Question

The radioisotope technetium-99 is often used as a radiotracer to detect disorders of the body. It has a half-life of 6.00 hours. If a patient received a 20.0 mg dose of this isotope during a medical procedure, how much would remain 24.0 hours after the dose was given?

Answer 1

Answer:

$0.25\ \text{mg}$

Explanation:

$t_{1/2}$ = Half-life of technetium-99 = 6 hours

$N_0$ = Initial mass of sample = 20 mg

$t$ = Time elapsed = 24 hours

Amount of mass remaining is given by

$N=N_0e^{-\dfrac{\ln 2}{t_{1/2}}t}\\\Rightarrow N=20e^{-\dfrac{\ln 2}{6}\times 24}\\\Rightarrow N=0.25\ \text{mg}$

The amount of the sample that would remain is $0.25\ \text{mg}$.