radioactivity - 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?

Answer

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 $