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Suppose that 50 mg of a radioactive substance, having a half-life of three years, is initially present. More of this material is to be added at a constant rate so that 100 mg of the substance is present at the end of 2 years. At what constant rate must this radioactive material be added?

I'm really confused by this question, as I have never seen a radioactive decay differential asking for the addition of a constant.

I know that half life is denoted by $k\tau=\ln(2)$ so $k=\frac{\ln(2)}{3}$.

A standard radioactive decay problem can be expressed by the differential equation:

$$Q'(t)=-kQ(t)$$ $$Q'(t)=-\frac{\ln(2)}{3}Q(t)$$

But clearly we must be adding a constant to offset the fact we are loosing mass (half the original mass in 3 years) and then some. Let's call that $A$. So I think we would have:

$$Q'(t)=-\frac{\ln(2)}{3}Q(t) + A$$ which, I think, could be solved by introducing the initial condition $Q(2)=100$ but there's no $t$ in the equation to substitute in.

Any ideas on how to approach this problem? Any help would be greatly appreciated! Thanks!

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The differential equation that describes the quantity of material is:

$$\frac{dQ(t)}{dt} = -kQ(t) + r$$

Note the addition of the constant $r$ on the RHS, which represents the addition of material at a constant rate. The initial condition can be written as $Q(0) = Q_0$.

Separating the variables and solving,

$$\int_{Q_0}^{Q(T)}\frac{dQ(t)}{-kQ(t) + r} = \int_0^Tdt$$

$$-\frac{1}{k}\ln\frac{r-kQ(T)}{r-kQ_0} = T$$

Can you proceed with the remaining algebra and substitution to get the final answer? You're given $Q_0, T$ and $Q(T)$ (and you can work out $k$ based on the half-life). You need to solve for $r$.

I'm a little out of time at the moment. If you're having further difficulty, please comment.

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