Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Radioactive Radium has a half-life of approximately 1600 years. What percentage of the present amount remains after 100 years?

share|cite|improve this question
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Julian Kuelshammer Oct 30 '12 at 15:22

The formula for exponential decay is: $$\frac{dN}{dt}=-N\lambda$$

Solving the differential equation, we get the following:

$$N(t)=N(0)\cdot \rm{e}^{-\lambda t}$$

To get the percentage, we will start off with $N(0)=100$, solving for $t=100$ and $\lambda=\frac{\ln{2}}{1600}$ (which we find from the definition of half-life: $t_{\frac{1}{2}}=\frac{\ln{2}}{\lambda}$), we get:


So $95.76\%$ remains after $100$ years.

share|cite|improve this answer
Actually, the definition of half-life should be $N(t_{1/2}) = N(0)/2$. Then your equation $t_{1/2} = (\ln 2)/\lambda$ follows from that. – GEdgar Oct 30 '12 at 15:00
@GEdgar Sorry, yes it should be, I cut that out for brevity; and went straight to the equation for $t_{\frac{1}{2}}$ itself, do you think I should include the derivation of the half-life equation then? – Shaktal Oct 30 '12 at 15:02
I think these two comments should be enough. – GEdgar Oct 30 '12 at 15:04
Actually: probably for someone asking this question, solving the differential equation would be out of his depth, and you would start with your second display, which is probably in his textbook. But your solution goes beyond that to benefit more advanced readers as well. Which is good. – GEdgar Oct 30 '12 at 15:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.