Given a rock that contains 10^20 atoms of a particular substance, each atom has an exponentially distributed lifetime with a half-life of one century. How many centuries must pass before:

a) It is most likely 100 atoms remain

b) There is a 50% chance one atom remains

A solution I came across for a)

$ \lambda = \frac{ln(2)}{0.5} = 1.3862$

$ P(T>t) = e^{- 1.3862t} = \frac{100}{10^{20}} $

$ t= 29.89 $

If $P(T>t)$ means the probability of the thing surviving until past time t, how does the expression used to calculate $P(T>t)$ equal the proportion of atoms remaining to the original amount of atoms?

Also, how would I approach problem b)? Where would the 50% fit in for one atom remaining?

  • $\begingroup$ In my opinion the value for $\lambda$ is not right. The equation should be $0.5=e^{-\lambda \cdot 1}\Rightarrow \lambda=ln(2)$ $\endgroup$ Nov 10, 2015 at 18:21

1 Answer 1


As stated in the comments, there is an error in $\lambda$ (unless the halflife is supposed to be half a century and you made a typo).

"Exponentially distributed lifetime" means that the number of atoms in the rock follows an evolution of the form $$ N(t) = N_0 e^{-\lambda t}, $$ where $N$ is the number of atoms at time $t$ and $N_0=N(0)$. To find $\lambda$, recall that after one halflife ($t=1$ centuries) half of the initial number of atoms remain, thus $$ \frac{N_0}{2} = N_0 e^{-\lambda} \Leftrightarrow \lambda = \log{(2)}. $$

(a) Put $N(t)=100$, $N_0 = 10^{20}$, $\lambda = \log{(2)}$ and solve for $t$. You will obtain the same equation as you have written in the question.

(b) Recall that a halflife is the time during which the probability of decay of one atom is 50%. Thus, set $N(t)=1/2$, $N_0=10^{20}$, $\lambda={(2)}$ and solve for $t$.


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