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How exactly was the half life equation $\frac{ln2}{\lambda}$ derived? My book states something about integrated rate laws but doesn't really explain anything. Sorry, this is my first post in Physics stack-exchange. Thanks!

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    $\begingroup$ Its mathematical derivation comes from the rate law for the radioactive decay $\endgroup$ Commented Mar 22, 2019 at 8:26
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    $\begingroup$ Do you know any calculus? Wikipedia explains the relation between mean lifetime and half life, but you'll need to be familiar with the derivative of the natural exponentiation function. $\endgroup$
    – PM 2Ring
    Commented Mar 22, 2019 at 8:32
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    $\begingroup$ It involves integration of the first order rate law, hence called integrated rate laws.. $\endgroup$ Commented Mar 22, 2019 at 8:33
  • $\begingroup$ My answer shows the integrated rate law, mentioned by Sidharth. $\endgroup$
    – Nikita
    Commented Mar 22, 2019 at 8:40
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    $\begingroup$ If we were interested in "third lives" rather than half lives then the equivalent expression would involve $\ln 3$ instead of $\ln 2$. $\endgroup$
    – gandalf61
    Commented Mar 22, 2019 at 17:44

3 Answers 3

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Welcome to Physics Stackexchange.

The empirical formula of reaction rate for equation: - $$aA+bB\rightarrow cC+dD$$ can be represented as, $Rate\rightarrow k[A]^{m}[B]^{n}$ where proportionality constant $k$ is called rate constant, and $m+n$ is called reaction order.

Consider a simple first order reaction, $A\rightarrow B$

$$\text{Rate}\Rightarrow k[A]^{1}$$ $$-\frac{d[A]}{dt}=k[A]$$ $$\int_{[A]_{0}}^{[A]_{t}}\frac{d[A]}{[A]}=-\int_{0}^{t}kdt$$ $$\ln\left (\frac{[A]_{t}}{[A]_{0}} \right )=-kt$$ $$[A]_{t}=[A]_{0}e^{kt}$$ In order to calculate half-life the concentration has to be halfed. Let, $$[A]_{t}=\frac{[A]_{0}}{2}$$ $$e^{-kt}=\frac{1}{2}$$ Solving the above equation for $t$, you get $$T_{1/2}=\frac{\ln 2}{k}.$$ Does that look familiar? Let me know about any queries in the comments below :)

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    $\begingroup$ Thanks! I take reaction rates are dependent on concentration of reactants and products then? $\endgroup$
    – user226159
    Commented Mar 22, 2019 at 8:59
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    $\begingroup$ Yes, Rate of reaction = Change in concentration/Time $\endgroup$
    – Nikita
    Commented Mar 22, 2019 at 8:59
  • $\begingroup$ Is k affected by environmental factors or is unique to each set of reaction? $\endgroup$
    – user226159
    Commented Mar 22, 2019 at 9:02
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    $\begingroup$ Each reaction has its unique rate of reaction equation which are determines empirically and k is affected by various factors, including environmental conditions. $\endgroup$
    – Nikita
    Commented Mar 22, 2019 at 9:03
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    $\begingroup$ @Prograxer The given rate law is basically equivalent to the assumption that a reaction that requires $n_i$ of each species $i$ has some probability of occurring when those components are in reactive-distance, which each particle has an equal likelihood of being in at any given moment. Since this tends to be a pretty idealistic mechanism, it's common to describe retreats from it using activity coefficients. When activities are an insufficient remedy, we typically have to do more simulation-like calculations. $\endgroup$
    – Nat
    Commented Mar 23, 2019 at 11:59
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In terms of half life $\tau$ the number of undecayed nuclei after a time $t$ is given by $N(t)=N(0)\left(\frac 12\right)^{\frac t \tau}$

In terms of decay constant $\lambda$ the number of undecayed nuclei after a time $t$ is given by $N(t)=N(0)e^{-\lambda t}$.

Equating gives $\left(\frac 12\right)^{\frac t \tau}= 2 ^{-\frac t \tau}= e^{-\lambda t} \Rightarrow \ln (2^{-\frac t \tau} )= \ln (e^{-\lambda t} ) \Rightarrow -\frac t \tau \ln 2 = -\lambda t \ln e \Rightarrow \tau \lambda = \ln 2$

Update as the result of some comments.
In general the half life or the decay constant relating to radioactive decay are not affected by any chemical or physical changes because the energy scales involved in such processes are so much smaller than those involved in reactions involving the nucleus.

The relationship $N(t)=N(0)\left(\frac 12\right)^{\frac t \tau}$ can be written as $N(t)=N(0)2^{-\frac{t} {\tau_{1/2}}}$ where $\tau_{1/2}$ is the half life, the average time it takes $\frac 12 $ of the unstable nuclei to decay.

The relationship $N(t)=N(0)e^{-\lambda t}$ can be written as $N(t)=N(0)e^{-\frac{t} {\tau_{1/e}}}$ where $\tau_{1/e}= \frac 1 \lambda $ is the average time it takes $\frac 1 e$ of the unstable nuclei to decay. ($\frac 1 \lambda$ also is the average lifetime of the unstable nuclei.)

What I hope that you can see is that bases $2$ and $e$ have been chosen although any base could have been used eg $N(t)=N(0)7^{-\frac{t} {\tau_{1/7}}}$ but those other bases would not have been as "user friendly" as the choice of $2$ and $e$.

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The fundamental assumption is that a radioactive particle is as likely to decay at any moment as any other. And that's it. There's no more physics involved – no more assumptions, no more laws, nothing! It's just math after this.

So, if we have $n$ radioactive particles, then the number decaying, $-\frac{\mathrm{d}n}{\mathrm{d}t} ,$ is going to be proportional to the number of radioactive particles: $$ \underbrace{-\frac{\mathrm{d}n}{\mathrm{d}t} \phantom{-}}_{\text{loss rate}} ~{\propto}~n \tag{1} \,.$$Then the math's:

  1. Rewrite $\operatorname{Eq.\,}\left(1\right)$ as$$ \frac{\mathrm{d}n}{\mathrm{d}t}~=~-\lambda n \tag{2} \,,$$where $\lambda$ is just a proportionality constant.

  2. Solve $\operatorname{Eq.\,}\left(2\right)$ using WolframAlpha, finding:$$ n~=~c e^{-\lambda t} \tag{3} \,.$$

  3. Define $n_0$ at the number of particles at $t=0 ,$ such that$$ \require{cancel} n_0 ~=~c e^{-\lambda \cancelto{0}{t}} \qquad \Rightarrow \qquad n_0 = c \,,$$which basically just means that we can describe $c$ as $n_0$, which is more useful since $`` c "$ was just an arbitrary constant while $`` n_0 "$ has a useful physical interpretation, allowing us to rewrite $\operatorname{Eq.\,}\left(3\right)$ as$$ n~=~n_0 \, e^{-\lambda t} \tag{4} \,.$$

  4. Rewrite $\operatorname{Eq.\,}\left(4\right)$ in terms of the ratio of particles remaining at time $t$ as$$ \frac{n}{~n_0} ~=~ e^{-\lambda t} \tag{5} \,.$$

  5. Find when half of the particles have decayed,$$ \frac{n}{~n_0} ~=~ \frac{1}{2} ~=~ e^{-\lambda t} \tag{6} \,,$$then solve for $t$ to find$$ t ~=~ -\frac{\ln{\left(\frac{1}{2}\right)}}{\lambda} ~=~ \frac{\ln{\left(2\right)}}{\lambda} \tag{7} \,.$$ And that's it!


The main point I wanted to make here is that this immediately follows from the assumption that radioactive decay is as likely at any moment as any other. After that, it's just some math. No need to use any empirical correlations or anything.

Of course, if you find some physics that suggests that particles become more or less likely to decay over time, then you'd just have to rewrite $\operatorname{Eq.\,}\left(1\right)$ as$$ -\frac{\mathrm{d}n}{\mathrm{d}t} ~{\propto}~ \left[ \begin{array}{c} \text{whatever the physics} \\[-10px] \text{you discovered says} \end{array} \right] _{,}$$then redo the math.

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