Answer: d) 312.5 mg
Explanation:
This problem can be solved using the Radioactive Half Life Formula:
[tex]A=A_{o}.2^{\frac{-t}{h}}[/tex]
Where:
[tex]A[/tex] is the remaining amount of Actinium-226
[tex]A_{o}=10 g[/tex] is the initial amount of Actinium-226
[tex]t=145 h[/tex] is the time elapsed
[tex]h=29 h[/tex] is the half life of Actinium-226
Knowing this, let's find [tex]A[/tex]:
[tex]A=(10 g) (2)^{\frac{-145 h}{29 h}}[/tex]
[tex]A=0.3125 g \frac{1000 mg}{1 g}=312.5 mg[/tex]
