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!


1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .