The number of radioactive atoms, $N$, in a sample of a sodium isotope after time $t$ hours can be modelled by $$N = N_0 e^{-kt}$$ where $N_0$ is the initial number of radioactive atoms in the sample and $k$ is a positive constant - AQA - A-Level Maths Mechanics - Question 5 - 2020 - Paper 3

We need to find $t$ for when 90% of the atoms have decayed, meaning 10% remain:

$N = 0.1N_0$

Using the model: $0.1N_0 = N_0 e^{-kt}$

Dividing by $N_0$: $0.1 = e^{-kt}$

Taking the natural logarithm: $-kt = \ln(0.1)$

Thus: $t = -\frac{\ln(0.1)}{k}$ Substituting $k = 0.0436$ yields:

$t \approx -\frac{\ln(0.1)}{0.0436} \approx 52.8 \text{ hours}$

To convert hours to days, we use: $\text{Days} = \frac{52.8}{24} \approx 2.2\text{ days}$

Thus, the number of days required for at least 90% of the original number of atoms to decay is approximately 2.2 days.