The particular solution to the ordinary differential equation that describes the one-step disintegration of the isotope is equal to the function [tex]A(t) = 10\cdot e^{-\frac{t}{1.443} }[/tex].
In this question we have a case of simple disintegration of an isotope, that is, that the radioactive isotope has only one stage of disintegration. This process is represented by the following ordinary differential equation:
[tex]\frac{dA}{dt} = -\frac{t}{\tau}[/tex] (1)
Where:
The solution of this differential equation is:
[tex]A(t) = A_{o}\cdot e^{-\frac{t}{\tau} }[/tex] (2)
The time constant can be calculated from the half-life:
[tex]\tau = \frac{t_{1/2}}{\ln 2}[/tex] (3)
If we know that [tex]t_{1/2} = 1\,min[/tex] and [tex]A_{o} = 10\,kg[/tex], then the particular solution of the differential equation is:
τ = (1 min)/ln 2
τ = 1.443 min
[tex]A(t) = 10\cdot e^{-\frac{t}{1.443} }[/tex]
The particular solution to the ordinary differential equation that describes the one-step disintegration of the isotope is equal to the function [tex]A(t) = 10\cdot e^{-\frac{t}{1.443} }[/tex].
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