Introduction
Many quantities decrease over time: a chemical sample may lose a fixed percentage of its amount in each time period, or a radioactive substance may be described by the time it takes for half of its quantity to remain. These are examples of exponential decay.
In an exponential decay model, the amount lost during a time period is proportional to the amount present at the start of that period. As a result, the same fraction is retained from one period to the next. A half-life is one useful way to describe that repeated proportional decrease.
Decay by a Repeated Percentage
For example, if 8% is lost per year, then \(r=0.08\), and the fraction retained per year is \(1-0.08=0.92\). The model uses \(0.92\) as its factor, not \(0.08\): the factor represents what remains, not what is lost.
For whole-number \(t\), this formula represents repeated multiplication by the same retention factor. The exponential model can also be used for real-valued \(t\), such as \(t=2.5\), provided fractional periods are part of the model being assumed. The base \(1-r\) is positive, so the real power is defined.
If \(A_0>0\), each full period leaves a smaller amount than the preceding period because \(0<1-r<1\). If \(A_0=0\), the model remains zero in every period. The amount does not decrease further from zero.
Why the Repeated-Percentage Formula Works
Let \(A_n\) denote the amount after \(n\) whole periods. Each period retains the fraction \(1-r\) of the amount at its start, so \(A_{n+1}=(1-r)A_n\). Starting with \(A_0\), repeated application gives the model for whole-number \(n\).
At \(n=0\), the amount is \(A_0=A_0(1-r)^0\), since \((1-r)^0=1\).
Suppose \(A_n=A_0(1-r)^n\).
Then \(A_{n+1}=(1-r)A_n=(1-r)A_0(1-r)^n=A_0(1-r)^{n+1}\).
This proves by induction that the formula holds for every whole number of periods. It does not, by itself, prove a formula for fractional periods. To use the same expression for real-valued time, the model must additionally assume exponential decay between the counted periods.
Worked Example: A Percentage Lost Each Year
Worked Example: A Percentage Lost Each Year
A sample has an initial mass of \(500\) mg and loses 12% of its mass per year. Use the repeated-percentage model to find its mass after 3 years.
The lost fraction is \(r=0.12\), so the retained fraction is \(1-r=1-0.12=0.88\). Here \(A_0=500\) mg and \(t=3\) years:
Calculate the powers step by step:
Therefore, \(A(3)=500(0.681472)=340.736\) mg. The intermediate amounts also follow the same rule: after year 1, \(500(0.88)=440\) mg; after year 2, \(440(0.88)=387.2\) mg; after year 3, \(387.2(0.88)=340.736\) mg. The final value agrees with the formula, so the calculation checks.
Half-Life Models
A half-life is the time required for a positive quantity to fall to half its value, under a specified decay model. If the half-life is \(h>0\) units of time, the amount remaining after time \(t\geq 0\) is modeled by
The ratio \(t/h\) counts the number of half-lives elapsed. For example, if \(t=12\) hours and \(h=4\) hours, then \(t/h=3\): three half-lives have elapsed. The units must match before dividing. If time is given in minutes but \(h\) is in hours, convert one of them first.
The formula has the required half-life property: when \(t=h\), it gives \(A(h)=A_0(1/2)^1=A_0/2\). More generally, applying it after an additional half-life multiplies the amount by one half:
Thus the model halves the amount over every interval of length \(h\), not just the first one. If \(A_0=0\), the formula gives zero for all \(t\); interpreting a half-life as the time for a positive amount to become half its value is most meaningful when \(A_0>0\).
Worked Example: Calculating a Fractional Number of Half-Lives
Worked Example: Calculating a Fractional Number of Half-Lives
A substance has a mass of \(160\) g and a half-life of 6 hours. Under the exponential half-life model, find the mass after 15 hours. Round the final answer to three decimal places.
The elapsed time is \(t=15\) hours, so the number of half-lives is \(t/h=15/6=2.5\). Since this is fractional, we are using the model’s exponential assumption between whole half-lives:
Separate the two whole half-lives from the remaining half of a half-life:
So \(A(15)=40/\sqrt{2}=20\sqrt{2}\) g, which is approximately \(28.284\) g to three decimal places. To check, after 12 hours, two whole half-lives have reduced the mass from \(160\) g to \(40\) g. The additional 3 hours are half of the stated 6-hour half-life; the model’s factor over that fraction is \(1/\sqrt{2}\), giving \(40/\sqrt{2}=20\sqrt{2}\) g as calculated.
Connecting a Decay Rate and a Half-Life
A repeated-percentage model and a half-life model can describe the same decay when their parameters are consistent. Suppose the unit period in \(A(t)=A_0(1-r)^t\) is the same time unit used to measure \(t\), and the model is being used for real-valued time. A half-life \(h\), measured in those units, means that the amount after \(h\) units is half the starting amount. Substituting \(t=h\) gives
For \(A_0>0\), divide both sides by \(A_0\) to obtain \((1-r)^h=1/2\). Thus a stated half-life determines the per-unit retention factor, and therefore the per-unit lost fraction, through
This relationship relies on the exponential model when \(h\) is not a whole number of the chosen periods. It is not a claim that a percentage rate and a half-life can be combined without checking their time units.
Worked Example: Checking Consistency with a Per-Day Rate
Worked Example: Checking Consistency with a Per-Day Rate
A sample has a half-life of 2 days. Under the exponential model, what fraction is retained per day, and what percentage is lost per day? Give the percentage to two decimal places.
Here \(h=2\) days. The per-day retention factor is \((1/2)^{1/h}\), so
Therefore \(r=1-1/\sqrt{2}\approx 0.2928932\), which is a loss of approximately \(29.29\%\) per day to two decimal places. Check the half-life: retaining \(1/\sqrt{2}\) on each of two days gives \((1/\sqrt{2})^2=1/2\). Thus the amount is halved after 2 days, as required.
Common Mistakes and Model Checks
A common error is to use the percentage lost as the multiplier. For a 12% loss per period, a wrong calculation for a starting amount of \(500\) mg after one period is \(500(0.12)=60\) mg. This is the amount lost, not the amount remaining. The correct retention factor is \(1-0.12=0.88\), so the remaining amount is \(500(0.88)=440\) mg. The difference \(500-440=60\) mg confirms that 60 mg was lost.
Another mistake is to treat the number of half-lives as \(h/t\) rather than \(t/h\). If \(t=15\) hours and \(h=6\) hours, then \(15/6=2.5\) half-lives have elapsed; \(6/15\) is not the elapsed number of half-lives. Also check that the computed amount is between zero and the initial amount for positive \(A_0\) and nonnegative time. If a decay calculation gives a larger amount than \(A_0\), check whether the retained factor, time, or model has been entered incorrectly.
Why These Models Matter
The two formulas provide different ways to express a proportional decrease. A per-period percentage is useful when the loss is specified directly, such as a fixed fraction per year. A half-life is useful when the time to halve is known or reported. In either case, the model makes its assumptions visible: the same proportional rule is applied throughout the time interval, and fractional periods use an exponential model rather than following solely from repeated whole-period calculations.
| Information given | Model to use | Key check |
|---|---|---|
| Fraction \(r\) lost per period | \(A(t)=A_0(1-r)^t\) | Use \(1-r\) as the retained fraction. |
| Half-life \(h\) | \(A(t)=A_0(1/2)^{t/h}\) | Use matching time units and recognize the exponential assumption for fractional half-lives. |
Check Your Understanding
For each question, identify the model assumptions and keep the time units consistent.
- A quantity loses 5% per month. What fraction is retained each month?
- A \(240\) g sample loses 25% per year. Use the repeated-percentage model to find the amount after 2 years, and check by applying the yearly factor twice.
- A substance has a half-life of 8 hours. How many half-lives elapse in 20 hours, and what fraction remains under the exponential half-life model?
- Why does the argument based on repeated halving establish the half-life formula at whole-number multiples of \(h\), but not by itself at fractional multiples?
- If \(A_0=0\) in a repeated-percentage decay model, what amount does the model give after each period?
