Respuesta :
Answer:
See explanation below.
Explanation:
For this case we atart fom the proportional model given by the following differential equation:
[tex]\frac{dA}{dt}= kA[/tex]
And if we rewrite this expression we got:
[tex] \frac{dA}{A} = k dt[/tex]
If we integrate both sides we got:
[tex] ln|A| = kt +C[/tex]
And using exponential on both sides we got:
[tex] A(t) = e^{kt} e^C = A_o e^{kt}[/tex]
Where [tex] A_o [/tex] represent the initial amount for the isotope and t the time in years and A the amount remaining.
If we want to apply a model for the half life we know that after some time definfd [tex] t_{1/2}[/tex] the amount remaining is the hal, so if we apply this we got:
[tex] \frac{A_o}{2} = A_o e^{kt_{1/2}}[/tex]
We can cancel [tex] A_o[/tex] and we got:
[tex] \frac{1}{2}= e^{kt_{1/2}}[/tex]
If we solve for k we can apply natural log on both sides and we got:
[tex]ln (\frac{1}{2}) = kt_{1/2}[/tex]
[tex] k = \frac{ln(1/2)}{t_{1/2}}[/tex]
And that would be our proportional constant on this case.
If we replace this value for k int our model we will see that:
[tex] A(t) =A_o e^{ \frac{ln(1/2)}{t_{1/2}} t}[/tex]
And using properties of logs we can rewrite this like that:
[tex]A(t) = A_o (\frac{1}{2})^{\frac{t}{t_{1/2}}}[/tex]
And thats the common formula used for the helf life time.
