Explanation:
This problem can be solved using the Radioactive Half Life Formula:
[tex]A=A_{o}.2^{\frac{-t}{h}}[/tex] (1)
Where:
[tex]A=10g[/tex] is the final amount of the material
[tex]A_{o}[/tex] is the initial amount of the material (the quantity we are asked to find)
[tex]t=15.78y[/tex] is the time elapsed
[tex]h=5.26y[/tex] is the half life of cobalt
Knowing this, let's find [tex]A_{o}[/tex] from (1):
[tex]A_{o}=\frac{A}{2^{\frac{-t}{h}}}[/tex]
[tex]A_{o}=A.2^{-\frac{-t}{h}}[/tex]
This is the same as:
[tex]A_{o}=A.2^{\frac{t}{h}}[/tex] (2)
[tex]A_{o}=(10g)(2)^{\frac{15.78y}{5.26y}}[/tex]
[tex]A_{o}=(10g)(2)^{3}[/tex]
Finally:
[tex]A_{o}=80g[/tex]>>> This is the amount of grams in the original sample
