Answer:
[tex]t_{1/2}=6 h[/tex]
Explanation:
Let's use the decay equation.
[tex]A=A_{0}e^{-\lambda t}[/tex]
Where:
- A is the activity at t time
- A₀ is the initial activity
- λ is the decay constant
We know that [tex]\lambda=\frac{ln(2)}{t_{1/2}}[/tex]
So we have:
[tex]\lambda=\frac{ln(A/A_{0})}{t}[/tex]
[tex]\frac{ln(2)}{t_{1/2}}=\frac{ln(A/A_{0})}{t}[/tex]
[tex]t_{1/2}=\frac{t*ln(2)}{ln(A/A_{0})}[/tex]
[tex]t_{1/2}=6 h[/tex]
Therefore, the half-life of the source is 6 hours.
I hope it helps you!
