Question

The radioactivity of a radionuclide with decay constant \(3.22 \times 10^{-5}\) s\(^{-1}\) is 6 mCi at 10:30 AM. The radioactivity at 4:30 PM the same day will be \(\underline{\hspace{2cm}}\) mCi. (rounded off to two decimal places)

Hint:

Radioactivity decreases exponentially over time with a half-life determined by the decay constant.

Answer 1

Correct Answer: 2.96

Radioactive decay formula: \[ A(t) = A_0 e^{-\lambda t} \] Where: \[ A_0 = 6\ \text{mCi}, \lambda = 3.22 \times 10^{-5}\ \text{s}^{-1}, t = 6 \text{hours} = 21600\ \text{seconds} \] Thus: \[ A(t) = 6 \times e^{-3.22 \times 10^{-5} \times 21600} = 6 \times e^{-0.696} \] \[ A(t) = 6 \times 0.498 = 2.988\ \text{mCi} \] Thus: \[ \boxed{2.99} \]