# Radioactive decay - the difference between "lambda" and "k"

1. Oct 29, 2016

### wolf1728

From what I have interpreted, k = ln (.5) / half-life
whereas λ = ln (2) / half-life.
Have I defined these correctly?
If this is so, the difference between the two is slight.
When putting these into equations, we see:
Nt = N0*ekt
Nt = N0*e-λt
The only difference seems to be that when using lambda, you have to remember to put the negative sign in the exponent.
It just seems confusing to me that some authors use λ whereas others use "k".

2. Oct 29, 2016

### Simon Bridge

Well $\ln (2) = -\ln (1/2)$ ...notice that $\ln (1/2)<0$?
Basically you are correct, $k=-\lambda$

... the standard relations is actually $N(t)=N_0e^{t/\tau}$ where $\tau$ is the mean life.
The expressions are usually derived from the model where the rate of decay is proportional to the amount of material present ... so $\dot N = -kN: N(0)=N_0$, where k>0 is the constant of proportionality. This is a 1st order homogeneous DE and an initial value problem ... solve it by proposing solutions of form $N(t)=e^{\lambda t}$ ... which is just the standard way of solving DEs.
This is where the $\lambda$ and $k$ comes from.

Their relationship to the half-life is down to the definition of "half life".

3. Oct 30, 2016

Simon Bridge