Question

Cesium- 131 is the latest tool of nuclear medicine. It is used to treat malignant tumors by implanting Cs-131 directly into the tumor site. Its first- order half-life is 9.7 days. If a patient is implanted with \(20.0 \mathrm{mg}\) of Cs-131, how long will it take for \(33 \%\) of the isotope to remain in his system?

Short answer

Answer: Approximately 14.36 days.

Step-by-step solution

Determine the final amount of Cs-131

To determine the final amount of Cs-131 remaining in the system, we need to find 33% of the initial amount, which is 20.0 mg. Final amount of Cs-131 = (\(33 \% \space of \space 20.0 \mathrm{mg}\)) Final amount of Cs-131 = \((0.33 \times 20.0 \mathrm{mg})\) Final amount of Cs-131 = \(6.6 \mathrm{mg}\)

Use the decay formula

The decay formula for a radioactive substance is defined as: \(N_t = N_0 e^{-\lambda t}\), where \(N_t\) is the amount remaining at time t, \(N_0\) is the initial amount, \(\lambda\) is the decay constant, and t is the time in the same units as the half-life (days in our case). The decay constant \(\lambda\) is related to the half-life through the expression: \(\lambda = \frac{\ln2}{\text{half-life}}\)

Calculate the decay constant

Plug in the given half-life into the expression for decay constant \(\lambda\): \(\lambda = \frac{\ln2}{9.7 \space \text{days}}\) \(\lambda \approx 0.0712 \space \mathrm{day^{-1}}\)

Calculate the time

Plug the given values into the decay formula and solve for t: \(6.6 \mathrm{mg} = 20.0 \mathrm{mg} \times e^{-0.0712 t}\) Divide both sides by 20.0 mg: \(\frac{6.6}{20.0} = e^{-0.0712 t}\) Now solve for t: \(t = \frac{\ln (\frac{6.6}{20.0})}{-0.0712}\) Calculate the value of t: \(t \approx 14.36 \space \mathrm{days}\)

Answer the question

It will take approximately 14.36 days for 33% of the Cs-131 isotope to remain in the patient's system.