Potassium-40 is a significant source of radioactivity in the human body - Leaving Cert Physics - Question b - 2017

One commonly used device to detect beta-radiation is the Geiger-Müller tube, which operates on the principle of ionization. When beta particles enter the tube, they collide with the gas inside, causing ionization. This results in a flow of current that produces a detectable electrical signal, indicating the presence of beta radiation.

The nuclear equation for the decay of potassium-40 can be written as:

$K^{19}_{40} \rightarrow Ca^{20}_{40} + e^{-1}$

Given the activity (A) of the human body due to potassium-40 is 5400 Bq, we can start by calculating the decay constant ($\lambda$) using the formula:

$\lambda = \frac{\ln(2)}{t_{1/2}}$

where $t_{1/2} = 1.25 \times 10^9$ years. We first convert this into seconds: $1.25 \times 10^9 \text{ years} = 3.942 \times 10^{16} \text{ seconds}$

Then, substituting the half-life into the decay constant formula:

$\lambda = \frac{\ln(2)}{3.942 \times 10^{16}} \approx 1.76 \times 10^{-17} \text{ s}^{-1}$

Next, we can use the relationship between activity, the decay constant, and the number of nuclei (N): $A = -\lambda N$

Rearranging this gives: $N = \frac{A}{\lambda}$

Substituting the provided values: $N = \frac{5400}{1.76 \times 10^{-17}} \approx 3.07 \times 10^{20}$

Thus, the number of potassium-40 nuclei in this person is approximately $3.07 \times 10^{20}$.