Question

In a typical human body, the amount of radioactive \(^{40}K\) is \(3.24 \times 10^{-5}\) percent of its mass. The activity due to \(^{40}K\) in a human body of mass 70 kg is .................. kBq.

Hint:

For calculating activity, use the relationship between the number of atoms, the decay constant, and the half-life of the substance.

Answer 1

Correct Answer: 5.95

Step 1: Determine the mass of \(^{40}K\) in the human body.
The total mass of \(^{40}K\) in the body is given by the fraction of its mass in the body. For a body of mass 70 kg: \[ \text{Mass of }^{40}K = 70 \, \text{kg} \times 3.24 \times 10^{-5} = 0.002268 \, \text{kg} \]
Step 2: Calculate the number of moles of \(^{40}K\).
The molar mass of \(^{40}K\) is approximately 39.1 g/mol. Converting kg to grams: \[ \text{Mass of }^{40}K = 0.002268 \, \text{kg} = 2.268 \, \text{g} \] Now, calculate the number of moles: \[ \text{Number of moles} = \frac{2.268}{39.1} = 0.0581 \, \text{mol} \]
Step 3: Use the activity formula.
The activity \(A\) is related to the number of moles by the formula: \[ A = \frac{N}{t_{1/2}} \times \lambda \] where \(N\) is the number of atoms, \(t_{1/2} = 3.942 \times 10^{16}\) s is the half-life, and \(\lambda = \frac{\ln 2}{t_{1/2}}\) is the decay constant. Using \(N = n \times N_A\) where \(N_A = 6.022 \times 10^{23}\): \[ A = \frac{0.0581 \times 6.022 \times 10^{23}}{3.942 \times 10^{16}} = 0.33 \, \text{kBq} \]
Step 4: Conclusion.
Thus, the activity due to \(^{40}K\) in the human body is approximately 0.33 kBq.