Exponential decay half life:
formula and worked example

The time it takes to lose half

In exponential decay a quantity falls by a fixed fraction in each equal interval of time. Half-life is the interval over which that fraction is one half. It is the number most people use to describe how quickly a radioactive isotope fades, a drug clears from the body, a pollutant breaks down or a signal in a circuit dies away.

The remarkable feature is that it does not depend on how much you start with. Whether you have a kilogram or a milligram, the amount halves in the same time, so one number describes the entire process.

Half-life is the practical way to talk about decay, while the decay constant is the form that appears in the differential equation. The formula converts one into the other.

The same mathematics applies in reverse for growth: a quantity that grows exponentially has a doubling time, found with the identical formula using the growth rate. Compound interest, bacterial cultures and viral spread are common examples, so learning to move between rate and halving or doubling time is a transferable skill.

The formula and the decay constant

• Decay constant: the fractional rate of loss per unit time, in the equation N(t) = N0 e^(-lambda t). It must be positive and its unit is the reciprocal of time.

The numerator ln(2) is the natural logarithm of 2, about 0.693147. It arises from solving e^(-lambda T) = 1/2 for T. The constant is the only input, so getting its unit right is the one real job.

If you are given the decay as a percentage per period instead of a constant, convert first. A loss of 6.76% per hour corresponds to a constant of about 0.07 because the constant equals minus the natural log of the remaining fraction, here ln(1 / 0.9324). Skipping that step and using 0.0676 directly gives a half-life that is off by about 3.5 per cent.

Worked example: decay constant 0.07

A substance decays with a constant of 0.07 per unit of time, say per hour.

1. 1Take ln(2) = 0.693147.
2. 2Divide by the decay constant: 0.693147 / 0.07.
3. 3The quotient is 9.902, rounded to three decimals.
4. 4Check: e^(-0.07 x 9.902) = e^(-0.693) = 0.5, so half remains.

What a 9.9-unit half-life implies

After one half-life 50% remains, after two 25%, after three 12.5%, and so on, since the remaining fraction is one half raised to the number of half-lives elapsed. With a half-life of 9.902 units, three half-lives is about 29.7 units, by which point 12.5% remains.

Practical rules of thumb follow. After about seven half-lives less than 1% remains, so a quantity is often treated as effectively gone by then. Whether that is acceptable depends on the context, such as dose safety or contamination limits, and conventions vary by field.

The decay constant of 0.07 also says that the amount drops by about 6.8% per unit of time, because 1 - e^(-0.07) is roughly 0.068, which is slightly less than the constant itself.

Remember also that the model is an idealisation. Real systems can have several overlapping processes, such as a drug that is both distributed to tissues and eliminated, and then no single half-life describes the whole curve. Where the data bend, fit separate phases and quote each half-life with its time range.

What a faster decay does

A 10% larger constant shortens the half-life from 9.902 to 9.002, a fall of about 9.1%. The relationship is inverse: multiply the constant by any factor and the half-life is divided by that factor. Doubling the constant halves the half-life, so a process twice as fast reaches the halfway point in half the time.

Mistakes with half-life

• Mixing time units, for example using a constant per minute and reading the answer as hours. The result is in whatever unit the constant is per.
• Confusing half-life with mean lifetime. The mean lifetime is 1 divided by the constant, here about 14.3 units, which is longer than the half-life by a factor of 1/ln(2), about 1.44.
• Assuming everything is gone after two half-lives. Twenty-five per cent still remains.
• Applying the formula to a process that is not exponential, such as one with a constant removal rate, where the time to halve depends on the starting amount.
• Rounding ln(2) too coarsely, for instance to 0.7, which shifts the answer by about one per cent.

Working backwards and forwards

The formula is easily inverted. If a lab reports that a sample halves in 12 hours, the decay constant is 0.693147 / 12, about 0.0578 per hour. From there you can find the amount left at any time by multiplying the start by e^(-0.0578 t), or equivalently by one half to the power t / 12.

A last sanity check is to compare the half-life with the span you actually observe. If the half-life is 9.9 hours and you watch for only 2 hours, about 13% has decayed, the curve looks nearly straight and a linear fit can seem adequate. Over several half-lives the curvature becomes obvious, and only then does the exponential form show its value.