Exhibit 19-1
The following question(s) pertain to a 0.25 gram sample of the ³²P nuclide which has a half life of 14.28 days and a specific activity of 1.07×10¹⁶ Bq·g^-¹.
Refer to Exhibit 19-1. How long will it take for the activity of the 0.25 g sample to fall below 1 Ci? (1 Ci=3.7×10¹⁰ Bq)

Question

Quizplus · Accepted Answer

To find the time for the activity to fall below 1 Ci, use the decay formula: $ A = A_0 \times (1/2)^{t/T_{1/2}} $. Start with the initial activity: $ A_0 = 0.25 \, \text{g} \times 1.07 \times 10^{16} \, \text{Bq/g} = 2.675 \times 10^{15} \, \text{Bq} $. Convert 1 Ci to Bq: $ 1 \, \text{Ci} = 3.7 \times 10^{10} \, \text{Bq} $. Set $ A = 3.7 \times 10^{10} \, \text{Bq} $ and solve for $ t $ using the half-life $ T_{1/2} = 14.28 \, \text{days} $. Solving gives approximately 231 days.

Exhibit 19-1 the Following Question(s) Pertain to a 0.25 Gram Sample of Sample