# Radioactive decay, why such unintuitive formula?

When talking of exponential decay, as with radioactive decay, the formula used (e.g. Wikipedia and my textbook) is:

$$N(t) = N_0e^{-\lambda t}$$

This formula, with the decay constant $\lambda$ makes little intuitive sense. It is the ratio between the amount of radioactive material and the decay at any time. It might lead one to believe that after one time unit, the amount of radioactive material has been decreased by a factor $1/\lambda$, but that is not even the case.

A much more intuitive form would be like the formula of exponential growth:

$$N_{wrong}(t) = N_0*(1-k)^t, k= 1-e^{-\lambda}$$

One only needs to look at that formula for a second to get an intuitive understanding of the rate of the decay.

I got curious about this, and I want to ask why mathematicians or physicists have chosen the first mentioned formula. Did I miss something clever here?

• $e^{-\lambda t} = (1- (1-e^{-\lambda \epsilon}))^{t\over\epsilon}$ and $(1-e^{-\lambda\epsilon})\approx \lambda\epsilon$. This is intuitive after you take calculus, the discrete exponential decay limits to the continuous one for small time steps. It's like compounding interest in infinitesimal increment at a bank. – Ron Maimon Jul 28 '12 at 5:16
• – dmckee May 9 '13 at 19:29

The first answer (form Shaktal)is in my opinion perfect from the mathematical point of view (solving the differential equation) as well as from the physical point of view (having all units in order).

However your probably conceive your approach more intuitive because you know this from calculating things with Interest i.e.

$N_{Money}=N_{0,Money}\times (1\pm Interest)^{t(in\;years)}$

Aside from the units being messed up, physicists like functions like $e^{something}$ because you can easily calculate with them. I.e. try to take $\frac{d}{dt}N_{wrong}$.

$\frac{d}{dt}N_{wrong}=\frac{d}{dt}N_0\times(1-k)^t=\frac{d}{dt}N_0 e^{ln(1-k)t}$

gives you just

$\frac{d}{dt}N_0 e^{-\lambda t}$

so why the detour?

The derivate is called the activity and is very important so you have to calculate it quite often. You may also want to integrate $N_{wrong}$ sometimes which would also be really annoying.

Further considering the answer from Samuel Hapak: The half-life (not the great game) as the "real" half-life of an element, as in $N_{1/2}=N_0\times\left(\frac{1}{2}\right)^{t_{1/2}/\lambda}\;\;t_{1/2}\text{: half-life, }\lambda:\frac{1}{\text{(half-life)}}$

gives you a quick overview over the decay rate with an SI-unit. So it can easily be given an order of Magnitude one can relate to. So it is indeed as intuitive as it can get in my opinion. One can of course easily switch to the mathematical expression resulting from solving the differential equation shown in Shaktal's answer:

$N_t=N_0 e^{-\tilde\lambda t}$

by setting $\tilde\lambda = ln(2)\lambda$

We come to the first formula by considering the differential equation which we can experimentally measure:

$$\frac{dN}{dt}=-\lambda N\tag{1}$$

We can solve differential equation $(1)$ by rewriting as follows:

$$\frac{dN}{N}=-\lambda\cdot dt$$

We then integrate:

$$\int{\frac{dN}{N}}=\int{-\lambda\:dt}\implies\ln{N}=-\lambda t+c_{1}$$

Exponentiating both sides (with base $\rm e$), gives:

$$N(t)={\rm e}^{-\lambda t+c_{1}}={\rm e}^{\lambda t}{\rm e}^{c_{1}}$$

We can rewrite ${\rm e}^{c_{1}}=C$ as it is an arbitrary multiplicative constant. So we have:

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

It just so happens that this constant $C=N_{0}$. This is why we choose to write it in the form:

$$N(t)=N_{0}{\rm e}^{-\lambda t}\tag{2}$$

I hope this helps!

• Thanks, but I do understand the derivation. I just don't see why the former formula is preferred in textbooks when the latter to me (and my co-students) seem much more intuitive. In fact even my teacher accidentally said that after one time unit a fraction of $\lambda$ will have decayed, which is only true if the time unit is defined to be sufficiently small. – Lucy Brennan Jul 28 '12 at 8:36

"It might lead one to believe that after one time unit...": bad. What is "one time unit" supposed to be? How much will be gone depends on its length in a nontrivial way, so any formula of the type you propose actually depends on a particular choice of this unit time.

This problem does not happen when you use an infinitesimally short time interval, which is the typical physicist's interpretation of $\mathrm{d}t$. During such an interval, the rate of decay stays constant, so doubling it will just give twice as many decays, which is trivial. This is what the differential equation $$\frac{\mathrm{d}\:N}{\mathrm{d}t} = -\lambda\cdot N$$ expresses: for any infinitesimal time $\mathrm{d}t$, the ratio of decays taken part and time past is constant, namely $-\lambda\cdot N$. We then don't depend on any particular choice of $\mathrm{d}t$, it only needs to be sufficiently small.

(Note that the notion of $\mathrm{d}t$ as a short, finite time interval is not really mathematically sound)

• Yes, k will depend on the size of the time unit used. But so does $\lambda$, since $\lambda = -ln(1-k)$. Right? – Lucy Brennan Jul 28 '12 at 8:27
• @LucyBrennan: this formula isn't even well-defined. $\lambda$ has the physical dimension of $1/{\text{time}}$, so calculating $e^{-\lambda}$ on its own doesn't make any sense mathematically. You can only calculate the exponential of dimensionless quantities, like $4$, or $\pi$, or $\lambda\cdot t$, or $t/T$. So to make your formula correct you need to put in such a $T$ constant explicitly. That ain't nice! – leftaroundabout Jul 28 '12 at 19:34
• I'm afraid that I don't understand it, and I guess I just don't have the prerequisites to. But thanks in any case :) – Lucy Brennan Jul 28 '12 at 20:04

If we write the differential equation for $N$ in a discrete notation we obtain: $$N_{k+1} = (1-\Delta t\lambda)N_k$$ So if at time $k=1$ there are N radioactive atoms, at time $k=2$ there will be $1-\Delta t \lambda$ less.

This is like your money in the bank, but in reverse: the more you have, the more you loose.

Edit: this is also closely related to what is called a Poisson process. The probability of seeing $k$ random atoms decay in a given time period is distributed according to a Poisson distribution, very much like the probability that you receive $k$ e-mails in (say) an hour. Then, it turns out (mathematically I mean), that the time until the next e-mail is distributed according to an exponential distribution, very much like the formula you originally posted.

• hat is only true for sufficiently small $\Delta t$. I understand the concept of exponential growth, as in the bank. But in fact the bank uses the formula that I prefer – Lucy Brennan Jul 28 '12 at 8:51
• There are many many many atoms, but each atom is very very unlikely to decay at a given instant of time: this justifies the use of a continuous equation even though, by design, the system is itself discretized. – Gabriel Landi Jul 28 '12 at 11:51
• I think you misunderstood me, because I already understand what you tell me in your comment. I restate my point: If $N_0 = 10$, $\lambda = 0.1$ and $\Delta t=1$, then you get 9.048. Whereas $(1- \Delta t \Lambda) = 0.9$. Those two numbers are simply not the same. With higher values of $\Delta t$ the difference is greater. – Lucy Brennan Jul 28 '12 at 16:30

Well, I think the formula is pretty intuitive.

Let us T be the half life. Then number of particles after time t is

$$N = N_0 \cdot 2^{-t/T}$$

This is really intuitive, after time T, you have got half. After another time T, you have only one quarter, etc, etc ...

Now, you know, we could compute not half life, but third life and we would have formula:

$$N = N_0 \cdot 3^{-t/T_3}$$

We can do so, we can choose any base we want. And for one very important reason, it is good idea to choose the base of $e$. So we write

$$N = N_0 \cdot e^{-t/T_e}$$

Or, if we call $\lambda = 1/T_e$ decay constant, we have $$N = N_0 \cdot e^{-\lambda t}$$