Cobalt-60 has a half-life of 5.27 years - AQA - A-Level Physics - Question 30 - 2017 - Paper 2

Using Avogadro's number ($N_A = 6.022 imes 10^{23} ext{ atoms/mol}$): $N = n \times N_A$ Substituting the values: $N = 0.01667 ext{ moles} \times 6.022 \times 10^{23} ext{ atoms/mol} \approx 1.0 \times 10^{22} ext{ atoms}$

The decay constant is calculated using the formula: ext{λ} = rac{ ext{ln}(2)}{t_{1/2}} where $t_{1/2} = 5.27 ext{ years} = 5.27 \times 3.154 \times 10^7 ext{ s}$. Thus: $ext{λ} = \frac{0.693}{5.27 \times 3.154 \times 10^7} \approx 4.2 \times 10^{-9} ext{ s}^{-1}$

The activity ($A$) can be calculated with: $A = ext{λ} \times N$ Substituting our values: $A = (4.2 \times 10^{-9} ext{ s}^{-1}) \times (1.0 \times 10^{22} ext{ atoms}) \approx 4.2 \times 10^{13} ext{ Bq}$