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A radioactive isotope has a half life of 1 min. The amount of material, A(t), at t = 0 is 10 kg.

Write the particular solution to the differential equation which will describe the amount of isotope after time t, where t is measured in minutes.

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The particular solution to the ordinary differential equation that describes the one-step disintegration of the isotope is equal to the function $$A(t) = 10\cdot e^{-\frac{t}{1.443} }$$.

How to derive the equation of a radioactive decay

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:

$$\frac{dA}{dt} = -\frac{t}{\tau}$$ (1)

Where:

- A - Amount of material, in kilograms
- t - Time, in minutes
- τ - Time constant, in minutes

The solution of this differential equation is:

$$A(t) = A_{o}\cdot e^{-\frac{t}{\tau} }$$ (2)

The time constant can be calculated from the half-life:

$$\tau = \frac{t_{1/2}}{\ln 2}$$ (3)

If we know that $$t_{1/2} = 1\,min$$ and $$A_{o} = 10\,kg$$, then the particular solution of the differential equation is:

τ = (1 min)/ln 2

τ = 1.443 min

$$A(t) = 10\cdot e^{-\frac{t}{1.443} }$$

The particular solution to the ordinary differential equation that describes the one-step disintegration of the isotope is equal to the function $$A(t) = 10\cdot e^{-\frac{t}{1.443} }$$.