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=3.13mg[/tex] is the final amount of the material
[tex]A_{o}=50mg[/tex] is the initial amount of the material
[tex]t=19.7days[/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]3.13mg=(50mg)2^{\frac{-19.7days}{h}}[/tex] (2)
[tex]\frac{3.13mg}{50mg}=2^{\frac{-19.7days}{h}}[/tex] (3)
Applying natural logarithm in both sides:
[tex]ln(\frac{3.13mg}{50mg})=ln(2^{\frac{-19.7days}{h}})[/tex] (4)
[tex]-2.77=-\frac{19.7days}{h}ln(2)[/tex] (5)
Clearing [tex]h[/tex]:
[tex]h=\frac{-19.7days}{-2.77}(0.693)[/tex] (6)
Finally:
[tex]h=4.928days[/tex]
