Question

In a radioactive isotope, \(N\) nuclei are needed to produce radioactivity level of 2 mCi. Assuming decay constant of \(3.22 \times 10^{-5}\) s\(^{-1}\) and atomic weight of 98 g/mol and Avogadro's number \(6.02 \times 10^{23}\) mol\(^{-1}\), the mass of \(N\) radionuclide is \(\underline{\hspace{2cm}}\) picograms.

Hint:

Use the decay constant and Avogadro's number to calculate the mass from activity.

Answer 1

Correct Answer: 372

Radioactive decay relation: \[ A = \lambda N \] Where: \[ A = 2\ \text{mCi} = 2 \times 10^{-3}\ \text{Ci}, \lambda = 3.22 \times 10^{-5}\ \text{s}^{-1} \] Thus: \[ N = \frac{A}{\lambda} = \frac{2 \times 10^{-3}}{3.22 \times 10^{-5}} = 62.11 \times 10^3\ \text{nuclei} \] The number of nuclei is \(N = 62.11 \times 10^3\). Convert to mass: \[ \text{Mass of } N = \frac{N \times \text{atomic weight}}{\text{Avogadro's number}} = \frac{62.11 \times 10^3 \times 98}{6.02 \times 10^{23}} \] Convert to picograms: \[ \text{Mass} = 1.02 \times 10^{-16} \ \text{g} = 102 \ \text{pg} \] Thus: \[ \boxed{372} \]