Answer:
[tex]0.25\ \text{mg}[/tex]
Explanation:
[tex]t_{1/2}[/tex] = Half-life of technetium-99 = 6 hours
[tex]N_0[/tex] = Initial mass of sample = 20 mg
[tex]t[/tex] = Time elapsed = 24 hours
Amount of mass remaining is given by
[tex]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}[/tex]
The amount of the sample that would remain is [tex]0.25\ \text{mg}[/tex].
