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Half-life formula

A radioactive sample loses a fixed fraction of whatever is left in each equal stretch of time, never a fixed amount. The half-life formula says that twice, in two notations that look unrelated and are not:

$$N = N_0 e^{-\lambda t} \qquad\qquad N = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$$

$N_0$ is the number of undecayed nuclei at $t = 0$, $N$ is the number still undecayed at time $t$, $\lambda$ is the decay constant, and $T_{1/2}$ is the half-life. The first form comes straight from the physics, which is that each surviving nucleus has a fixed probability per unit time of decaying, so $-dN/dt = \lambda N$. The second form is the first one rewritten so that the arithmetic is easy when you are counting half-lives.

The two forms are one formula

Put $t = T_{1/2}$ into the exponential form, where by definition $N = N_0/2$:

$$\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}$$

Cancel $N_0$, take logs of both sides, and the minus signs collapse:

$$T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.6931}{\lambda}$$

That is the whole connection. Now substitute it back. Since $\lambda = \ln 2 / T_{1/2}$,

$$e^{-\lambda t} = e^{-(\ln 2)\,t/T_{1/2}} = \left(e^{\ln 2}\right)^{-t/T_{1/2}} = 2^{-t/T_{1/2}} = \left(\frac{1}{2}\right)^{t/T_{1/2}}$$

The two formulas are not two models of decay that happen to agree. They are the same expression with the base changed from $e$ to 2, and the factor of $\ln 2$ is the price of the change. Which one to use is purely a question of what the problem hands you: a decay constant or a half-life.

Activity, and the two units it is quoted in

Nobody counts nuclei. What a detector measures is the rate at which they decay, called the activity:

$$A = \left|\frac{dN}{dt}\right| = \lambda N = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}$$

Activity is proportional to $N$, so it decays on exactly the same curve with exactly the same half-life. The SI unit is the becquerel: 1 Bq is one decay per second. The older unit is the curie, defined as the activity of one gram of radium-226, and fixed at $1\ \text{Ci} = 3.70\times10^{10}$ Bq.

Strontium-90 makes the size of these units concrete. Its half-life is 28.8 years, so

$$\lambda = \frac{0.6931}{28.8\ \text{y}} = 0.02407\ \text{y}^{-1} = \frac{0.02407}{3.16\times10^{7}\ \text{s/y}} = 7.616\times10^{-10}\ \text{s}^{-1}$$

One gram of it, at an atomic mass of 89.91 g/mol, contains

$$N_0 = \frac{1.00}{89.91} \times 6.022\times10^{23} = 6.698\times10^{21}\ \text{nuclei}$$

$$A_0 = \lambda N_0 = 7.616\times10^{-10} \times 6.698\times10^{21} = 5.10\times10^{12}\ \text{Bq}$$

Five terabecquerels from one gram. In the legacy unit that is $5.10\times10^{12}/3.70\times10^{10} = 138$ Ci, and the fact that a gram of a fairly long-lived isotope comes to 138 curies is exactly why the curie fell out of use: it is an enormous unit tied to one particular substance, and the becquerel is neither.

Running it forwards, and backwards

When the elapsed time is a whole multiple of the half-life, use the base-2 form and do not touch a calculator. Five half-lives of strontium-90 is $5 \times 28.8 = 144$ years, and

$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^5 = \frac{1}{32} = 0.03125$$

so 31.25 mg of the original gram is left. Check it on the exponential: $\lambda t = 0.02407 \times 144 = 3.466$, and $e^{-3.466} = 0.03125$. The same number, as it must be.

The awkward case is the one that is not a whole number of half-lives, and it is the same calculation. After 100 years,

$$\frac{t}{T_{1/2}} = \frac{100}{28.8} = 3.472\ \text{half-lives}$$

$$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{3.472} = 0.0901$$

90.1 mg left. Bracket it before believing it: 3 half-lives is 86.4 years and leaves 12.5 percent, 4 half-lives is 115.2 years and leaves 6.25 percent, and 100 years sits between those two, so the answer has to sit between 6.25 and 12.5 percent. It does.

Run the same formula backwards and it becomes a clock. A living organism exchanges carbon with its surroundings and holds the atmospheric fraction of carbon-14 while it does. When it dies the exchange stops and the carbon-14 it contains starts running down with a half-life of 5730 years, so what is left of it dates the death. Invert the half-life formula for $t$:

$$t = \frac{1}{\lambda}\ln\frac{N_0}{N} = T_{1/2}\,\log_2\frac{N_0}{N}$$

Take a piece of charcoal measured at 22.0 percent of the modern carbon-14 level. Using the decay constant, $\lambda = 0.6931/5730 = 1.210\times10^{-4}$ y$^{-1}$ and $\ln(1/0.220) = 1.5141$, so

$$t = \frac{1.5141}{1.210\times10^{-4}} = 12{,}510\ \text{years}$$

Using half-lives instead, $\log_2(1/0.220) = 2.184$, so $t = 2.184 \times 5730 = 12{,}510$ years. Round to 12,500 years. Sanity check it against the whole half-lives: two half-lives leaves 25 percent at 11,460 years and three leaves 12.5 percent at 17,190 years, and 22 percent is just below 25 percent, so a date a little past 11,460 years is the only place the answer could have landed.

One honest caveat on that number. It is a radiocarbon age, computed on the assumption that the atmospheric carbon-14 fraction has always been what it is today, and it has not been - it varies with solar activity and with the carbon released by burning fossil fuels. Real dates are corrected against tree-ring and other calibration records. The exponential is exact; the input to it is the part that needs care.

Half gone, then all gone

The failure mode is treating decay as linear. Half the sample goes in one half-life, so all of it goes in two. That reasoning gives zero strontium-90 after 57.6 years, and the right answer is 250 mg of the original gram, still running at 34.5 Ci. Extend it: ten half-lives, 288 years, leaves a fraction $2^{-10} = 1/1024$, which is 0.098 percent - small, and not zero. No finite time makes it zero. The curve removes a fixed fraction per interval, and a fixed fraction of something positive is positive.

The reason it is a fixed fraction is worth stating plainly, because it is the physical content of the whole thing. Radioactive decay is memoryless. A nucleus that has sat there for 288 years has precisely the same chance of decaying in the next second as one created a moment ago. Nuclei do not age, wear out, or accumulate damage, and there is no internal clock counting down to a scheduled event. If there were, $\lambda$ would depend on how long the nucleus had already survived and the curve would steepen with age, which is what wear-out processes look like and what this one conspicuously does not.

Memorylessness has one consequence that catches people out. The average lifetime of a nucleus is not half the half-life. It is

$$\bar{t} = \frac{1}{\lambda} = \frac{T_{1/2}}{\ln 2} = 1.443\,T_{1/2}$$

which for strontium-90 is 41.5 years - longer than the 28.8 year half-life, not shorter. Half the nuclei are gone by 28.8 years, but the survivors have a long tail ahead of them and the tail is what drags the mean up.

The nuclear decay problems alternate between handing you $\lambda$ and handing you $T_{1/2}$, which is the one thing that decides which form of the formula to reach for. The physics simulations run the decay curve so you can see the fixed-fraction behaviour rather than take it on the algebra, and the wider practice catalogue has the binding-energy and nuclear-reaction problems that this leads into.

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