Answer:
[tex]3t_{1/2}[/tex]
Explanation:
To find the half-lifes of the isotope we need to use the following equation:
[tex] N_{t} = N_{0}2^{-\frac{t}{t_{1/2}}} [/tex] (1)
where Nt: is the amount of the isotope that has not yet decayed after a time t, N₀: is the initial amount of the isotope, t: is the time and [tex]t_{1/2}[/tex]: is the half-lifes.
By solving equation (1) for t we have:
[tex]\frac{t}{t_{1/2}} = - \frac{Ln(Nt/N_{0})}{Ln(2)}[/tex]
Having that:
Nt = 450
N₀ = 3150 + 450 = 3600,
The half-lifes of the isotope is:
[tex]t = - \frac{Ln(450/3600)}{Ln(2)} \cdot t_{1/2} = 3t_{1/2}[/tex]
Therefore, 3 half-lives of the isotope passed since the rock was formed.
I hope it helps you!
