds: 2.7 × 86400 = 2.33 × 10⁵ s. Then λ = 0.693/(2.33 × 10⁵ s) = 2.97 × 10⁻⁶ s⁻¹.

   A = λ N = 2.97 × 10⁻⁶ × 4.55 × 10¹⁸ ≈ 1.35 × 10¹³ Bq

3. Convert to curie. Use 1 Ci = 3.7 × 10¹⁰ Bq. Then A = (1.35 × 10¹³)/(3.7 × 10¹⁰) ≈ 365 Ci.
4. So. The activity comes out at about 360 Ci. The closest option is (b) 357 Ci, and the small difference comes from rounding the constants. The other options are much further away.

The idea to keep. For activity, find N from the mass, the molar mass and N_A. Use λ = 0.693/T_(1/2) with T_(1/2) in seconds. Then A = λ N in Bq, and divide by 3.7 × 10¹⁰ to get curie.

Answer from the NTA final answer key. Solution by PrepWiser.