Respuesta :
Answer:
18.42 years.
Step-by-step explanation:
We have been that the half-life of a certain substance is 25 years. We are asked to find time taken by substance to decay to 60% of its original amount.
We will use formula [tex]A=A_0\cdot e^{kt}[/tex] to solve our problem.
First of all, we will find value of k using [tex]A=\frac{1}{2}[/tex], [tex]A_0=1[/tex] and [tex]t=25[/tex].
[tex]\frac{1}{2}=1\cdot e^{k\cdot 25}[/tex]
[tex]0.5=e^{25k}[/tex]
Now we will take natural log of both sides of equation.
[tex]\text{ln}(0.5)=\text{ln}(e^{25k})[/tex]
[tex]\text{ln}(0.5)=25k\text{ln}(e)[/tex]
[tex]\text{ln}(0.5)=25k(1)[/tex]
[tex]k=\frac{\text{ln}(0.5)}{25}[/tex]
Our function would be [tex]A=A_0\cdot e^{\frac{\text{ln}(0.5)}{25}t}[/tex].
Now we have [tex]A=0.60[/tex] and [tex]A_0=1[/tex].
[tex]0.60=1\cdot e^{\frac{\text{ln}(0.5)}{25}t}[/tex]
[tex]0.60=e^{\frac{-0.6931471805599}{25}t}[/tex]
[tex]0.60=e^{-0.02772588722239t}[/tex]
Now we will take natural log on both sides.
[tex]\text{ln}(0.60)=\text{ln}(e^{-0.02772588722239t})[/tex]
[tex]\text{ln}(e^{-0.02772588722239t})=\text{ln}(0.60)[/tex]
[tex]-0.02772588722239t\cdot \text{ln}(e)=\text{ln}(0.60)[/tex]
[tex]-0.02772588722239t\cdot 1=\text{ln}(0.60)[/tex]
[tex]\frac{-0.02772588722239t}{-0.02772588722239}=\frac{\text{ln}(0.60)}{-0.02772588722239}[/tex]
[tex]t=\frac{-0.5108256237659907}{-0.02772588722239}[/tex]
[tex]t=18.424139854[/tex]
[tex]t\approx 18.42[/tex]
Therefore, it will take approximately 18.42 years for a sample of this substance to decay to 60% of its original amount
