Answer:
60.0 years must pass to reduce a 24 mg of cesium 237 to 6.0 mg
Explanation:
For radioactive decay of a radioactive nuclide-
[tex]N_{t}=N_{0}(\frac{1}{2})^{(\frac{t}{t_{\frac{1}{2}}})}[/tex]
Where, [tex]N_{t}[/tex] is amount of radioactive nuclide after "t" time , N_{0} is initial amount of radioactive nuclide and [tex]t_{\frac{1}{2}}[/tex] is half-life of radioactive nuclide
Here N_{0} = 24 mg, N_{t} = 6.0 mg and [tex]t_{\frac{1}{2}}[/tex] = 30.0 years
So, [tex]6.0mg=24mg\times (\frac{1}{2})^{(\frac{t}{30.0years})}[/tex]
or, [tex]t=60.0years[/tex]
So 60.0 years must pass to reduce a 24 mg of cesium 237 to 6.0 mg