Intuition for half life formula?

Question

Intuitively what does this equation for half life mean if this is the formula for something to be decayed?

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

Answer 1

If you have a very large number $\ N\ $ of some entity, whose remaining lifetimes independently follow an identical exponential distribution with parameter $\ \lambda\ $ (i.e. the probability that any one of the individual entities surviving for at most another $\ t\ $ units of time is $\ 1-e^{-\lambda t}\ $), then at time $\ t_{1/2}\ $ the number of the entities surviving will be very close to $\ \frac{N}{2}\ $ with probability very close to $1$.

The number $\ n\ $ of entities still alive at time $\ t_{1/2}\ $ in fact follows a binomial distribution with parameters $\ N\ $ and $\ p = e^{-\lambda t_{1/2}}=\frac{1}{2}\ $ (i.e. the probability that $\ n\ $ of the entities are still alive at time $\ t_{1/2}\ $ is ${N\choose n}\ 2^{-n}\ $). The mean number alive at time $\ t_{1/2}\ $ is $\ \frac{N}{2}\ $, and the standard deviation of that number is $\ \frac{\sqrt{N}}{2}\ $. When $\ N\ $ is very large, the ratio of the standard deviation to the mean, $\ \frac{1}{\sqrt{N}}\ $, will be very small, so the actual number of the entities surviving at time $\ t_{1/2}\ $ is highly likely to be very close to the mean of $\ \frac{N}{2}\ $.

If $\ N=1,000,000\ $, for instance, the probability that the actual number surviving at time $\ t_{1/2}\ $ is within $\ 0.2\%\ $ of $\ \frac{N}{2}\ $ is $\ 0.954\ $.

I should note that "very close to" in the above account (except when referring to probabilities) means "very close to" in relative terms, not absolute terms. Typically, the difference between $\ \frac{N}{2}\ $ and the number of entities still surviving at $\ t_{1/2}\ $ will be of magnitude around the order of $\ \sqrt{N}\ $. While $\ \sqrt{N}\ $ will be very large if $\ N\ $ is very large, its size will only be a very small fraction of that of $\ N\ $.

Answer 2

Rewrite it in the form $$\lambda \cdot t_{1/2} = \ln (2)$$ or, even better, $$e^{-\lambda \cdot t_{1/2}} = 1/2$$

This says precisely that when you take the formula $f(t) = A e^{-\lambda t}$ and set the time equal to $t_{1/2}$, the result is $$f(t_{1/2}) = A \cdot \frac 12$$ So after one half-life, you have half of the original amount remaining.

Want to keep going? Set $t$ equal to $2\cdot t_{1/2}$. Then $$f(2\cdot t_{1/2}) = A \cdot \left( \frac 12 \right)^2=A\cdot \frac 14$$ So after two half-lives, you have one-fourth of the original amount remaining.

Answer 3

Recall that something which experiences exponential decay follows the formula:

$$f(t)=f_0e^{-\lambda t}$$

This is by definition of what it means to have exponential decay.

Now, let $t_{1/2}$ be the value of $t$ such that $f(t_{1/2})=\frac{1}{2}f(0)$

That is to say, $f_0e^{-\lambda t_{1/2}}=\frac{1}{2}f_0$ or that $\frac{1}{2}=e^{-\lambda t_{1/2}}$

Solving for $t_{1/2}$, we take the natural logarithm of each side:

$\ln(\frac{1}{2})=\ln(e^{-\lambda t_{1/2}})$ which simplifies as $\ln(2) = \lambda t_{1/2}$ and moving things over we get to:

$$t_{1/2}=\dfrac{\ln(2)}{\lambda}$$

Reworded, if an amount follows a pattern of exponential decay with rate $\lambda$ then after $t_{1/2}$ amount of time there will be only half of the original amount remaining.