This problem can be solved using the Radioactive Half Life Formula:
[tex]A=A_{o}.2^{\frac{-t}{h}}[/tex] (1)
Where:
[tex]A=2.7g[/tex] is the final amount of the material
[tex]A_{o}=10g[/tex] is the initial amount of the material
[tex]t[/tex] is the time elapsed (the quantity we are asked to find)
[tex]h=432y[/tex] is the half life of americium-241
Knowing this, let's find [tex]t[/tex] from (1):
[tex]2.7g=(10g).2^{\frac{-t}{432y}}[/tex]
[tex]\frac{2.7g}{10g}=2^{\frac{-t}{432y}}[/tex]
[tex]0.27g=2^{\frac{-t}{432y}}[/tex]
Applying natural logarithm in both sides:
[tex]ln(0.27g)=ln(2^{\frac{-t}{432y}})[/tex]
[tex]-1.309=-\frac{t}{432y}ln(2)[/tex]
[tex]-t=\frac{(-1.309)(432y)}{0.693}[/tex]
Finally:
[tex]t=816y[/tex]
