Radioactive decay formula

Question

We know that

dN/dt=-Nλ

where λ= probability of the atom decaying per second.

So surely 1-λ = the probability of the atom not decaying.

So to calculate half life is it ok if I do

N = $ (1-λ)^{t_{1/2}} N_{0} $

$\frac{1}{2} N_{0}$ = $ (1-λ)^{t_{1/2}} N_{0} $

$\frac{1}{2} $ = $ (1-λ)^{t_{1/2}} $

Then $ t_{1/2}=$ $\frac{-ln 2}{ln (1-λ)}$

We know $ t_{1/2}=$ $\frac{ln 2}{λ}$

So -λ = ln (1-λ)??

I know this doesn't really work but I can't figure out why is that?

Answer 1

I think you need to use an exponential Ansatz to solve this, because $ N(t)=N_{0} \cdot\lambda^{t} $ does not satisfy the differential equation $\frac{d}{dt}N(t) = -\lambda\cdot N(t) $:

$$ N(t) = N_{0} \cdot e^{-\lambda \cdot t} $$

Then

$$ N(t_{1/2})= \frac{N_{0}}{2} =N_{0} \cdot e^{-(1-\lambda)\cdot t_{1/2}} $$ $$ \frac{1}{2} = e^{-(1-\lambda) \cdot t_{1/2}} $$ $$ -log(2) = -(1-\lambda) \cdot t_{1/2} $$ $$ t_{1/2} = \frac{log(2)}{1-\lambda} $$.

Using $\lambda$ instead of $(1-\lambda)$, you would obtain

$$ t_{1/2} = \frac{log(2)}{\lambda} $$

which is fine considering that $\lambda = (1-\lambda)$.

Answer 2

$\lambda$ is not the probability that the atom decays in one second. For a very short period $\Delta t$ the probability that the atom decays is $\lambda \Delta t$. The chance the atom survives a full second is $e^{-\lambda}$. When $\lambda$ is small, this will be just about $1-\lambda$ In this approximation $\log (1-\lambda)$ is just about $-\lambda$. We are neglecting terms of order $\lambda^2$

Answer 3

The probability of decay within a time $t>0$ is $1-e^{-\lambda t}$. You've misunderstood a concept called "probability rate": $\lambda$ isn't the probability of decaying within a unit time, but rather the $t\to0^+$ limit of the decay-within-$t$ probability divided by $t$, i.e. $\lim_{t\to0^+}\frac{1-e^{-\lambda t}}{t}=\lambda$. Note that $\lambda$ has units of inverse time; it's not dimensionless, so expressions such as $\ln(1-\lambda)$ (or for that matter $1-\lambda$) aren't even defined.