Answer to Question #160170 in Atomic and Nuclear Physics for Hillary

Question

The radioactive isotope 198Au has a half-life of 64.8h. A sample containing this isotope has an initial activity of 40.0 Bq. Calculate the number of nuclei that decay in the time interval between in 2h.

Answer 1

$t_{1/2}=64.8 \;h = 64.8 \times 3600 = 233280 \;sec$

Initial activity = 40.0 Bq

$t = 2 \;h \\ λ = \frac{ln(2)}{t_{1/2}} \\ = \frac{ln(2)}{233280} \\ = 2.97 \times 10^{-6} \;s^{-1} \\ A_0 = λN_0 \\ N_0 = \frac{A_0}{λ} \\ = \frac{40}{2.97 \times 10^{-6}} \\ = 1.346 \times 10^7 \\ N = N_0e^{-λt}$

$= 1.346 \times 10^7 \times e^{-2.97 \times 10^{-6} \times 2 \times 3600} \\ = 1.346 \times 10^7 \times e^{-0.02138} \\ = 1.346 \times 10^7 \times 0.9788 \\ = 1.317 \times 10^7$

The number of nuclei that decay

$= N_0 – N \\ = 1.346 \times 10^7 - 1.317 \times 10^7 \\ = 0.029$