Answer: 0.333 h
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=0.375 g[/tex] is the final amount of the material
[tex]A_{o}=3 g[/tex] is the initial amount of the material
[tex]t=1 h[/tex] is the time elapsed
[tex]H[/tex] is the half life of the material (the quantity we are asked to find)
Knowing this, let's substitute the values and find [tex]h[/tex] from (1):
[tex]0.375 g=(3 g)2^{\frac{-1h}{H}}[/tex] (2)
[tex]\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}}[/tex] (3)
Applying natural logarithm in both sides:
[tex]ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}})[/tex] (4)
[tex]-2.079=-\frac{1 h}{H}ln(2)[/tex] (5)
Clearing [tex]H[/tex]:
[tex]H=\frac{-1h}{-2.079}(0.693)[/tex] (6)
Finally:
[tex]h=0.333 h[/tex] This is the half-life of the Bismuth-218 isotope
