Answer:
The half-life is [tex] t_{1/2} = 1.005 h[/tex]
Explanation:
Using the decay equation we have:
[tex]A=A_{0}e^{-\lambda t}[/tex]
Where:
We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that [tex]A = \frac{A_{0}}{2}[/tex]
[tex]\frac{A_{0}}{2}=A_{0}e^{-\lambda*1 h}[/tex]
[tex]0.5=e^{-\lambda*1 h}[/tex]
Taking the natural logarithm on each side we have:
[tex]ln(0.5)=-\lambda[/tex]
[tex]\lambda=0.69 h^{-1}[/tex]
Now, the relationship between the decay constant λ and the half-life t(1/2) is:
[tex]\lambda = \frac{ln(2)}{t_{1/2}}[/tex]
[tex] t_{1/2} = \frac{ln(2)}{\lambda}[/tex]
[tex] t_{1/2} = \frac{ln(2)}{0.69}[/tex]
[tex] t_{1/2} = 1.005 h[/tex]
I hope it helps you!
