# Half-Life Exponential Decay using base e?

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

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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:

$$N(100)=100\cdot\rm{e}^{-\frac{100\ln{2}}{1600}}=95.76$$

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

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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